Question

The minimum value of 4 cos x – 3 sin x + 7 is _________.

Answer 1

$$\begin{gathered} 4\cos x-3\sin x+7 \\ \mathrm{Express}4\cos x-3\sin x\mathrm{as}a\cos (x+A) \\ \mathrm{i}.\mathrm{e}.4\cos x-3\sin x=a\left[\cos x\cos A-\sin x\sin A\right] \\ \mathrm{on}\mathrm{compairing}\mathrm{coefficients}\mathrm{of}\sin x\mathrm{and}\cos x, \\ \mathrm{we}\mathrm{get}, \\ 4=a\cos A\mathrm{and}-3=-a\sin A \\ \mathrm{i}.\mathrm{e}.a\cos A=4\mathrm{and}a\sin A=3 \\ \mathrm{i}.\mathrm{e}.\frac{a\sin A}{a\cos A}=\frac{3}{4} \\ \mathrm{i}.\mathrm{e}.\tan A=\frac{3}{4} \\ \mathrm{i}.\mathrm{e}.A={\tan }^{-1}\left(\frac{3}{4}\right) \\ \mathrm{and}{\left(a\sin A\right)}^2+{(a\cos A)}^2=9+16 \\ {a}^2\left({\sin }^{2}A+{\cos }^{2}A\right)=25 \\ \mathrm{i}.\mathrm{e}.a^2=25 \\ \mathrm{i}.\mathrm{e}.a=5 \end{gathered}$$
$$\begin{gathered} \therefore 4\cos x-3\sin x+7=5\cos \left(x+{\tan }^{-1}\left(\frac{3}{4}\right)+7\right) \\ \mathrm{Since}\cos \theta =-1\mathrm{is}\mathrm{minimum}\mathrm{value}\mathrm{of}\cos \\ \Rightarrow 4\cos x-3\sin x+7=5(-1)+7 \\ \mathrm{Minimum}\mathrm{values}\mathrm{for}4\cos -3\sin x+7=2. \end{gathered}$$