Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keenam

$\begin{array}{l}\\ 26.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\sin (x-y)}& & & \text{d}.& \cos (x-y)\\ \text{b}.& {\displaystyle -\tan (x-y)}& & & \text{e}.& {\displaystyle \tan (x-y)}\\ \text{c}.& {\displaystyle \sin (x+y)}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 2x-\cos 2y}{\sin 2x+\sin 2y}}\\ & ={\displaystyle \frac{-2\sin (x+y)\sin (x-y)}{2\sin (x+y)\cos (x-y)}}\\ & =-\tan (x-y)\end{array}\end{array}$.

$\begin{array}{l}\\ 27.& \text{Nilai dari}\\ & 8\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 4(\sqrt{3}+\sqrt{2})}& & & \text{d}.& 2(\sqrt{3}-\sqrt{2})\\ \text{b}.& {\displaystyle 4(\sqrt{3}-\sqrt{2})}& & & \text{e}.& {\displaystyle \sqrt{3}-\sqrt{2}}\\ \text{c}.& {\displaystyle 2(\sqrt{3}+\sqrt{2})}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 8\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & =4\times 2\cos 82,5^{\circ }\sin 37,5^{\circ }\\ & =4\times (\sin (82,5^{\circ }+37,5^{\circ })-\sin (82,5^{\circ }-37,5^{\circ }))\\ & =4\times (\sin {120}^{\circ }-\sin {45}^{\circ })\\ & =4\times (\sin ({180}^{\circ }-{60}^{\circ })-\sin {45}^{\circ })\\ & =4\times (\sin {60}^{\circ }-\sin {45}^{\circ })\\ & =4\times \left({\displaystyle \frac{1}{2}\sqrt{3}-{\displaystyle \frac{1}{2}\sqrt{2}}}\right)\\ & =2(\sqrt{3}-\sqrt{2})\end{array}\end{array}$.

$\begin{array}{l}\\ 28.& \text{Bentuk lain dari}\\ & -2\cos 5A.\cos 7A\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\cos 6A-\cos A}& & & \\ \text{b}.& {\displaystyle -\cos 6A+\cos A}& & & \\ \text{c}.& {\displaystyle \cos 12A-\cos 2A}\\ \text{d}.& -\cos 12A+\cos 2A\\ \text{e}.& {\displaystyle -\cos 12A-\cos 2A}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& -2\cos 5A.\cos 7A\\ & =-(2\cos 5A.\cos 7A)\\ & =-(\cos 12A+\cos (-2A))\\ & =-(\cos 12A+\cos 2A)\\ & =-\cos 12A-\cos 2A\end{array}\end{array}$.

$\begin{array}{l}\\ 29.& \text{Bentuk sederhana dari}\\ & 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle 2+2\sin 2x}& & & \text{d}.& 2+2\sin x\\ \text{b}.& {\displaystyle 2+\sin 2x}& & & \text{e}.& 2+\sin x\\ \text{c}.& {\displaystyle 2\sin 2x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& 4\sin \left({\displaystyle \frac{1}{4}\pi +x}\right)\cos \left({\displaystyle \frac{1}{4}\pi -x}\right)\\ & =2(\sin \left({\displaystyle \frac{1}{2}\pi }\right)+\sin (2x))\\ & =2(1+\sin 2x)\\ & =2+2\sin 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 30.& \text{Nilai dari}\\ & \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -{\displaystyle \frac{3}{8}}}& & & \text{d}.& {\displaystyle \frac{3}{8}}\\ \\ \text{b}.& {\displaystyle -\frac{1}{8}}& & & \text{e}.& {\displaystyle \frac{5}{8}}\\ \text{c}.& {\displaystyle \frac{1}{8}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sqrt{3}\sin {80}^{\circ }\sin {160}^{\circ }\sin {320}^{\circ }\\ & =\sqrt{3}\sin {80}^{\circ }\sin {20}^{\circ }(-\sin {40}^{\circ })\\ & =-\sqrt{3}\sin {80}^{\circ }\sin {40}^{\circ }\sin {20}^{\circ }\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{2}(\cos {60}^{\circ }-\cos {20}^{\circ })})\\ & =-\sqrt{3}\sin {80}^{\circ }(-{\displaystyle \frac{1}{4}+{\displaystyle \frac{\cos {20}^{\circ }}{2}}})\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{2}\sqrt{3}\sin {80}^{\circ }\cos {20}^{\circ }}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {100}^{\circ }+\sin {60}^{\circ })}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}(\sin {80}^{\circ }+{\displaystyle \frac{1}{2}\sqrt{3}})}}\\ & ={\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }-{\displaystyle \frac{1}{4}\sqrt{3}\sin {80}^{\circ }+{\displaystyle \frac{1}{8}\sqrt{9}}}}\\ & ={\displaystyle \frac{3}{8}}\end{array}\end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kelima

$\begin{array}{l}\\ 21.& \text{Nilai}\cos {\displaystyle \frac{5}{12}\pi -\cos {\displaystyle \frac{1}{12}\pi \text{adalah}....}}\\ & \begin{array}{l}\\ \text{a}.& -{\displaystyle \frac{1}{2}\sqrt{6}}& & & \text{d}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \\ \text{b}.& -{\displaystyle \frac{1}{2}\sqrt{3}}& \text{c}.& -{\displaystyle \frac{1}{2}\sqrt{2}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{6}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \cos {\displaystyle \frac{5}{12}\pi -\cos {\displaystyle \frac{1}{12}\pi }}\\ & =-2\sin \left({\displaystyle \frac{{\displaystyle \frac{5}{12}\pi +{\displaystyle \frac{1}{12}\pi }}}{2}}\right)\sin \left({\displaystyle \frac{{\displaystyle \frac{5}{12}\pi -{\displaystyle \frac{1}{12}\pi }}}{2}}\right)\\ & =-2\sin \left({\displaystyle \frac{{\displaystyle \frac{6}{12}\pi }}{2}}\right)\sin \left({\displaystyle \frac{{\displaystyle \frac{4}{12}\pi }}{2}}\right)\\ & =-2\sin \left({\displaystyle \frac{1}{4}\pi }\right)\sin \left({\displaystyle \frac{1}{6}\pi }\right)\\ & =-2\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle -\frac{1}{2}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 22.& \text{Bentuk}\sin (2x-{\displaystyle \frac{3}{2}\pi })-\sin (4x+{\displaystyle \frac{1}{2}\pi })\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -2\sin 3x.\sin x}& & & \text{d}.& {\displaystyle 2\sin 3x.\sin x}\\ \text{b}.& {\displaystyle -2\cos 3x.\sin x}& & & \text{e}.& {\displaystyle 2\cos 3x.\sin x}\\ \text{c}.& {\displaystyle 2\sin 2(x-\pi )}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \sin (2x-{\displaystyle \frac{3}{2}\pi })-\sin (4x+{\displaystyle \frac{1}{2}\pi })\\ & =2\cos \left({\displaystyle \frac{2x-{\displaystyle \frac{3}{2}\pi +4x+{\displaystyle \frac{1}{2}\pi }}}{2}}\right)\\ & \times \sin \left({\displaystyle \frac{2x-{\displaystyle \frac{3}{2}\pi -(4x+{\displaystyle \frac{1}{2}\pi })}}{2}}\right)\\ & =2\cos (3x-{\displaystyle \frac{1}{2}\pi )\sin (-x-\pi )}\\ & =2\cos \left({\displaystyle \frac{1}{2}\pi -3x}\right)(-\sin (\pi +x))\\ & =2(\sin 3x)(-(-\sin x))\\ & =2\sin 3x.\sin x\end{array}\end{array}$.

$\begin{array}{l}\\ 23.& \text{Bentuk}{\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & \text{senilai dengan}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\tan 6x}& & & \text{d}.& 6\cot x\\ \text{b}.& {\displaystyle -\cot 6x}& & & \text{e}.& {\displaystyle \tan 6x}\\ \text{c}.& {\displaystyle 6\tan x}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos 3x-\sin 6x-\cos 9x}{\sin 9x-\cos 6x-\sin 3x}}\\ & ={\displaystyle \frac{\cos 3x-\cos 9x-\sin 6x}{\sin 9x-\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{-2\sin 6x\sin (-3x)-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{2\sin 6x\sin 3x-\sin 6x}{2\cos 6x\sin 3x-\cos 6x}}\\ & ={\displaystyle \frac{\sin 6x(2\sin 3x-1)}{\cos 6x(2\sin 3x-1)}}\\ & =\tan 6x\end{array}\end{array}$.

$\begin{array}{l}\\ 24.& \text{Nilai dari}\\ & {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan 2x}& & \text{d}.& {\displaystyle \tan 8x}\\ \text{b}.& {\displaystyle \tan 4x}& & \text{e}.& {\displaystyle \tan 16x}\\ \text{c}.& {\displaystyle {\displaystyle \tan 6x}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin x+\sin 3x+\sin 5x+\sin 7x}{\cos x+\cos 3x+\cos 5x+\cos 7x}}\\ & ={\displaystyle \frac{\sin 7x+\sin x+\sin 5x+\sin 3x}{\cos 7x+\cos x+\cos 5x+\cos 3x}}\\ & ={\displaystyle \frac{2\sin 4x\cos 3x+2\sin 4x\cos x}{2\cos 4x\cos 3x+2\cos 4x\cos x}}\\ & ={\displaystyle \frac{2\sin 4x(\cos 3x+\cos x)}{2\cos 4x(\cos 3x+\cos x)}}\\ & =\tan 4x\end{array}\end{array}$.

$\begin{array}{l}\\ 25.& \text{Bentuk sederhana dari}\\ & {\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \tan A}& & & \text{d}.& 2\cos 2A\\ \text{b}.& {\displaystyle 2\tan A}& & & \text{e}.& {\displaystyle 2\tan 2A}\\ \text{c}.& {\displaystyle 2\sin 2A}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& {\displaystyle \frac{\cos A+\sin A}{\cos A-\sin A}-\frac{\cos A-\sin A}{\cos A+\sin A}}\\ & ={\displaystyle \frac{(\cos A+\sin A{)}^2-(\cos A-\sin A{)}^2}{(\cos A-\sin A)(\cos A+\sin A)}}\\ & ={\displaystyle \frac{({\cos }^{2}A+2\cos A\sin A+{\sin }^{2}A)-({\cos }^{2}A-2\cos A\sin A+{\sin }^{2}A)}{{\cos }^{2}A-{\sin }^{2}A}}\\ & ={\displaystyle \frac{4\cos A\sin A}{{\cos }^{2}A-{\sin }^{2}A}}\\ & ={\displaystyle \frac{2\sin 2A}{\cos 2A}}\\ & =2\tan 2A\end{array}\end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Keempat

$\begin{array}{l}\\ 16.& \text{Himpunan penyelesaian dari persamaan}\\ & 3\tan (2x-{\displaystyle \frac{1}{3}\pi })=-\sqrt{3}\\ & \text{untuk}0\le x\le \pi \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{12}\pi ,{\displaystyle \frac{9}{12}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{3}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{3}{12}\pi ,{\displaystyle \frac{9}{12}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{5}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& 3\tan (2x-{\displaystyle \frac{1}{3}\pi })=-\sqrt{3}\\ & \tan (2x-{\displaystyle \frac{1}{3}\pi })=-\frac{\sqrt{3}}{3}\\ & (\text{kuadran IV, karena Y negatif, X positif})\\ & \tan (2x-{\displaystyle \frac{1}{3}\pi })=-\tan {\displaystyle \frac{1}{6}\pi ,\mathbf{\text{menjadi}}}\\ & \tan (2x-{\displaystyle \frac{1}{3}\pi })=\tan (2\pi -{\displaystyle \frac{1}{6}\pi })=\tan {\displaystyle \frac{11}{6}\pi }\\ & (2x-{\displaystyle \frac{1}{3}\pi })={\displaystyle \frac{11}{6}\pi }\\ & \iff 2x={\displaystyle \frac{1}{3}\pi +{\displaystyle \frac{11}{6}\pi +k.\pi ={\displaystyle \frac{13}{6}\pi +k.\pi }}}\\ & \iff x={\displaystyle \frac{13}{12}\pi +{\displaystyle \frac{k.\pi }{2}}}\\ & k=0\Rightarrow x={\displaystyle \frac{13}{12}\pi ={\displaystyle \frac{1}{12}\pi (\text{mm})}}\\ & k=1\Rightarrow x={\displaystyle \frac{13}{12}\pi +{\displaystyle \frac{1}{2}\pi ={\displaystyle \frac{19}{12}\pi ={\displaystyle \frac{7}{12}\pi (\text{mm})}}}}\\ & k=2\Rightarrow x={\displaystyle \frac{13}{12}\pi +\pi \text{tidak memenuhi}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{12}\pi ,{\displaystyle \frac{7}{12}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 17.& \text{Salah satu nilai}x\text{yang memenuhi}\\ & \text{persamaan}\cos x+\sin x={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{24}\pi }& & & \text{d}.& {\displaystyle \frac{1}{8}\pi }\\ \\ \text{b}.& {\displaystyle \frac{1}{15}\pi }& \text{c}.& {\displaystyle \frac{1}{12}\pi }& \text{e}.& {\displaystyle \frac{1}{6}\pi }\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \text{Diketahui bahwa}\\ & \sin x+\cos x={\displaystyle \frac{1}{2}\sqrt{6}(\mathbf{\text{ingat}}:a=1,b=1)}\\ & \sin x+\cos x=k\cos (x-\theta )={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \{\begin{array}{ll}k& =\sqrt{1^2+1^2}=\sqrt{2}\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{1}{1}=1\Rightarrow \theta ={45}^{\circ }={\displaystyle \frac{1}{4}\pi }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran I, karena}a,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& \sin x+\cos x=k\cos (x-\theta )={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \iff \sqrt{2}\cos (x-{\displaystyle \frac{1}{4}\pi })={\displaystyle \frac{1}{2}\sqrt{6}}\\ & \iff \cos (x-{\displaystyle \frac{1}{4}\pi })={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{6}}}{\sqrt{2}}={\displaystyle \frac{1}{2}\sqrt{3}}}\\ & \iff \cos (x-{\displaystyle \frac{1}{4}\pi })=\cos {\displaystyle \frac{1}{6}\pi }\\ & \iff x-{\displaystyle \frac{1}{4}\pi =\pm {\displaystyle \frac{1}{6}\pi +k.2\pi }}\\ & \iff x={\displaystyle \frac{1}{4}\pi \pm {\displaystyle \frac{1}{6}\pi +k.2\pi }}\\ & \iff x_1={\displaystyle \frac{5}{12}\pi +k.2\pi \mathbf{\text{atau}}}\\ & x_2={\displaystyle \frac{1}{12}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{5}{12}\pi (\text{memenuhi})}\\ & \Rightarrow x_2={\displaystyle \frac{1}{12}\pi (\text{memenuhi})}\\ & \text{Langkah berikutnya tidak diperlukan}\\ & \text{karena jawaban sudah kita dapatkan}\\ & \text{yaitu}:{\displaystyle \frac{1}{12}\pi }\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 18.& \text{Himpunan penyelesaian persamaan}\\ & \cos x^{\circ }-\sqrt{3}\sin x^{\circ }=1\\ & \text{untuk}0\le x<360\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \{0,240\}}& & & \text{d}.& {\displaystyle \{180,240\}}\\ \\ \text{b}.& {\displaystyle \{150,270\}}& \text{c}.& {\displaystyle \{180,300\}}& \text{e}.& {\displaystyle \{210,270\}}\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \text{Diketahui dari soal bahwa}\\ & \cos x^{\circ }-\sqrt{3}\sin x^{\circ }=1,\\ & \text{lalu kita ubah posisinya menjadi}\\ \\ & -\sqrt{3}\sin x+\cos x=1(\mathbf{\text{ingat}}:a=-\sqrt{3},b=1)\\ & -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \{\begin{array}{ll}k& =\sqrt{{(-\sqrt{3})}^2+{\left(1\right)}^2}=\sqrt{4}=2\\ \tan & \theta ={\displaystyle \frac{a}{b}={\displaystyle \frac{-\sqrt{3}}{1}=-\sqrt{3}\Rightarrow \theta ={300}^{\circ }}}\end{array}\\ & \text{sudut}\theta \text{di kuadran IV, karena}a<0,b>0\\ & \text{selanjutnya}\\ & \begin{array}{rl}& -\sqrt{3}\sin x+\cos x=k\cos (x-\theta )=1\\ & \iff 2\cos (x-{300}^{\circ })=1\\ & \iff \cos (x-{300}^{\circ })={\displaystyle \frac{1}{2}}\\ & \iff \cos (x-{300}^{\circ })=\cos {60}^{\circ }\\ & \iff x-{300}^{\circ }=\pm {60}^{\circ }+k{.360}^{\circ }\\ & \iff x={300}^{\circ }\pm {60}^{\circ }+k{.360}^{\circ }\\ & k=0\Rightarrow x_1={300}^{\circ }+{60}^{\circ }={360}^{\circ }=0^{\circ }(\text{mm})\\ & \text{atau}\\ & x_2={300}^{\circ }-{60}^{\circ }={240}^{\circ }(\text{mm})\\ & k=1\Rightarrow x={300}^{\circ }\pm {60}^{\circ }+{360}^{\circ }(\text{tm})\end{array}\\ & \mathbf{\text{HP}}=\{0^{\circ },{240}^{\circ }\}\end{array}\end{array}$

$\begin{array}{l}\\ 19.& \text{Diketahui}\alpha -\beta ={\displaystyle \frac{\pi }{3}\text{dan}\sin \alpha \sin \beta ={\displaystyle \frac{1}{4}}}\\ & \text{dengan}\alpha \text{dan}\beta \text{adalah sudut}\mathbf{\text{lancip}}\\ & \text{Nilai dari}\cos (\alpha +\beta )\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 1& & & \text{d}.& {\displaystyle \frac{1}{4}}\\ \text{b}.& {\displaystyle \frac{3}{4}}& \text{c}.& {\displaystyle \frac{1}{2}}& \text{e}.& {\displaystyle 0}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \bullet \alpha -\beta ={\displaystyle \frac{\pi }{3}\text{dan}\sin \alpha \sin \beta ={\displaystyle \frac{1}{4}}}\\ & \bullet \text{dengan}\alpha \text{dan}\beta \text{sudut}\mathbf{\text{lancip}}\\ & \text{akibatnya semua sudut dikuadran I}\\ & \text{sehingga}\{\begin{array}{ll}\sin & =+\\ \cos & =+\\ \tan & =+\end{array}\\ & \text{ditanya}\cos (\alpha +\beta ),\text{maka}\\ & \text{sebagai langkah awal kita adalah}:\\ & \cos (\alpha -\beta )=\cos \left({\displaystyle \frac{\pi }{3}}\right)\\ & \iff \cos \alpha \cos \beta +\sin \alpha \sin \beta ={\displaystyle \frac{1}{2}}\\ & \iff \cos \alpha \cos \beta +{\displaystyle \frac{1}{4}={\displaystyle \frac{1}{2}}}\\ & \iff \cos \alpha \cos \beta ={\displaystyle \frac{1}{2}-{\displaystyle \frac{1}{4}={\displaystyle \frac{1}{4}}}}\\ & \mathbf{\text{Selanjutnya nilai dari}}\\ & \cos (\alpha +\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & ={\displaystyle \frac{1}{4}-{\displaystyle \frac{1}{4}=0}}\end{array}\end{array}$.

$\begin{array}{l}\\ 20.& \text{Nilai}\sin {75}^{\circ }-\sin {165}^{\circ }\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{4}\sqrt{2}}& & & \text{d}.& {\displaystyle \frac{1}{2}\sqrt{2}}\\ \\ \text{b}.& {\displaystyle \frac{1}{4}\sqrt{3}}& \text{c}.& {\displaystyle \frac{1}{4}\sqrt{6}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{6}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \sin {75}^{\circ }-\sin {165}^{\circ }\\ & =2\cos \left({\displaystyle \frac{{75}^{\circ }+{165}^{\circ }}{2}}\right)\sin \left({\displaystyle \frac{{75}^{\circ }-{165}^{\circ }}{2}}\right)\\ & =2\cos {\displaystyle \frac{{240}^{\circ }}{2}\sin (-{\displaystyle \frac{{90}^{\circ }}{2}})}\\ & =2\cos {120}^{\circ }\sin (-{45}^{\circ })\\ & =2(-\cos {60}^{\circ })(-\sin {45}^{\circ })\\ & =2(-{\displaystyle \frac{1}{2}})(-{\displaystyle \frac{1}{2}\sqrt{2}})\\ & ={\displaystyle \frac{1}{2}\sqrt{2}}\end{array}\end{array}$.

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Ketiga

$\begin{array}{l}\\ 11.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\cos x}\\ \\ & \text{b}.{\displaystyle y=\cos 2x}\\ \\ & \text{c}.{\displaystyle y=\cos {\displaystyle \frac{1}{2}x}}\\ \\ & \text{d}.{\displaystyle y=2\cos 2x}\\ \\ & \text{e}.{\displaystyle y=2\cos {\displaystyle \frac{1}{2}x}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\\ & \mathbf{\text{atas dan ke bawah}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{2}={180}^{\circ }=\pi ,}\\ & \text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\cos 2x\end{array}\end{array}$.

$\begin{array}{l}\\ 12.& \text{Grafik fungsi trigonometri pada gambar}\\ & \text{berikut adalah}....\end{array}$.

$.\begin{array}{l}\\ & \text{a}.{\displaystyle y=2\sin (2x+\pi )}\\ \\ & \text{b}.{\displaystyle y=\sin (2x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{c}.{\displaystyle y=2\sin (2x-{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{d}.{\displaystyle y=\sin (2x+{\displaystyle \frac{1}{2}\pi })}\\ \\ & \text{e}.{\displaystyle y=2\sin {\displaystyle (x+{\displaystyle \frac{1}{2}\pi })}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}& \text{Dari grafik tampak jelas bahwa}\\ & \text{gambar di atas adalah garfik}\\ & \text{fungsi cosinus di geser ke}\mathbf{\text{kiri}}\\ & \text{dengan}\mathbf{\text{amplitudo}}2\\ & \text{dan}\mathbf{\text{periodenya}}{\displaystyle \frac{{360}^{\circ }}{1}={360}^{\circ }=2\pi ,}\\ & \text{maka}\\ & \text{bentuk persamaan}\mathbf{\text{grafik fungsinya}}\\ & y=2\sin (x+{\displaystyle kx})\\ & \text{dengan}+k\text{adalah}\\ & \mathbf{\text{besar geseran ke kiri}}{\displaystyle \frac{1}{2}\pi \text{atau}{90}^{\circ }}\\ & \text{Jadi},y=2\sin (x+{\displaystyle \frac{1}{2}\pi {)}^{\circ }}\end{array}\end{array}$.

$\begin{array}{l}\\ 13.& \text{Himpunan penyelesaian dari persamaan}\\ & \sin x=\sin {\displaystyle \frac{2}{10}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{22}{10}\pi ,{\displaystyle \frac{28}{10}\pi }}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{12}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \sin x=\sin {\displaystyle \frac{2}{10}\pi }\\ & \iff x_1={\displaystyle \frac{2}{10}\pi +k.2\pi \text{atau}}\\ & \iff x_2=(\pi -{\displaystyle \frac{2}{10}\pi })+k.2\pi ={\displaystyle \frac{8}{10}\pi +k.2\pi }\\ & k=0\Rightarrow x_1={\displaystyle \frac{2}{10}\pi (\text{mm})\text{atau}}\\ & x_2={\displaystyle \frac{8}{10}\pi (\text{mm})}\\ & k=1\Rightarrow x_{1,2}=....+2\pi (\text{tidak memenuhi})\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{2}{10}\pi ,{\displaystyle \frac{8}{10}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 14.& \text{Himpunan penyelesaian dari persamaan}\\ & \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi \text{untuk}{0}^{\circ }\le x\le {360}^{\circ }}\\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \left\{{\displaystyle \frac{1}{3}\pi ,\pi ,{\displaystyle \frac{5}{3}\pi ,{\displaystyle \frac{7}{3}\pi }}}\right\}}\\ \\ & \text{b}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{5}\pi ,\frac{5}{4}\pi ,\frac{8}{5}\pi }}\right\}}\\ \\ & \text{c}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{6}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{d}.{\displaystyle \left\{{\displaystyle \frac{2}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\pi ,{\displaystyle \frac{7}{4}\pi }}}\right\}}\\ \\ & \text{e}.{\displaystyle \left\{{\displaystyle \frac{1}{4}\pi ,\frac{3}{4}\pi ,\frac{5}{4}\pi ,{\displaystyle \frac{7}{4}\pi }}\right\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \tan (2x-{\displaystyle \frac{1}{4}\pi })=\tan {\displaystyle \frac{1}{4}\pi }\\ & \iff 2x-{\displaystyle \frac{1}{4}\pi ={\displaystyle \frac{1}{4}\pi +k.\pi }}\\ & \iff 2x={\displaystyle \frac{2}{4}\pi +k.\pi }\\ & \iff x={\displaystyle \frac{1}{4}\pi +k.\frac{\pi }{2}}\\ & k=0\Rightarrow x={\displaystyle \frac{1}{4}\pi (\text{mm})}\\ & k=1\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{\pi }{2}={\displaystyle \frac{3}{4}\pi (\text{mm})}}\\ & k=2\Rightarrow x={\displaystyle \frac{1}{4}\pi +\pi ={\displaystyle \frac{5}{4}\pi (\text{mm})}}\\ & k=3\Rightarrow x={\displaystyle \frac{1}{4}\pi +\frac{3\pi }{2}={\displaystyle \frac{7}{4}\pi (\text{mm})}}\\ & k=4\Rightarrow x={\displaystyle \frac{1}{4}\pi +2\pi ={\displaystyle \frac{9}{4}\pi (\text{tidak memenuhi})}}\end{array}\\ & \mathbf{\text{HP}}=\left\{{\displaystyle \frac{1}{4}\pi ,{\displaystyle \frac{3}{4}\pi ,\frac{5}{4}\pi ,\frac{7}{4}\pi }}\right\}\end{array}\end{array}$.

$\begin{array}{l}\\ 15.& \text{Himpunan penyelesaian dari persamaan}\\ & \cos 2x-2\cos x=-1\text{untuk}0<x<2\pi \\ & \text{adalah}....\\ \\ & \text{a}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{3}{2}\pi ,2\pi }}\}}\\ \\ & \text{b}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{2}{3}\pi ,2\pi }}\}}\\ \\ & \text{c}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,\pi ,{\displaystyle \frac{3}{2}\pi }}\}}\\ \\ & \text{d}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,{\displaystyle \frac{2}{3}\pi }}\}}\\ \\ & \text{e}.{\displaystyle \{0,{\displaystyle \frac{1}{2}\pi ,\pi }\}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{l}\\ & \begin{array}{rl}& \cos 2x-2\cos x=-1\\ & \iff \cos 2x-2\cos x+1=0\\ & \iff (2{\cos }^{2}x-1)-2\cos x+1=0\\ & \iff 2\cos x(\cos x-1)=0\\ & \iff \cos x=0\text{atau}\cos x=1\\ & \iff \cos x=\cos {\displaystyle \frac{1}{2}\pi \text{atau}\cos x=\cos 0}\\ & \iff x_{1,2}=\pm {\displaystyle \frac{1}{2}\pi +k.2\pi \text{atau}{x}_3=k.2\pi }\\ & \text{maka}\\ & k=0\Rightarrow x_1=-{\displaystyle \frac{1}{2}\pi (\text{tm})\text{atau}}\\ & x_2={\displaystyle \frac{1}{2}\pi (\text{mm})}\\ & x_3={\displaystyle 0(\text{mm})}\\ & k=1\Rightarrow x_1={\displaystyle \frac{3}{2}\pi (\text{mm})}\\ & x_2={\displaystyle \frac{5}{2}\pi (\text{tm})}\\ & x_3={\displaystyle 2\pi (\text{mm})}\end{array}\end{array}\end{array}$

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Kedua

$\begin{array}{l}\\ 6.& \text{Nilai dari}{\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& 0,143\\ \text{b}.& {\displaystyle -0,321}& \text{c}.& 0& \text{e}.& {\displaystyle 0,321}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{\sin {49}^{\circ }}{\cos {41}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin {73}^{\circ }}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\cos ({90}^{\circ }-{49}^{\circ })}-{\displaystyle \frac{\cos {17}^{\circ }}{\sin ({90}^{\circ }-{17}^{\circ })}}}\\ & ={\displaystyle \frac{\sin {49}^{\circ }}{\sin {49}^{\circ }}-{\displaystyle \frac{\cos {17}^{\circ }}{\cos {17}^{\circ }}}}\\ & =1-1\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Nilai dari}\\ & p=r\sin \alpha \cos \beta \\ & q=r\sin \alpha \sin \beta \\ & s=r\cos \alpha \\ & \text{maka pernyataan berikut yang}\\ & \text{tepat adalah}....\\ & \begin{array}{l}\\ \text{a}.& p^2+t^2+s^2=r^2& & & \\ \text{b}.& p^2-t^2+s^2=r^2\\ \text{c}.& p^2+t^2-s^2=r^2& \\ \text{d}.& -p^2+t^2+s^2=r^2& \\ \text{e}.& -p^2-t^2+s^2=r^2\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Saat}\\ & p^2+q^2\text{maka hasilnya adalah}\\ & \begin{array}{l}\\ p^2& =r^2{\sin }^2\alpha {\cos }^2\beta & \\ q^2& =r^2{\sin }^2\alpha {\sin }^2\beta & +\\ & =r^2{\sin }^2\alpha ({\cos }^2\beta +{\sin }^2\beta )\\ & =r^2{\sin }^2\alpha (1)\\ & =r^2{\sin }^2\alpha \end{array}\\ & \text{Dan saat}\\ & p^2+q^2+s^2\text{akan diperoleh hasil}\\ & \begin{array}{l}\\ p^2+q^2& =r^2{\sin }^2\alpha & \\ s^2& =r^2{\cos }^2\alpha & +\\ & =r^2{\sin }^2\alpha +r^2{\cos }^2\alpha \\ & =r^2({\sin }^2\alpha +{\cos }^2\alpha )\\ & =r^2(1)\\ & =r^2\end{array}\end{array}\end{array}$.

$\begin{array}{l}\\ 8.& \text{Nilai dari}\\ & {\displaystyle \frac{\cos ({90}^{\circ }+\theta )\sec (2\pi -\theta )\tan (\pi -\theta )}{\sec (\theta -2\pi )\sin ({540}^{\circ }+\theta )\cot (\theta -{90}^{\circ })}}\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -1& & & \text{d}.& -\tan \theta \\ \text{b}.& {\displaystyle 0}& \text{c}.& 1& \text{e}.& {\displaystyle \tan \theta }\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Ingat kembali sudut-sudut}\\ & \text{yang berelasi dari kudran selain I}\\ & \text{ke kuadran I beserta tandanya}\\ & {\displaystyle \frac{\cos ({90}^{\circ }+\theta )\sec (2\pi -\theta )\tan (\pi -\theta )}{\sec (\theta -2\pi )\sin ({540}^{\circ }+\theta )\cot (\theta -{90}^{\circ })}}\\ & ={\displaystyle \frac{(-\sin \theta ).\sec \theta .(-\tan \theta )}{\sec \theta .(-\sin \theta ).(-\tan \theta )}}\\ & =1\end{array}\end{array}$.

$\begin{array}{l}\\ 9.& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \text{maka nilai dari}\\ & {\sin }^3\theta +{\cos }^3\theta \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{1}{2}}& & & \text{d}.& {\displaystyle \frac{5}{8}}\\ \\ \text{b}.& {\displaystyle \frac{3}{4}}& \text{c}.& {\displaystyle \frac{9}{15}}& \text{e}.& {\displaystyle \frac{11}{16}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}\\ & \sin \theta +\cos \theta ={\displaystyle \frac{1}{2}}\\ & \iff {(\sin \theta +\cos \theta )}^2={\displaystyle \frac{1}{4}}\\ & \iff {\sin }^2\theta +{\cos }^2\theta +2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 1+2\sin \theta \cos \theta ={\displaystyle \frac{1}{4}}\\ & \iff 2\sin \theta \cos \theta =-{\displaystyle \frac{3}{4}}\\ & \iff \sin \theta \cos \theta =-{\displaystyle \frac{3}{8}}\\ & \mathbf{\text{Selanjutnya}}\\ & {\sin }^3\theta +{\cos }^3\theta \\ & =(\sin \theta +\cos \theta )({\sin }^2\theta -\sin \theta \cos \theta +{\cos }^2\theta )\\ & =\left({\displaystyle \frac{1}{2}}\right)(1-(-{\displaystyle \frac{3}{8}}))\\ & ={\displaystyle \frac{1}{2}\times {\displaystyle \frac{11}{8}}}\\ & ={\displaystyle \frac{11}{16}}\end{array}\end{array}$.

$\begin{array}{l}\\ 10.& \text{Jika diketahui}{\displaystyle \frac{3}{2}\pi <x<2\pi }\\ & \text{dan}\tan x=m,\\ & \text{maka nilai dari}\sin x\cos x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle -\frac{1}{m^2+1}}& & & \text{d}.& {\displaystyle -\frac{m}{m^2-1}}\\ \\ \text{b}.& {\displaystyle -\frac{m}{m^2+1}}& \text{c}.& {\displaystyle \frac{m}{m^2+1}}& \text{e}.& {\displaystyle \frac{m}{m^2-1}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& \text{Diketahui bahwa}{\displaystyle \frac{3}{2}\pi <x<2\pi }\\ & \text{ini daerah Kwadran IV, akibatnya adalah nilai}\\ & \{\begin{array}{ll}\sin x& =-\\ \cos x& =+\\ \tan x& =-\end{array}\\ & \mathbf{\text{Selanjutnya ada pernyataan}}\tan x=m\\ & \text{ini artinya}\tan x={\displaystyle \frac{m}{1}}\\ & \mathbf{\text{Perhatikanlah ilustrasi gambar berikut}}\end{array}\end{array}$.

$.\begin{array}{rl}& \text{maka nilai dari}\\ & \sin x\cos x\left(\text{ingat yang diminta di Kwadran IV}\right)\\ & =(-{\displaystyle \frac{m}{\sqrt{m^2+1}}})\times (+{\displaystyle \frac{1}{\sqrt{m^2+1}}})\\ & =-{\displaystyle \frac{m}{m^2+1}}\end{array}$

Soal dan jawaban Persiapan Semester gasal Kelas XI Matematika Peminatan Bagian Pertama

$\begin{array}{l}\\ 1.& \text{Nilai}{105}^{\circ }\text{jika dinyatakan ke radian}\\ & \text{adalah}....\text{radian}\\ \\ & \text{a}.{\displaystyle \frac{1}{3}\pi }\\ \\ & \text{b}.{\displaystyle \frac{5}{6}\pi }\\ \\ & \text{c}.{\displaystyle \frac{5}{12}\pi }\\ \\ & \text{d}.{\displaystyle \frac{7}{12}\pi }\\ \\ & \text{e}.{\displaystyle \frac{9}{12}\pi }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Diketah}& \text{ui bahwa}\\ {180}^{\circ }& =\pi radian\\ 1^{\circ }& ={\displaystyle \frac{\pi }{180}radian}\\ 105\times 1^{\circ }& =105\times {\displaystyle \frac{\pi }{180}radian}\\ {105}^{\circ }& ={\displaystyle \frac{7}{12}\pi radian}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nilai}\tan {240}^{\circ }\text{adalah}....\\ \\ & \text{a}.{\displaystyle \sqrt{3}}\\ \\ & \text{b}.{\displaystyle \frac{1}{3}\sqrt{3}}\\ \\ & \text{c}.{\displaystyle -\frac{1}{3}\sqrt{3}}\\ \\ & \text{d}.{\displaystyle \frac{1}{2}\sqrt{3}}\\ \\ & \text{e}.{\displaystyle -\sqrt{3}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {240}^{\circ }& =\tan ({180}^{\circ }+{60}^{\circ })\\ & =\tan {60}^{\circ }\\ & =\sqrt{3}\\ \mathbf{\text{catatan}}& :\text{ingat sudut berelasi}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Perhatikanlah gambar berikut}\end{array}$.

$.\begin{array}{l}\\ & \text{Pada gambar di atas perbandingan}\\ & \sin \theta \text{adalah}....\\ & \text{a}.\sqrt{{\displaystyle \frac{a^2-d^2}{f^2+g^2}}}\\ & \text{b}.\sqrt{{\displaystyle \frac{a^2+b^2}{f^2+g^2}}}\\ & \text{c}.\sqrt{{\displaystyle \frac{a^2-b^2}{f^2-g^2}}}\\ & \text{d}.\sqrt{{\displaystyle \frac{a^2+b^2}{f^2-g^2}}}\\ & \text{e}.\sqrt{{\displaystyle \frac{a^2-b^2}{f^2+g^2}}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{Dari so}& \text{al diketahui bahwa}\\ \sin \theta & ={\displaystyle \frac{c}{e}={\displaystyle \frac{\sqrt{a^2-b^2}}{\sqrt{f^2+g^2}}}}\\ & =\sqrt{{\displaystyle \frac{a^2-b^2}{f^2+g^2}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Nilai dari}({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 0& & & \text{d}.& 1\\ \text{b}.& {\displaystyle \frac{1}{3}}& \text{c}.& {\displaystyle \frac{2}{\sqrt{3}}}& \text{e}.& {\displaystyle \frac{1}{2}\sqrt{3}}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& ({\cos }^2{17}^{\circ }-{\sin }^2{73}^{\circ })\\ & =({\cos }^2{17}^{\circ }-{(\sin {73}^{\circ })}^2)\\ & =({\cos }^2{17}^{\circ }-{(\sin ({90}^{\circ }-{17}^{\circ }))}^2)\\ & =({\cos }^2{17}^{\circ }-{\cos }^2{17}^{\circ })\\ & =0\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Jika diketahui}\\ & {\displaystyle \frac{x{\csc }^2{30}^{\circ }{\sec }^2{45}^{\circ }}{8{\cos }^2{45}^{\circ }{\sin }^2{60}^{\circ }}={\tan }^2{60}^{\circ }-{\tan }^2{30}^{\circ },}\\ & \text{maka nilai}x\text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& -2& & & \text{d}.& 1\\ \text{b}.& {\displaystyle -1}& \text{c}.& 0& \text{e}.& {\displaystyle 2}\end{array}\\ \\ & \text{Jawab}:\\ & \begin{array}{rl}& {\displaystyle \frac{x{\csc }^2{30}^{\circ }{\sec }^2{45}^{\circ }}{8{\cos }^2{45}^{\circ }{\sin }^2{60}^{\circ }}={\tan }^2{60}^{\circ }-{\tan }^2{30}^{\circ }}\\ & {\displaystyle \frac{x\left(4\right)\left({\displaystyle \frac{4}{2}}\right)}{8\left({\displaystyle \frac{2}{4}}\right)\left({\displaystyle \frac{3}{4}}\right)}=3-\left({\displaystyle \frac{1}{3}}\right)}\\ & {\displaystyle \frac{8x}{3}={\displaystyle \frac{8}{3}}}\\ & x=1\end{array}\end{array}$

Lanjutan 6 Materi Rumus Jumlah dan Selisih Sinus dan Cosinus

Untuk rumus jum lah dan selisih sinus dan cosinus untuk sudut berupa X dan Y adalah sebagai berikut:

$\{\begin{array}{ll}\sin X+\sin Y& =2\sin \frac{1}{2}(X+Y)\cos \frac{1}{2}(X-Y)\\ \sin X-\sin Y& =2\cos \frac{1}{2}(X+Y)\sin \frac{1}{2}(X-Y)\\ \cos X+\cos Y& =2\cos \frac{1}{2}(X+Y)\cos \frac{1}{2}(X-Y)\\ \cos X-\cos Y& =-2\sin \frac{1}{2}(X+Y)\sin \frac{1}{2}(X-Y)\end{array}$.

Berikut contoh proses pembuktian rumus no.1, yaitu:

$\begin{array}{l}\\ \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta & & \\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta & +& \\ \sin (\alpha +\beta )+\sin (\alpha -\beta )=2\sin \alpha \cos \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \sin X+\sin Y=2\sin {\displaystyle \frac{(X+Y)}{2}\cos {\displaystyle \frac{(X-Y)}{2}}}\end{array}$.

Dan berikut proses pembuktian rumus no. 2, yaitu:

$\begin{array}{l}\\ \sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta & & \\ \sin (\alpha -\beta )=\sin \alpha \cos \beta -\cos \alpha \sin \beta & -& \\ \sin (\alpha +\beta )-\sin (\alpha -\beta )=2\cos \alpha \sin \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \sin X-\sin Y=2\cos {\displaystyle \frac{(X+Y)}{2}\sin {\displaystyle \frac{(X-Y)}{2}}}\end{array}$

Berikut untuk proses pembktian rumus no. 3, yaitu:

$\begin{array}{l}\\ \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta & & \\ \cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta & +& \\ \cos (\alpha +\beta )+\cos (\alpha -\beta )=2\cos \alpha \cos \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \cos X+\cos Y=2\cos {\displaystyle \frac{(X+Y)}{2}\cos {\displaystyle \frac{(X-Y)}{2}}}\end{array}$

Dan berikut tyang terakhir untuk bukti pada rumus terakhir no.4, yaitu:

$\begin{array}{l}\\ \cos (\alpha +\beta )=\cos \alpha \cos \beta -\sin \alpha \sin \beta & & \\ \cos (\alpha -\beta )=\cos \alpha \cos \beta +\sin \alpha \sin \beta & -& \\ \cos (\alpha +\beta )-\cos (\alpha -\beta )=-2\cos \alpha \sin \beta & & \\ \\ \text{saat}X=\alpha +\beta X=\alpha +\beta & & \\ Y=\alpha -\beta +Y=\alpha -\beta & -& \\ X+Y=2\alpha X-Y=2\beta & & \\ 2\alpha =X+Y2\beta =X-Y& & \\ \alpha ={\displaystyle \frac{X+Y}{2}\beta ={\displaystyle \frac{X-Y}{2}}}\\ \mathbf{\text{Sehingga rumus akan menjadi}}& & \\ \cos X-\cos Y=-2\sin {\displaystyle \frac{(X+Y)}{2}\sin {\displaystyle \frac{(X-Y)}{2}}}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Sederhanakanlah bentuk berikut ini}\\ & \begin{array}{l}\\ \text{a}.& \sin (2x+{60}^{\circ })-\sin (2x-{60}^{\circ })\\ \text{b}.& \sin (x+{\displaystyle \frac{1}{4}m})+\sin (x-{\displaystyle \frac{1}{4}m})\\ \text{c}.& \cos (2x+4y)-\cos (2x-4y)\\ \text{d}.& \cos \left({\displaystyle \frac{5}{4}m+3x}\right)+\cos \left({\displaystyle \frac{5}{4}m-3x}\right)\end{array}\\ \\ & \mathbf{\text{Jawab}}:\\ \\ & \begin{array}{rl}\text{a}.& \sin (2x+{60}^{\circ })-\sin (2x-{60}^{\circ })\\ & =2\cos {\displaystyle \frac{(2x+{60}^{\circ })+(2x-{60}^{\circ })}{2}\sin {\displaystyle \frac{(2x+{60}^{\circ })-(2x-{60}^{\circ })}{2}}}\\ & =2\cos {\displaystyle \frac{4x}{2}\sin {\displaystyle \frac{{120}^{\circ }}{2}}}\\ & =2\cos 2x\sin {60}^{\circ }\\ & =2\cos 2x.\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & =\sqrt{3}\cos 2x\end{array}\\ & \begin{array}{rl}\text{b}.& \sin (x+{\displaystyle \frac{1}{4}m})+\sin (x-{\displaystyle \frac{1}{4}m})\\ & =2\sin {\displaystyle \frac{(x+{\displaystyle \frac{1}{4}m+x-\frac{1}{4}m})}{2}\cos {\displaystyle \frac{(x+{\displaystyle \frac{1}{4}m-(x-\frac{1}{4}m)})}{2}}}\\ & =2\sin {\displaystyle \frac{2x}{2}\cos {\displaystyle \frac{{\displaystyle \frac{2}{4}m}}{2}}}\\ & =2\sin x\cos {\displaystyle \frac{1}{4}m}\end{array}\\ & \begin{array}{rl}\text{c}.& \cos (2x+4y)-\cos (2x-4y)\\ & =-2\sin {\displaystyle \frac{(2x+4y)+(2x-4y)}{2}\sin {\displaystyle \frac{(2x+4y)-(2x-4y)}{2}}}\\ & =-2\sin {\displaystyle \frac{4x}{2}\sin {\displaystyle \frac{8y}{2}}}\\ & =-2\sin 2x\sin 4y\end{array}\\ & \begin{array}{rl}\text{d}.& \cos \left({\displaystyle \frac{5}{4}m+3x}\right)+\cos \left({\displaystyle \frac{5}{4}m-3x}\right)\\ & =2\cos {\displaystyle \frac{\left({\displaystyle \frac{5}{4}m+3x+{\displaystyle \frac{5}{4}m-3x}}\right)}{2}\cos {\displaystyle \frac{\left({\displaystyle \frac{5}{4}m+3x-\left({\displaystyle \frac{5}{4}m-3x}\right)}\right)}{2}}}\\ & =2\cos {\displaystyle \frac{5m}{2}\cos {\displaystyle \frac{6x}{2}}}\\ & =2\cos {\displaystyle \frac{5x}{2}\cos 3x}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Buktikanlah bahwa}\\ & \begin{array}{l}\\ {\displaystyle \frac{\sin 3\gamma +\sin \gamma }{\cos 3\gamma +\cos \gamma }=\tan 2\gamma }\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{|ll|}\hline \begin{array}{rl}{\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }}& ={\displaystyle \frac{2\sin {\displaystyle \frac{(3\alpha +\alpha )}{2}\cos {\displaystyle \frac{(3\alpha -\alpha )}{2}}}}{2\cos {\displaystyle \frac{(3\alpha +\alpha )}{2}\cos {\displaystyle \frac{(3\alpha -\alpha )}{2}}}}}\\ & ={\displaystyle \frac{\sin 2\alpha }{\cos 2\alpha }}\\ & =\tan 2\alpha ◼\\ & \\ & \\ & \mathbf{\text{atau}}\Rightarrow \\ & \text{mungkin lumayan rumit}\\ & \\ & \end{array}& \begin{array}{rl}{\displaystyle \frac{\sin 3\alpha +\sin \alpha }{\cos 3\alpha +\cos \alpha }}& ={\displaystyle \frac{3\sin \alpha -4{\sin }^3\alpha +\sin \alpha }{4{\cos }^3\alpha -3\cos \alpha +\cos \alpha }}\\ & ={\displaystyle \frac{4\sin \alpha -4{\sin }^3\alpha }{4{\cos }^3\alpha -2\cos \alpha }}\\ & ={\displaystyle \frac{4\sin \alpha (1-{\sin }^2\alpha )}{2\cos \alpha (2{\cos }^2\alpha -1)}={\displaystyle \frac{4\sin \alpha }{2\cos \alpha }\times {\displaystyle \frac{{\cos }^2\alpha }{\cos 2\alpha }}}}\\ & ={\displaystyle \frac{2\tan \alpha {\cos }^2\alpha }{\cos 2\alpha }}\\ & ={\displaystyle \frac{2\tan \alpha {\cos }^2\alpha }{{\cos }^2\alpha -{\sin }^2\alpha }}\\ & ={\displaystyle \frac{2\tan \alpha }{{\displaystyle \frac{{\cos }^2\alpha }{{\cos }^2\alpha }-{\displaystyle \frac{{\sin }^2\alpha }{{\cos }^2\alpha }}}}={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}}\\ & =\tan 2\alpha ◼\end{array}\\ \hline\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Buktikanlah bahwa}\\ & \begin{array}{l}\\ {\displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta }=\tan \theta }\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}{\displaystyle \frac{\sin 5\theta -\sin 3\theta }{\cos 3\theta +\cos 5\theta }}& ={\displaystyle \frac{2\cos {\displaystyle \frac{(5\theta +3\theta )}{2}\sin {\displaystyle \frac{(5\theta -3\theta )}{2}}}}{2\cos {\displaystyle \frac{(3\theta +5\theta )}{2}\cos {\displaystyle \frac{(3\theta -5\theta )}{2}}}}}\\ & ={\displaystyle \frac{\sin \theta }{\cos (-\theta )}={\displaystyle \frac{\sin \theta }{\cos \theta }}}\\ & =\tan \theta ◼\\ \\ \mathbf{\text{Catatan}}:& \text{ingat bahwa saat}\cos \text{sudut}\\ & -\theta =\theta ,\text{sehingga}\cos (-\theta )=\cos \theta \end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Buktikanlah bahwa}\\ & \begin{array}{l}\\ {\displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta }=\cot 2\beta }\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rl}{\displaystyle \frac{\sin 3\beta -\sin \beta }{\cos \beta -\cos 3\beta }}& ={\displaystyle \frac{2\cos {\displaystyle \frac{(3\beta +\beta )}{2}\sin {\displaystyle \frac{(3\beta -\beta )}{2}}}}{-2\sin {\displaystyle \frac{(\beta +3\beta )}{2}\sin {\displaystyle \frac{(\beta -3\beta )}{2}}}}}\\ & ={\displaystyle \frac{\cos 2\beta \times \sin \beta }{\sin 2\beta (-\sin (-\beta ))}={\displaystyle \frac{\cos 2\beta }{\sin 2\beta }}}\\ & =\cot 2\beta ◼\\ \\ \mathbf{\text{Catatan}}:& \text{ingat bahwa saat}\sin \text{sudut}\\ & -\beta =-\beta ,\text{sehingga}\sin (-\beta )=-\sin \beta \\ & \text{dan}-\sin (-\beta )=-(-\sin \beta )=\sin \beta \end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Tentukan nilai dari}\\ & \begin{array}{l}\\ \text{a}.& \cos {105}^{\circ }+\cos {15}^{\circ }\\ \text{b}.& \sin {105}^{\circ }-\sin {15}^{\circ }\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{a}.& \cos {105}^{\circ }+\cos {15}^{\circ }\\ & =2\cos \left({\displaystyle \frac{{105}^{\circ }+{15}^{\circ }}{2}}\right)\cos \left({\displaystyle \frac{{105}^{\circ }-{15}^{\circ }}{2}}\right)\\ & =2\cos {\displaystyle \frac{{120}^{\circ }}{2}\cos {\displaystyle \frac{{90}^{\circ }}{2}}}\\ & =2\cos {60}^{\circ }\cos {45}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}}\end{array}\\ & \begin{array}{rl}\text{b}.& \sin {105}^{\circ }-\sin {15}^{\circ }\\ & =2\cos \left({\displaystyle \frac{{105}^{\circ }+{15}^{\circ }}{2}}\right)\sin \left({\displaystyle \frac{{105}^{\circ }-{15}^{\circ }}{2}}\right)\\ & =2\cos {\displaystyle \frac{{120}^{\circ }}{2}\sin {\displaystyle \frac{{90}^{\circ }}{2}}}\\ & =2\cos {60}^{\circ }\sin {45}^{\circ }\\ & =2\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{2}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Tentukan nilai dari}\\ & \begin{array}{l}\\ \text{a}.& {\displaystyle \frac{\cos {75}^{\circ }+\cos {15}^{\circ }}{\sin {75}^{\circ }-\sin {15}^{\circ }}}\\ \\ \text{b}.& {\displaystyle \frac{\cos {195}^{\circ }-\cos {105}^{\circ }}{\sin {105}^{\circ }-\sin {15}^{\circ }}}\end{array}\\ \\ & \mathbf{\text{Jawab}}\\ & \begin{array}{rl}\text{a}.& {\displaystyle \frac{\cos {75}^{\circ }+\cos {15}^{\circ }}{\sin {75}^{\circ }-\sin {15}^{\circ }}}\\ & ={\displaystyle \frac{2\cos \left({\displaystyle \frac{{75}^{\circ }+{15}^{\circ }}{2}}\right)\cos \left({\displaystyle \frac{{75}^{\circ }-{15}^{\circ }}{2}}\right)}{2\cos \left({\displaystyle \frac{{75}^{\circ }+{15}^{\circ }}{2}}\right)\sin \left({\displaystyle \frac{{75}^{\circ }-{15}^{\circ }}{2}}\right)}}\\ & ={\displaystyle \frac{\cos {\displaystyle \frac{{90}^{\circ }}{2}\cos {\displaystyle \frac{{60}^{\circ }}{2}}}}{\cos {\displaystyle \frac{{90}^{\circ }}{2}\sin {\displaystyle \frac{{60}^{\circ }}{2}}}}}\\ & ={\displaystyle \frac{\cos {30}^{\circ }}{\sin {30}^{\circ }}=\cot {30}^{\circ }}\\ & =\sqrt{3}\end{array}\\ & \begin{array}{rl}\text{b}.& {\displaystyle \frac{\cos {195}^{\circ }-\cos {105}^{\circ }}{\sin {105}^{\circ }-\sin {15}^{\circ }}}\\ & ={\displaystyle \frac{-2\sin \left({\displaystyle \frac{{195}^{\circ }+{105}^{\circ }}{2}}\right)\sin \left({\displaystyle \frac{{195}^{\circ }-{105}^{\circ }}{2}}\right)}{2\cos \left({\displaystyle \frac{{105}^{\circ }+{15}^{\circ }}{2}}\right)\sin \left({\displaystyle \frac{{105}^{\circ }-{15}^{\circ }}{2}}\right)}}\\ & =-{\displaystyle \frac{\sin {\displaystyle \frac{{300}^{\circ }}{2}\sin {\displaystyle \frac{{90}^{\circ }}{2}}}}{\cos {\displaystyle \frac{{120}^{\circ }}{2}\sin {\displaystyle \frac{{90}^{\circ }}{2}}}}}\\ & =-{\displaystyle \frac{\sin {150}^{\circ }}{\cos {60}^{\circ }}=-{\displaystyle \frac{\sin ({180}^{\circ }-{30}^{\circ })}{\cos {60}^{\circ }}}}\\ & =-{\displaystyle \frac{\sin {30}^{\circ }}{\cos {60}^{\circ }}}\\ & =-{\displaystyle \frac{{\displaystyle \frac{1}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =-1\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Noormandiri, B.K. 2017. Matematika untuk Kelas SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA
2. Sukino. 2017. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Lanjutan 5 Materi Rumus Perkalian Sinus dan Cosinus

C.   Rumus Perkalian Sinus dan Cosinus

Untuk sudut  $\alpha \text{dan}\beta$ berlaku rumus-rumus

$\begin{array}{rl}& \{\begin{array}{ll}2\sin \alpha \cos \gamma & =\sin (\alpha +\gamma )+\sin (\alpha -\gamma )\\ 2\cos \alpha \sin \gamma & =\sin (\alpha +\gamma )-\sin (\alpha -\gamma )\end{array}\\ & \{\begin{array}{ll}2\cos \alpha \cos \gamma & =\cos (\alpha +\gamma )+\cos (\alpha -\gamma )\\ -2\sin \alpha \sin \gamma & =\cos (\alpha +\gamma )-\cos (\alpha -\gamma )\end{array}\end{array}$.

Sebagai buktinya akan ditunjukkan pada tulisan berikut

Bukti untuk nomor pertama

$\begin{array}{rl}\sin & (\alpha +\beta )+\sin (\alpha -\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & +\sin \alpha \cos \beta -\cos \alpha \sin \beta \\ & =2\sin \alpha \cos \beta \end{array}$.

Selanjutnya untuk bukti baris kedua yaitu:

$\begin{array}{rl}\sin & (\alpha +\beta )-\sin (\alpha -\beta )\\ & =\sin \alpha \cos \beta +\cos \alpha \sin \beta \\ & -(\sin \alpha \cos \beta -\cos \alpha \sin \beta )\\ & =2\cos \alpha \sin \beta \end{array}$.

Dan bukti untuk rumus ketiga

$\begin{array}{rl}\cos & (\alpha +\beta )+\cos (\alpha -\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & +\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ & =2\cos \alpha \cos \beta \end{array}$.

Adapun bukti untuk rumus yang tertakhir adalah:

$\begin{array}{rl}\cos & (\alpha +\beta )-\cos (\alpha -\beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta \\ & -(\cos \alpha \cos \beta +\sin \alpha \sin \beta )\\ & =-2\sin \alpha \sin \beta \end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Nyatakanlah ke dalam bentuk jumlah}\\ & \text{atau selisih dari bentuk berikut}\\ & \text{a}.2\cos {80}^{\circ }\sin {50}^{\circ }\\ & \text{b}.2\sin {80}^{\circ }\cos {50}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\cos {80}^{\circ }\sin {50}^{\circ }\\ & =\sin ({80}^{\circ }+{50}^{\circ })-\sin ({80}^{\circ }-{50}^{\circ })\\ & =\sin {130}^{\circ }-\sin {30}^{\circ }\\ & =\sin {130}^{\circ }-{\displaystyle \frac{1}{2}}\\ \text{b}.& 2\sin {80}^{\circ }\cos {50}^{\circ }\\ & =\sin ({80}^{\circ }+{50}^{\circ })+\sin ({80}^{\circ }-{50}^{\circ })\\ & =\sin {130}^{\circ }+\sin {30}^{\circ }\\ & =\sin {130}^{\circ }+{\displaystyle \frac{1}{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Nyatakanlah ke dalam bentuk jumlah}\\ & \text{atau selisih dari bentuk berikut}\\ & \text{a}.2\cos {75}^{\circ }\cos {15}^{\circ }\\ & \text{b}.\sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\cos {75}^{\circ }\cos {15}^{\circ }\\ & =\cos ({75}^{\circ }+{15}^{\circ })+\cos ({75}^{\circ }-{15}^{\circ })\\ & =\cos {90}^{\circ }+\cos {60}^{\circ }\\ & =0+{\displaystyle \frac{1}{2}={\displaystyle \frac{1}{2}}}\\ \text{b}.& \sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right)\\ & =-{\displaystyle \frac{1}{2}(2\sin \left({\displaystyle \frac{3\pi }{8}}\right)\sin \left({\displaystyle \frac{\pi }{8}}\right))}\\ & =-{\displaystyle \frac{1}{2}(\cos \left({\displaystyle \frac{3\pi }{8}+\frac{\pi }{8}}\right)-\cos \left({\displaystyle \frac{3\pi }{8}-\frac{\pi }{8}}\right))}\\ & =-{\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{4\pi }{8}-\cos {\displaystyle \frac{2\pi }{8}}}\right)}\\ & =-{\displaystyle \frac{1}{2}\left(\cos {\displaystyle \frac{1}{2}\pi -\cos {\displaystyle \frac{1}{4}\pi }}\right)}\\ & =-{\displaystyle \frac{1}{2}(0-{\displaystyle \frac{1}{2}\sqrt{2}})}\\ & =-{\displaystyle \frac{1}{2}(-{\displaystyle \frac{1}{2}\sqrt{2}})}\\ & ={\displaystyle \frac{1}{4}\sqrt{2}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari bentuk berikut}\\ \\ & \text{a}.2\sin 37{\displaystyle {\frac{1}{2}}^{\circ }\cos 7{\displaystyle {\frac{1}{2}}^{\circ }}}\\ \\ & \text{b}.2\cos 82{\displaystyle {\frac{1}{2}}^{\circ }\sin 37{\displaystyle {\frac{1}{2}}^{\circ }}}\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\text{a}.& 2\sin 37{\displaystyle {\frac{1}{2}}^{\circ }\cos 7{\displaystyle {\frac{1}{2}}^{\circ }}}\\ & =\sin {\left(37{\displaystyle \frac{1}{2}+7{\displaystyle \frac{1}{2}}}\right)}^{\circ }+{\displaystyle \sin {\left(37{\displaystyle \frac{1}{2}-7{\displaystyle \frac{1}{2}}}\right)}^{\circ }}\\ & =\sin {45}^{\circ }+\sin {30}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{2}+{\displaystyle \frac{1}{2}}}\\ & ={\displaystyle \frac{1}{2}(\sqrt{2}+1)}\\ \text{b}.& 2\cos 82{\displaystyle {\frac{1}{2}}^{\circ }\sin 37{\displaystyle {\frac{1}{2}}^{\circ }}}\\ & =\sin {\left(82{\displaystyle \frac{1}{2}+37{\displaystyle \frac{1}{2}}}\right)}^{\circ }-{\displaystyle \sin {\left(82{\displaystyle \frac{1}{2}-37{\displaystyle \frac{1}{2}}}\right)}^{\circ }}\\ & =\sin {120}^{\circ }-\sin {45}^{\circ }\\ & =\sin ({180}^{\circ }-{60}^{\circ })-\sin {45}^{\circ }\\ & =\sin {60}^{\circ }-\sin {45}^{\circ }\\ & ={\displaystyle \frac{1}{2}\sqrt{3}-\frac{1}{2}\sqrt{2}}\\ & ={\displaystyle \frac{1}{2}(\sqrt{3}-\sqrt{2})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa}\\ & \text{a}.2\cos \left({\displaystyle \frac{1}{4}\pi +\theta }\right)\cos \left({\displaystyle \frac{3}{4}\pi -\theta }\right)=\sin 2\theta -1\\ & \text{b}.2\sin ({315}^{\circ }+B)\sin ({45}^{\circ }-B)=\sin 2B-1\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\text{a}.& 2\cos \left({\displaystyle \frac{1}{4}\pi +\theta }\right)\cos \left({\displaystyle \frac{3}{4}\pi -\theta }\right)\\ & =\cos \left({\displaystyle \frac{1}{4}\pi +\theta +{\displaystyle \frac{3}{4}\pi -\theta }}\right)+\cos \left({\displaystyle \frac{1}{4}\pi +\theta -\left({\displaystyle \frac{3}{4}\pi -\theta }\right)}\right)\\ & =\cos \left({\displaystyle \frac{1}{4}\pi +\theta +{\displaystyle \frac{3}{4}\pi -\theta }}\right)+\cos \left({\displaystyle \frac{1}{4}\pi +\theta -{\displaystyle \frac{3}{4}\pi +\theta }}\right)\\ & =\cos \pi +\cos (2\theta -{\displaystyle \frac{1}{2}\pi })\\ & =-1+\cos \left({\displaystyle \frac{1}{2}\pi -2\theta }\right)\\ & =-1+\sin 2\theta (\mathbf{\text{ingat sudut yang berelasi}})\\ & =\sin 2\theta -1◼\\ \text{b}.& 2\sin ({315}^{\circ }+B)\sin ({45}^{\circ }-B)\\ & =-(\cos ({315}^{\circ }+B+{45}^{\circ }-B)-\cos ({315}^{\circ }+B-({45}^{\circ }-B)))\\ & =-(\cos {360}^{\circ }-\cos ({315}^{\circ }-{45}^{\circ }+B+B))\\ & -=-(1-\cos ({270}^{\circ }+2B))\\ & =-(1-\sin 2B)(\mathbf{\text{ingat sudut yang berelasi}})\\ & =\sin 2B-1◼\end{array}\end{array}$

Lanjutan 4 Materi Rumus Sudut Paruh pada Trigonometri

Sebelumnya telah dituliskan bahwa

$\begin{array}{l}\\ \bullet & \sin \alpha =2\sin {\displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha }\\ \bullet & \cos \alpha ={\cos }^2{\displaystyle \frac{1}{2}\alpha -{\sin }^2{\displaystyle \frac{1}{2}\alpha }}\\ \bullet & \tan \alpha ={\displaystyle \frac{2\tan {\displaystyle \frac{1}{2}\alpha }}{1-{\tan }^2{\displaystyle \frac{1}{2}\alpha }}}\end{array}$.

 B. 2. 1   Rumus Sudut Paruh untuk Sinus

Perhatikanlah Identitas trigonometri berikut

Perhatikanlah ilsutrasi segitiga ABC berikut

• ${\sin }^2\alpha +{\cos }^2\alpha =1$
• ${\tan }^2\alpha -{\sec }^2\alpha =-1$
• ${\cot }^2\alpha -{\csc }^2\alpha =-1$

Serta pada pembahasan sebelumnya untuk sudut rangkap bahwa

• $\sin 2\alpha =2\sin \alpha \cos \alpha$
• $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$, dan 
• $\cos 2\alpha =2{\cos }^2\alpha -1$, serta
• $\cos 2\alpha =1-2{\sin }^2\alpha$

Maka rumus $\sin {\displaystyle \frac{1}{2}\alpha }$ adalah:

$\begin{array}{rl}\cos 2\alpha & =1-2{\sin }^2\alpha \\ 2{\sin }^2\alpha & =1-\cos 2\alpha \\ {\sin }^2\alpha & ={\displaystyle \frac{1-\cos 2\alpha }{2}}\\ \sin \alpha & =\sqrt{{\displaystyle \frac{1-\cos 2\alpha }{2}}}\\ \text{saat}& \text{sudut}\alpha \text{diganti}{\displaystyle \frac{1}{2}\alpha ,}\\ \text{maka}& \text{rumus akan menjadi}\\ \sin \left({\displaystyle \frac{1}{2}\alpha }\right)& =\sqrt{{\displaystyle \frac{1-\cos 2\left({\displaystyle \frac{1}{2}\alpha }\right)}{2}}}\\ \sin {\displaystyle \frac{1}{2}\alpha }& =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{2}}}\end{array}$.

B. 2. 2   Rumus Sudut Paruh untuk Cosinus

Sedangkan untuk rumus cosinusnya adalah:

$\begin{array}{rl}\cos 2\alpha & =2{\cos }^2\alpha -1\\ 2{\cos }^2\alpha & =1+\cos 2\alpha \\ {\cos }^2\alpha & ={\displaystyle \frac{1+\cos 2\alpha }{2}}\\ \cos \alpha & =\sqrt{{\displaystyle \frac{1+\cos 2\alpha }{2}}}\\ \text{saat}& \text{sudut}\alpha \text{diganti}{\displaystyle \frac{1}{2}\alpha ,}\\ \text{maka}& \text{rumus akan menjadi}\\ \cos \left({\displaystyle \frac{1}{2}\alpha }\right)& =\sqrt{{\displaystyle \frac{1+\cos 2\left({\displaystyle \frac{1}{2}\alpha }\right)}{2}}}\\ \cos {\displaystyle \frac{1}{2}\alpha }& =\pm \sqrt{{\displaystyle \frac{1+\cos \alpha }{2}}}\end{array}$

B. 2. 3   Rumus Sudut Paruh untuk Tangen

Adapun untuk rumus tangen adalah:

$\begin{array}{rl}\tan {\displaystyle \frac{1}{2}\alpha }& ={\displaystyle \frac{\sin {\displaystyle \frac{1}{2}\alpha }}{\cos {\displaystyle \frac{1}{2}\alpha }}}\\ & ={\displaystyle \frac{\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{2}}}}{\pm \sqrt{{\displaystyle \frac{1+\cos \alpha }{2}}}}}\\ & =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }}}\end{array}$

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\tan {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{1-\cos \alpha }{\sin \alpha }}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}\tan {\displaystyle \frac{1}{2}\alpha }& =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }}},\text{maka}\\ & =\pm \sqrt{{\displaystyle \frac{1-\cos \alpha }{1+\cos \alpha }\times {\displaystyle \frac{1-\cos \alpha }{1-\cos \alpha }}}}\\ & =\pm \sqrt{{\displaystyle \frac{{(1-\cos \alpha )}^2}{1-{\cos }^2\alpha }}}\\ & =\pm \sqrt{{\displaystyle \frac{{(1-\cos \alpha )}^2}{{\sin }^2\alpha }}}\\ & ={\displaystyle {\displaystyle \frac{1-\cos \alpha }{\sin \alpha }◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\tan {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{\sin \alpha }{1+\cos \alpha }}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}\tan {\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{1-\cos \alpha }{\sin \alpha },\text{maka}}}\\ & ={\displaystyle \frac{1-\cos \alpha }{\sin \alpha }\times \frac{1+\cos \alpha }{1+\cos \alpha }}\\ & ={\displaystyle \frac{1-{\cos }^2\alpha }{\sin \alpha (1+\cos \alpha )}}\\ & ={\displaystyle \frac{{\sin }^2\alpha }{\sin \alpha (1+\cos \alpha )}}\\ & ={\displaystyle {\displaystyle \frac{\sin \alpha }{1+\cos \alpha }◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa nilai}{\cos }^2{\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{1+\sec \alpha }{2\sec \alpha }}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}{\cos }^2{\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{1+\cos \alpha }{2},\text{maka}}}\\ & ={\displaystyle {\displaystyle \frac{1+\cos \alpha }{2}\times {\displaystyle \frac{{\displaystyle \frac{1}{\cos \alpha }}}{{\displaystyle \frac{1}{\cos \alpha }}}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{\cos \alpha }+{\displaystyle \frac{\cos \alpha }{\cos \alpha }}}}{{\displaystyle \frac{2}{\cos \alpha }}}}\\ & ={\displaystyle \frac{\sec \alpha +1}{2\sec \alpha }}\\ & ={\displaystyle {\displaystyle \frac{1+\sec \alpha }{2\sec \alpha }◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa nilai}\sin {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }}{2}}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \text{Sebelumnya diketahui bahwa}:\\ & \begin{array}{rl}{\sin }^2{\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{1-\cos \alpha }{2},\text{maka}}}\\ & ={\displaystyle \frac{1-\sqrt{{\cos }^2\alpha }}{2}}\\ & ={\displaystyle \frac{1-\sqrt{1-{\sin }^2\alpha }}{2}}\\ & ={\displaystyle \frac{1-\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{2}}\\ & ={\displaystyle \frac{1-\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{2}}\\ & ={\displaystyle \frac{2+\sin \alpha -\sin \alpha -2\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{4}}\\ & ={\displaystyle \frac{1+\sin \alpha +1-\sin \alpha -2\sqrt{(1+\sin \alpha ).(1-\sin \alpha )}}{4}}\\ & ={\displaystyle \frac{{(\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha })}^2}{4}}\\ & ={\left({\displaystyle \frac{(\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha })}{2}}\right)}^2\\ \sin {\displaystyle \frac{1}{2}\alpha }& ={\displaystyle {\displaystyle \frac{\sqrt{1+\sin \alpha }-\sqrt{1-\sin \alpha }}{2}◼}}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Dengan menggunakan rumus sudut}\\ & \text{paruh, tentukanlah nilai dari}\\ & \text{a}.\cos {\displaystyle \frac{\pi }{8},\text{b}.\sin {15}^{\circ }\text{c}.\tan {15}^{\circ }}\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Ingat baik sudut}{\displaystyle \frac{\pi }{8}\text{atau}{\displaystyle \frac{\pi }{4}\text{di kuadran I}}}\\ & \text{maka}\cos {\displaystyle \frac{\pi }{8}\text{bertanda positif, sehingga}}\\ & \begin{array}{rl}\text{a}.\cos {\displaystyle \frac{\pi }{8}}& =\sqrt{{\displaystyle \frac{1+\cos {\displaystyle \frac{\pi }{4}}}{2}}}=\sqrt{{\displaystyle \frac{1+{\displaystyle \frac{1}{2}\sqrt{2}}}{2}}}\\ & =\sqrt{{\displaystyle \frac{2+\sqrt{2}}{4}}}={\displaystyle \frac{1}{2}\sqrt{2+\sqrt{2}}}\end{array}\\ & \text{Adapun untuk sudut}{15}^{\circ }\text{ataupun}{30}^{\circ }\\ & \text{akan berada di kuadran I akibatnya }\\ & \text{tandanya positif, sehingga}\\ & \begin{array}{rl}\text{b}.\sin {15}^{\circ }& =\sqrt{{\displaystyle \frac{1-\cos {30}^{\circ }}{2}}}\\ & =\sqrt{{\displaystyle \frac{1-{\displaystyle \frac{1}{2}\sqrt{3}}}{2}}}=\sqrt{{\displaystyle \frac{2-\sqrt{3}}{4}}}\\ & ={\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}\end{array}\\ & \text{Dan untuk sudut}\tan {\displaystyle \frac{1}{2}\alpha ={\displaystyle \frac{1-\cos \alpha }{\sin \alpha }}}\\ & \begin{array}{rl}\text{c}.\tan {15}^{\circ }& ={\displaystyle \frac{1-\cos {30}^{\circ }}{\sin {30}^{\circ }}}\\ & ={\displaystyle \frac{1-{\displaystyle \frac{1}{2}\sqrt{3}}}{{\displaystyle \frac{1}{2}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{(2-\sqrt{3})}{2}}}{{\displaystyle \frac{1}{2}}}}\\ & =2-\sqrt{3}\end{array}\end{array}$.

DAFTAR PUSTAKA

1. Sukino. 2017. Matematika untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam. Jakarta: ERLANGGA.

Lanjutan 3 Materi Rumus Sudut Ganda(Rangkap) dan Sudut Paruh Pada Trigonometri

B. 1   Rumus Sudut Rangkap

Pada materi sebelumnya sudah dibahas tentang hal ini yaitu,

$\begin{array}{l}\\ \bullet & \sin 2\alpha =2\sin \alpha \cos \alpha \\ \bullet & \cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha \\ \bullet & \tan 2\alpha ={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\end{array}$.

B. 2   Rumus Sudut Rangkap Tiga

Berikut untuk rumus sudut rangkap tiga:

$\begin{array}{rl}\bullet \sin 3\alpha & =3\sin \alpha -4{\sin }^3\alpha \\ \bullet \cos 3\alpha & =4{\cos }^3\alpha -3\cos \alpha \\ \bullet \tan 3\alpha & ={\displaystyle \frac{3\tan x-{\tan }^3\alpha }{1-3{\tan }^2\alpha }}\end{array}$

B. 3   Rumus Sudut Paruh

Demikian juga untuk sudut paruhnya, juga sudah dibahas pada materi sebelumnya yang mana sebelumnya sudut paruh itu diperoleh dari aplikasi rumus jum lah dan selisih dua sudut ketika sudutnya berupa setengan sudut dan sama pula, yaitu:

$\begin{array}{l}\\ \bullet & \sin \alpha =2\sin {\displaystyle \frac{1}{2}\alpha \cos \frac{1}{2}\alpha }\\ \bullet & \cos \alpha ={\cos }^2{\displaystyle \frac{1}{2}\alpha -{\sin }^2{\displaystyle \frac{1}{2}\alpha }}\\ \bullet & \tan \alpha ={\displaystyle \frac{2\tan {\displaystyle \frac{1}{2}\alpha }}{1-{\tan }^2{\displaystyle \frac{1}{2}\alpha }}}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Diketahui}\sin A={\displaystyle \frac{4}{5},}\\ & \text{untuk}A\text{sudut tumpul. Tentukanlah}\\ & \text{a}.\sin 2A\\ & \text{b}.\cos 2A\\ & \text{c}.\tan 2A\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Dengan menggunakan identitas}\\ & {\sin }^2\alpha +{\cos }^2\alpha =1,\text{dapat diperoleh}\\ & \text{nilai}\cos A,\text{yaitu}:\\ & {\cos }^2\alpha =1-{\sin }^{2}A=1-{\left({\displaystyle \frac{4}{5}}\right)}^2\\ & =1-{\displaystyle \frac{16}{25}=\frac{9}{25}\Rightarrow \cos A=\pm {\displaystyle \frac{3}{5}}}\\ & \text{ambil}\cos A=-{\displaystyle \frac{3}{5},\text{karena di kuadran II}}\\ & \text{ingat sudut tumpul}=\text{sudut di kuadran II}\\ & \begin{array}{rl}\text{a}.\sin 2A& =2\sin A\cos A\\ & =2\left({\displaystyle \frac{4}{5}}\right)(-{\displaystyle \frac{3}{5}})\\ & =-{\displaystyle \frac{24}{25}}\\ \text{b}.\cos 2A& ={\cos }^{2}A-{\sin }^{2}A\\ & ={(-{\displaystyle \frac{3}{5}})}^2-{\left({\displaystyle \frac{4}{5}}\right)}^2\\ & ={\displaystyle \frac{9}{25}-\frac{16}{25}=-\frac{7}{25}}\\ \text{c}.\tan 2A& ={\displaystyle \frac{\sin 2A}{\cos 2A}}\\ & ={\displaystyle \frac{-{\displaystyle \frac{24}{25}}}{-{\displaystyle \frac{7}{25}}}={\displaystyle \frac{24}{7}}}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tentukanlah nilai dari}\\ & \text{a}.\sin {15}^{\circ }\\ & \text{b}.\cos {15}^{\circ }\\ & \text{c}.\tan {15}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \text{Banyak cara yang bisa digunakan}.\\ & \text{berikut salah satu cara itu, yaitu}\\ & \begin{array}{rlr}\text{a.}\cos \left(2\alpha \right)& =1-2{\sin }^2\alpha ,& \text{pilih untuk}\alpha ={15}^0\\ \cos (2\times {15}^0)& =1-2{\sin }^2{15}^0\\ \cos {30}^0& =1-2{\sin }^2{15}^0\\ {\displaystyle \frac{1}{2}\sqrt{3}}& =1-2{\sin }^2{15}^0\\ \sqrt{3}& =2-4{\sin }^2{15}^0\\ {\sin }^2{15}^0& ={\displaystyle \frac{2-\sqrt{3}}{4}}\\ \sin {15}^0& ={\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}\\ \text{b.}{\sin }^2\alpha +& {\cos }^2\alpha =1,& \text{rumus identitas}\\ {\sin }^2{15}^0+& {\cos }^2{15}^0=1\\ {\cos }^2{15}^0& =1-{\sin }^2{15}^0\\ & ={\displaystyle 1-\left(\frac{2-\sqrt{3}}{4}\right),}& \text{lihat jawaban poin a)}\\ & ={\displaystyle \frac{2+\sqrt{3}}{4}}\\ \cos {15}^0& ={\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}\\ \text{c.}\tan \alpha & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ \tan {15}^0& ={\displaystyle \frac{\sin {15}^0}{\cos {15}^0}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{2}\sqrt{2-\sqrt{3}}}}{{\displaystyle \frac{1}{2}\sqrt{2+\sqrt{3}}}}}\\ & =\sqrt{{\displaystyle \frac{2-\sqrt{3}}{2+\sqrt{3}}}}\times \sqrt{\frac{2-\sqrt{3}}{2-\sqrt{3}}},& \text{sekawan dari penyebut}\\ & =\sqrt{{\displaystyle \frac{4-2.2\sqrt{3}+3}{4-3}}}\\ & =\sqrt{7-4\sqrt{3}}\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tunjukkan bahwa nilai}\\ & {\displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }=\tan \alpha }\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}{\displaystyle \frac{1-\cos 2\alpha }{\sin 2\alpha }}& ={\displaystyle \frac{1-({\cos }^2\alpha -{\sin }^2\alpha )}{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{1-((1-{\sin }^2\alpha )-{\sin }^2\alpha )}{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{1-1+{\sin }^2\alpha +{\sin }^2\alpha }{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{2{\sin }^2\alpha }{2\sin \alpha \cos \alpha }}\\ & ={\displaystyle \frac{\sin \alpha }{\cos \alpha }}\\ & ={\displaystyle \tan \alpha ◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tunjukkan bahwa nilai}\\ & \text{a}.\cos 3\alpha =4{\cos }^3\alpha -3\cos \alpha \\ & \text{b}.\tan 3\alpha ={\displaystyle \frac{3\tan x-{\tan }^3\alpha }{1-3{\tan }^2\alpha }}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{|ll|}\hline \begin{array}{rl}\text{a}.\cos 3\alpha & =\cos (2\alpha +\alpha )\\ & =\cos 2\alpha \cos \alpha -\sin 2\alpha \sin \alpha \\ & =(2{\cos }^2\alpha -1)\cos \alpha -(2\sin \alpha \cos \alpha )\sin \alpha \\ & =2{\cos }^3\alpha -\cos \alpha -2{\sin }^2\alpha \cos \alpha \\ & =2{\cos }^3\alpha -\cos \alpha -2(1-{\cos }^2\alpha )\cos \alpha \\ & =2{\cos }^3\alpha -\cos \alpha -2\cos \alpha +2{\cos }^2\alpha \\ & =4{\cos }^3\alpha -3\cos \alpha ◼\\ & \\ & \\ & \\ & \end{array}& \begin{array}{rl}\text{b}.\tan 3\alpha & =\tan (2\alpha +\alpha )\\ & ={\displaystyle \frac{\tan 2\alpha +\tan \alpha }{1-\tan 2\alpha \tan \alpha }}\\ & ={\displaystyle \frac{\left({\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\right)+\tan \alpha }{1-\left({\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\right).\tan \alpha }}\\ & ={\displaystyle \frac{{\displaystyle \frac{2\tan \alpha +\tan \alpha -{\tan }^3\alpha }{1-{\tan }^2\alpha }}}{{\displaystyle \frac{(1-{\tan }^2\alpha )-2{\tan }^2\alpha }{1-{\tan }^2\alpha }}}}\\ & ={\displaystyle \frac{3\tan \alpha -{\tan }^3\alpha }{1-3{\tan }^2\alpha }◼}\end{array}\\ \hline\end{array}\end{array}$

Lanjutan 2 Materi Rumus-Rumus Trigonometri

 A. 3  Rumus  $\tan (\alpha +\beta )$ dan $\tan (\alpha -\beta )$.

Sebelumnya telah dibahas pada materi sebelumnya dan disertai pula dengan contoh soal rumus jumlah dan selisih dua sudut untuk sinus dan cosinus, yaitu:

Jika dua sudut yang dimaksud misalkan alfa dan beta, maka

$\{\begin{array}{ll}\sin (\alpha +\gamma )& =\sin \alpha \cos \gamma +\cos \alpha \sin \gamma \\ \sin (\alpha -\gamma )& =\sin \alpha \cos \gamma -\cos \alpha \sin \gamma \\ \cos (\alpha +\gamma )& =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \\ \cos (\alpha +\gamma )& =\cos \alpha \cos \gamma -\sin \alpha \sin \gamma \end{array}$.

Sengan menggunakan fakta yang ada-rumus yang telah diketahui-kita akan terbantu dalam menemukan rumus untuk tangen, yaitu:

$\begin{array}{rl}\tan (\alpha +\beta )& ={\displaystyle \frac{\sin (\alpha +\beta )}{\cos (\alpha +\beta )}}\\ & ={\displaystyle \frac{(\sin \alpha \cos \beta +\cos \alpha \sin \beta )\times \left({\displaystyle \frac{1}{\cos \alpha \cos \beta }}\right)}{(\cos \alpha \cos \beta -\sin \alpha \sin \beta )\times \left({\displaystyle \frac{1}{\cos \alpha \cos \beta }}\right)}}\\ & ={\displaystyle \frac{{\displaystyle \frac{\sin \alpha }{\cos \alpha }+{\displaystyle \frac{\sin \beta }{\cos \beta }}}}{1-{\displaystyle \frac{\sin \alpha \sin \beta }{\cos \alpha \cos \beta }}}}\\ & ={\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }.◼}\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\tan (\alpha +\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}\tan (\alpha +\beta )& ={\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta }}\\ & \text{dengan mengganti}\beta =-\beta \text{maka,}\\ & ={\displaystyle \frac{\tan \alpha +\tan (-\beta )}{1-\tan \alpha \tan (-\beta )}}\\ \tan (\alpha -\beta )& ={\displaystyle \frac{\tan \alpha -\tan \beta }{1+\tan \alpha \tan \beta }.◼}\end{array}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\tan {60}^{\circ }=\sqrt{3}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\tan {60}^{\circ }& =\tan ({30}^{\circ }+{30}^{\circ })\\ & ={\displaystyle \frac{\tan {30}^{\circ }+\tan {30}^{\circ }}{1-\tan {30}^{\circ }\tan {30}^{\circ }}}\\ & ={\displaystyle \frac{2\tan {30}^{\circ }}{1-{\tan }^2{30}^{\circ }}}\\ & ={\displaystyle \frac{2\left({\displaystyle \frac{1}{3}\sqrt{3}}\right)}{1-{\left({\displaystyle \frac{1}{3}\sqrt{3}}\right)}^2}={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{1-{\displaystyle \frac{1}{9}\sqrt{9}}}}}\\ & ={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{1-{\displaystyle \frac{3}{9}}}={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{{\displaystyle \frac{6}{9}}}={\displaystyle \frac{{\displaystyle \frac{2}{3}\sqrt{3}}}{{\displaystyle \frac{2}{3}}}}}}\\ & ={\displaystyle \sqrt{3}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\\ & \tan {90}^{\circ }=\text{Tidak Terdefinisi}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\tan {90}^{\circ }& =\tan ({45}^{\circ }+{45}^{\circ })\\ & ={\displaystyle \frac{\tan {45}^{\circ }+\tan {45}^{\circ }}{1-\tan {45}^{\circ }\tan {45}^{\circ }}}\\ & ={\displaystyle \frac{2\tan {45}^{\circ }}{1-{\tan }^2{45}^{\circ }}}\\ & ={\displaystyle \frac{2\left(1\right)}{1-{\left(1\right)}^2}={\displaystyle \frac{{\displaystyle 2}}{1-1}}}\\ & ={\displaystyle \frac{2}{0}}\\ & =\text{Tidak Terdefinisi}◼\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\tan {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {75}^{\circ }& =\tan ({30}^{\circ }+{45}^{\circ })\\ & ={\displaystyle \frac{\tan {30}^{\circ }+\tan {45}^{\circ }}{1-\tan {30}^{\circ }\tan {45}^{\circ }}}\\ & ={\displaystyle \frac{{\displaystyle \frac{1}{3}\sqrt{3}+1}}{1-\left({\displaystyle \frac{1}{3}\sqrt{3}}\right)\left(1\right)}}\\ & ={\displaystyle \frac{1+{\displaystyle \frac{1}{3}\sqrt{3}}}{1-{\displaystyle \frac{1}{3}\sqrt{3}}}={\displaystyle \frac{{\displaystyle \frac{1}{3}(3+\sqrt{3})}}{{\displaystyle \frac{1}{3}(3-\sqrt{3})}}}}\\ & ={\displaystyle \frac{3+\sqrt{3}}{3-\sqrt{3}}}\\ & ={\displaystyle \frac{3+\sqrt{3}}{3-\sqrt{3}}\times {\displaystyle \frac{3+\sqrt{3}}{3+\sqrt{3}}}}\\ & ={\displaystyle \frac{3^2+3\sqrt{3}+3\sqrt{3}+\sqrt{9}}{3^2-{\left(\sqrt{3}\right)}^2}}\\ & ={\displaystyle \frac{12+6\sqrt{3}}{9-3}={\displaystyle \frac{12+6\sqrt{3}}{6}}}\\ & ={\displaystyle \frac{6}{6}(2+\sqrt{3})}\\ & =2+\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\tan {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan {105}^{\circ }& =\tan ({45}^{\circ }+{60}^{\circ })\\ & ={\displaystyle \frac{\tan {45}^{\circ }+\tan {60}^{\circ }}{1-\tan {45}^{\circ }\tan {60}^{\circ }}}\\ & ={\displaystyle \frac{1+\sqrt{3}+1}{1-1.\left(\sqrt{3}\right)}}\\ & ={\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}}\\ & ={\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}}\\ & ={\displaystyle \frac{1+\sqrt{3}}{1-\sqrt{3}}\times {\displaystyle \frac{1+\sqrt{3}}{1+\sqrt{3}}}}\\ & ={\displaystyle \frac{1^2+\sqrt{3}+\sqrt{3}+{\left(\sqrt{3}\right)}^2}{1^2-{\left(\sqrt{3}\right)}^2}}\\ & ={\displaystyle \frac{4+2\sqrt{3}}{1-3}={\displaystyle \frac{4+2\sqrt{3}}{-2}}}\\ & =-2-\sqrt{3}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\tan ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan ({270}^{\circ }+A)& ={\displaystyle \frac{\tan {270}^{\circ }+\tan A}{1-\tan {270}^{\circ }\tan A}}\\ & ={\displaystyle \frac{\mathbf{\text{TD}}+\tan A}{1-\mathbf{\text{TD}}\tan A}}\\ & \text{dengan}\mathbf{\text{TD}}\text{adalah Tidak Terdefinisi}\\ & =\text{Bentuk yang harus dihindari}\\ \text{maka gunakan}& \text{bentuk berikut ini}\\ \tan ({270}^{\circ }+A)& ={\displaystyle \frac{\sin ({270}^{\circ }+A)}{\cos ({270}^{\circ }+A)}}\\ & ={\displaystyle \frac{-\cos A}{\sin A}}\\ & =-\cot A\\ \\ \mathbf{\text{Catatan}}:& \text{lihatlah materi sebelumnya}\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\tan ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\tan ({270}^{\circ }-A)& ={\displaystyle \frac{\sin ({270}^{\circ }-A)}{\cos ({270}^{\circ }-A)}}\\ & ={\displaystyle \frac{-\cos A}{-\sin A}}\\ & =\cot A\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tunjukkan bahwa}\\ & \begin{array}{l}\\ \text{a}.& \tan \left(2\alpha \right)={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }}\\ \text{b}.& \tan \alpha ={\displaystyle \frac{2\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}{1-\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}}\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rlrlr}\text{a.}\tan (\alpha +\beta )& ={\displaystyle \frac{\tan \alpha +\tan \beta }{1-\tan \alpha \tan \beta },}& \text{dan jika}\beta =\alpha ,& & \text{maka}\\ \tan (\alpha +\alpha )& ={\displaystyle \frac{\tan \alpha +\tan \alpha }{1-\tan \alpha \tan \alpha }}\\ \tan 2\alpha & ={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha }\alpha ◼}\\ \text{b.}\tan 2\alpha & ={\displaystyle \frac{2\tan \alpha }{1-{\tan }^2\alpha },}& \text{dan jika}\alpha ={\displaystyle \frac{1}{2}\alpha ,}& & \text{maka}\\ \tan 2\left({\displaystyle \frac{1}{2}\alpha }\right)& ={\displaystyle \frac{2\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}{1-{\tan }^2\left({\displaystyle \frac{1}{2}\alpha }\right)}}\\ \tan \alpha & ={\displaystyle \frac{2\tan \left({\displaystyle \frac{1}{2}\alpha }\right)}{1-{\tan }^2\left({\displaystyle \frac{1}{2}\alpha }\right)}◼}\end{array}\end{array}$.

Lanjutan 1 Materi Rumus-Rumus Trigonometri

A. 2  Rumus  $\cos (\alpha +\beta )$ dan $\cos (\alpha -\beta )$.

Dalam penentuan rumus $\cos (\alpha -\beta )$, pada uraian berikut akan ditunjukkan penentuan rumus yang dimaksud dengan bantuan segitiga ABC

Perhatikanlah ilustrai berikut

Jika diurai gambar di atas adalah 

Mungkin gambarnya ada tang kurang jelas, mari kita perjelas lagi gambar di atas

Perhatikan bahwa

dan

Sehingga

$\begin{array}{rl}\left[ABC\right]& =\left[ACD\right]+\left[BCD\right]\\ {\displaystyle \frac{1}{2}ab\sin ({90}^0-\alpha +\beta )}& ={\displaystyle \frac{1}{2}ab\cos \alpha \cos \beta +{\displaystyle \frac{1}{2}ab\sin \alpha \sin \beta }}\\ \sin ({90}^0-(\alpha -\beta ))& =\cos \alpha \cos \beta +\sin \alpha \sin \beta \\ \cos (\alpha -\beta )& =\cos \alpha \cos \beta +\sin \alpha \sin \beta .◼\end{array}$.

Selanjutnya dengan untuk mendapatkan rumus  $\cos (\alpha +\beta )$ adalah dengan mengganti  $\beta =-\beta$, maka

$\begin{array}{rl}& \cos (\alpha -(-\beta ))=\cos (\alpha +\beta )\\ & =\cos \alpha \cos (-\beta )+\sin \alpha \sin (-\beta )\\ & =\cos \alpha \cos \beta +\sin \alpha (-\sin \beta )\\ & =\cos \alpha \cos \beta -\sin \alpha \sin \beta ◼\end{array}$.

Catatan:

$\left[ABC\right]=\mathbf{\text{luas segitiga ABC}}$.

$\text{CONTOH SOAL}$.

$\begin{array}{l}\\ 1.& \text{Tunjukkan bahwa nilai}\cos {60}^{\circ }={\displaystyle \frac{1}{2}}\\ \\ & \mathbf{\text{Bukti}}:\\ & \begin{array}{rl}\cos {60}^{\circ }& =\cos ({30}^{\circ }+{30}^{\circ })\\ & =\cos {30}^{\circ }\cos {30}^{\circ }-\sin {30}^{\circ }\sin {30}^{\circ }\\ & ={\cos }^2{30}^{\circ }-{\sin }^2{30}^{\circ }\\ & ={\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)}^2-{\left({\displaystyle \frac{1}{2}}\right)}^2\\ & ={\displaystyle \frac{1}{4}\sqrt{9}-\frac{1}{4}={\displaystyle \frac{3}{4}-\frac{1}{4}={\displaystyle \frac{2}{4}}}}\\ & ={\displaystyle \frac{1}{2}◼}\end{array}\end{array}$.

$\begin{array}{l}\\ 2.& \text{Tunjukkan bahwa nilai}\cos {90}^{\circ }=0\\ \\ & \begin{array}{rl}\mathbf{\text{Bukti}}& \mathbf{\text{pertama}}\\ \cos {90}^{\circ }& =\cos ({60}^{\circ }+{30}^{\circ })\\ & =\cos {60}^{\circ }\cos {30}^{\circ }-\sin {60}^{\circ }\sin {30}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{3}-{\displaystyle \frac{1}{4}\sqrt{3}}}\\ & ={\displaystyle 0◼}\\ \mathbf{\text{Bukti}}& \mathbf{\text{kedua}}\\ \cos {90}^{\circ }& =\cos ({30}^{\circ }+{60}^{\circ })\\ & =....+....\\ & =....\\ & =....\\ \mathbf{\text{Bukti}}& \mathbf{\text{ketiga}}\\ \cos {90}^{\circ }& =\cos ({45}^{\circ }+{45}^{\circ })\\ & =....+....\\ & =....\\ & =....\end{array}\end{array}$.

$\begin{array}{l}\\ 3.& \text{Tentukan nilai dari}\cos {75}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos {75}^{\circ }& =\cos ({30}^{\circ }+{45}^{\circ })\\ & =\cos {30}^{\circ }\cos {45}^{\circ }-\sin {30}^{\circ }\sin {45}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)-\left({\displaystyle \frac{1}{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{6}-{\displaystyle \frac{1}{4}\sqrt{2}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{6}-\sqrt{2})}\end{array}\end{array}$.

$\begin{array}{l}\\ 4.& \text{Tentukan nilai dari}\cos {105}^{\circ }\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos {105}^{\circ }& =\cos ({45}^{\circ }+{60}^{\circ })\\ & =\cos {45}^{\circ }\cos {60}^{\circ }-\sin {45}^{\circ }\sin {60}^{\circ }\\ & =\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}}\right)-\left({\displaystyle \frac{1}{2}\sqrt{2}}\right)\left({\displaystyle \frac{1}{2}\sqrt{3}}\right)\\ & ={\displaystyle \frac{1}{4}\sqrt{2}-{\displaystyle \frac{1}{4}\sqrt{6}}}\\ & ={\displaystyle \frac{1}{4}(\sqrt{2}-\sqrt{6})}\end{array}\end{array}$.

$\begin{array}{l}\\ 5.& \text{Sederhanakan bentuk dari}\cos ({270}^{\circ }+A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos ({270}^{\circ }+A)& =\cos {270}^{\circ }\cos A-\sin {270}^{\circ }\sin A\\ & =\left(0\right)\cos A-(-1)\sin A\\ & =0+\sin A\\ & =\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 6.& \text{Sederhanakan bentuk dari}\cos ({270}^{\circ }-A)\\ \\ & \mathbf{\text{Jawab}}:\\ & \begin{array}{rl}\cos ({270}^{\circ }-A)& =\cos {270}^{\circ }\cos A+\sin {270}^{\circ }\sin A\\ & =\left(0\right)\cos A+(-1)\sin A\\ & =0-\sin A\\ & =-\sin A\end{array}\end{array}$.

$\begin{array}{l}\\ 7.& \text{Tunjukkan bahwa}\\ & \begin{array}{l}\\ \text{a}.& \cos \left(2\alpha \right)={\cos }^2\alpha -{\sin }^2\alpha \\ \text{b}.& \cos \alpha ={\cos }^2\left({\displaystyle \frac{1}{2}\alpha }\right)-{\sin }^2\left({\displaystyle \frac{1}{2}\alpha }\right)\end{array}\\ \\ & \mathbf{\text{Bukti}}\\ & \begin{array}{rlrlr}\text{a.}\cos (\alpha +\beta )& =\cos \alpha \cos \beta -\sin \alpha \sin \beta ,& \text{dan jika}\beta =\alpha ,& & \text{maka}\\ \cos (\alpha +\alpha )& =\cos \alpha \cos \alpha -\sin \alpha \sin \alpha \\ \cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha ◼\\ \text{b.}\cos 2\alpha & ={\cos }^2\alpha -{\sin }^2\alpha ,& \text{dan jika}\alpha ={\displaystyle \frac{1}{2}\alpha ,}& & \text{maka}\\ \cos 2\left({\displaystyle \frac{1}{2}\alpha }\right)& ={\cos }^2\left({\displaystyle \frac{1}{2}\alpha }\right)-{\sin }^2\left({\displaystyle \frac{1}{2}\alpha }\right)\\ \cos \alpha & ={\cos }^2\left({\displaystyle \frac{1}{2}\alpha }\right)-{\sin }^2\left({\displaystyle \frac{1}{2}\alpha }\right)◼\end{array}\end{array}$

DAFTAR PUSTAKA

1. Kanginan, M. 2007. Matematika untuk Kelas X Semester 2 Sekolah Menengah Atas. Bandung: GRAFINDO MEDIA PRATAMA.
2. Noormandiri, B.K. 2017. Matematika Jilid 2 untuk SMA/MA Kelas XI Kelompok Peminatan Matematika dan Ilmu-Ilmu Alam Berdasarkan Kurikulum 2013 Edisi Revisi 2016. Jakarta: ERLANGGA.