Question

Find the maximum value of
i) $4{\sin }^{2}x+3{\cos }^{2}x+\sin \frac{x}{2}+\cos \frac{x}{2}$,
ii) $5\cos \theta +3\cos (\theta +\frac{\pi }{3})+3$.

Answer 1

According to the question,

$\begin{array}{c}\left(i\right)f\left(x\right)=4si{n}^{2}x+3co{s}^{2}x+sin\frac{x}{2}+cos\frac{x}{2}\\ =4(1-{\cos }^{2}x)+3co{s}^{2}x+sin\frac{x}{2}+cos\frac{x}{2}\\ =4-4{\cos }^{2}x+3co{s}^{2}x+sin\frac{x}{2}+cos\frac{x}{2}\\ =(4-{\cos }^{2}x)+(sin\frac{x}{2}+cos\frac{x}{2})\\ \therefore 0\le {\cos }^{2}x\le 1\Rightarrow 3\le 4-{\cos }^{2}x\le 4\\ and,-\sqrt{2}\le sin\frac{x}{2}+cos\frac{x}{2}\le \sqrt{2}\\ \therefore max.value=4+\sqrt{2}\\ \left(ii\right)f\left(x\right)=5cos\theta +3cos(\theta +\frac{\pi }{3})+3\\ f\left(x\right)=5cos\theta +3(cos\theta \cos \frac{\pi }{3}-\sin \theta \sin \frac{\pi }{3})+3\\ =5cos\theta +3(cos\theta \frac{1}{2}-\sin \theta \frac{\sqrt{3}}{2})+3\\ =5cos\theta +\frac{3}{2}cos\theta -\frac{3\sqrt{3}}{2}\sin \theta +3\\ =\frac{13}{2}\cos \theta -\frac{3\sqrt{3}}{2}\sin \theta +3\\ Max.value=c+\sqrt{a^2+b^2}\\ =3+\sqrt{\frac{169}{4}+\frac{27}{4}}\\ \therefore 3+7=10\\ So,thatthe,\\ Min.value=4+\sqrt{2},andMax.value=10\end{array}$