Contoh Soal 13 Turunan Fungsi Trigonometri (Bagian 3)

 $\begin{array}{ll}\\ 61.&\textrm{(UM UNBRAW)}\\ &\textrm{Nilai maksimum dari fungsi}\\ &f(x)=4\cos ^{2}x+14\sin ^{2}x+24\sin x\cos x+10\\ &\textrm{adalah}\: ....\\ &\begin{array}{llll}\\ \textrm{a}.&6\\ \textrm{b}.&24\\ \textrm{c}.&26\\ \color{red}\textrm{d}.&32\\ \textrm{e}.&92 \end{array}\\\\ &\textrm{Jawab}:\quad \color{red}\textbf{d}\\ &\color{black}\begin{aligned}&f(x)=4\cos ^{2}x+14\sin ^{2}x+24\sin x\cos x+10\\ &f(x)=4\cos ^{2}x+4\sin ^{2}x+10\sin ^{2}x+12\sin 2x+10\\ &f(x)=4+5\left (1-\cos 2x \right )+12\sin 2x+10\\ &f(x)=19-5\cos 2x +12\sin 2x\\ &f(x)=19+12\sin 2x -5\cos 2x\\ &f(x)=19+\sqrt{12^{2}+(-5)^{2}}\cos \left ( 2x-\theta \right )\\ &f(x)=19+13\cos \left (2x -\theta \right )\\ &\textrm{Karena nilai}\: \: \cos \left ( 2x-\theta \right )=\pm 1,\: \textrm{maka}\\ &f(x)_{maks}=\color{red}19+13=32 \end{aligned} \end{array}$