Question

The maximum value of 3 cos x + 4 sin x + 5 is _________.

Answer 1

$$\begin{gathered} 3\cos x+4\sin x+5 \\ \mathrm{first}\mathrm{express}3\cos x+4\sin x\mathrm{as}a\cos \left(x+A\right) \\ \mathrm{i}.\mathrm{e}.3\cos x+4\sin x=a\left[\cos x\cos A-\sin x\sin A\right] \\ \mathrm{Now},\mathrm{equate}\mathrm{the}\mathrm{coefficients}\mathrm{of}\sin x\mathrm{and}\cos x \\ \mathrm{We}\mathrm{get}, \\ a\cos A=3 \\ -a\sin A=4 \\ \mathrm{i}.\mathrm{e}\frac{a\sin A}{a\cos A}=\frac{-4}{3}\mathrm{i}.\mathrm{e}\tan A=-\frac{4}{3} \\ \mathrm{i}.\mathrm{e}A={\tan }^{-1}\left(\frac{-4}{3}\right) \\ \mathrm{also}{\left(a\cos A\right)}^2+{\left(a\sin A\right)}^2=9+16=25 \\ {a}^2=25 \\ a=5 \\ \therefore 3\cos x+4\sin x=5\cos \left(x-{\tan }^{-1}\left(-\frac{4}{3}\right)\right) \\ \mathrm{i}.\mathrm{e}.3\cos x+4\sin x+5=5\cos \left(x-{\tan }^{-1}\left(-\frac{4}{3}\right)\right)+5 \\ \mathrm{Since}\mathrm{minimum}\mathrm{value}\mathrm{of}\cos \theta =-1 \\ \Rightarrow {\left(3\cos x+4\sin x+5\right)}_{\min }=5\left(-1\right)+5 \\ =-5+5=0 \\ \mathrm{i}.\mathrm{e}\mathrm{minimum}\mathrm{value}\mathrm{of}3\cos x+4\sin x+5=0 \end{gathered}$$