Contoh Soal 13 Turunan Fungsi Trigonometri (Bagian 3)

 $\begin{array}{l}\\ 61.& \text{(UM UNBRAW)}\\ & \text{Nilai maksimum dari fungsi}\\ & f(x)=4{\cos }^{2}x+14{\sin }^{2}x+24\sin x\cos x+10\\ & \text{adalah}....\\ & \begin{array}{l}\\ \text{a}.& 6\\ \text{b}.& 24\\ \text{c}.& 26\\ \text{d}.& 32\\ \text{e}.& 92\end{array}\\ \\ & \text{Jawab}:\mathbf{\text{d}}\\ & \begin{array}{rl}& f(x)=4{\cos }^{2}x+14{\sin }^{2}x+24\sin x\cos x+10\\ & f(x)=4{\cos }^{2}x+4{\sin }^{2}x+10{\sin }^{2}x+12\sin 2x+10\\ & f(x)=4+5(1-\cos 2x)+12\sin 2x+10\\ & f(x)=19-5\cos 2x+12\sin 2x\\ & f(x)=19+12\sin 2x-5\cos 2x\\ & f(x)=19+\sqrt{{12}^2+(-5{)}^2}\cos (2x-\theta )\\ & f(x)=19+13\cos (2x-\theta )\\ & \text{Karena nilai}\cos (2x-\theta )=\pm 1,\text{maka}\\ & f(x{)}_{maks}=19+13=32\end{array}\end{array}$