Question

The minimum value of 3 cos x + 4 sin x + 8 is
(a) 5
(b) 9
(c) 7
(d) 3

Answer 1

$$\begin{gathered} 3\cos x+4\sin x+8 \\ \mathrm{first}\mathrm{express}3\cos x+4\sin xasa\cos (x+A) \\ 3\cos x+4\sin x=a\left[\cos x\cos A-\sin x\sin A\right] \\ \mathrm{i}.\mathrm{e}3\cos x+4\sin x=a\cos x\cos A-a\sin x\sin A \\ \mathrm{Now},\mathrm{equate}\mathrm{the}\mathrm{coefficients}\mathrm{of}\sin xand\cos x \\ \mathrm{we}\mathrm{get}, \\ a\cos A=3 \\ \mathrm{and}-a\sin A=4 \\ \mathrm{i}.\mathrm{e}\frac{a\sin A}{a\cos A}=\frac{-4}{3} \\ \mathrm{i}.\mathrm{e}\tan A=\frac{-4}{3} \\ \mathrm{i}.\mathrm{e}A={\tan }^{-1}\left(\frac{-4}{3}\right). \\ \mathrm{also},{\left(a\cos A\right)}^2+{\left(a\sin A\right)}^2=9+16=25 \\ \mathrm{i}.\mathrm{e}{a}^2=25 \\ \mathrm{i}.\mathrm{e}a=5 \\ \Rightarrow 3\cos x+4\sin x=5\cos \left(x-{\tan }^{-1}\left(\frac{-4}{3}\right)\right) \\ \mathrm{i}.\mathrm{e}3\cos x+4\sin x+8=5\cos \left(x-{\tan }^{-1}\left(\frac{-4}{3}\right)\right)+8 \\ \mathrm{Since}\mathrm{minimum}\mathrm{value}\mathrm{of}\cos \theta =-1 \\ \Rightarrow \left(3\cos x+4\sin x+8\right)\min =5\left(-1\right)+8 \\ \mathrm{i}.\mathrm{e}\mathrm{Minimum}\mathrm{vlaue}\mathrm{of}3\cos x+4\sin x+8=3 \\ \mathrm{Hence},\mathrm{the}\mathrm{correct}\mathrm{answer}\mathrm{is}\mathrm{option}\mathrm{D}. \end{gathered}$$