Question

Prove: $\cos x+\sqrt{3}\sin x=2\cos (x-{60}^o)$ and hence find the maximum value of $\cos x+\sqrt{3}\sin x$.

Answer 1

We have, $\cos x+\sqrt{3}\sin x$

Multiplying and dividing the whole expression by $\sqrt{1^2+(\sqrt{3}{)}^2}=2$

We get, $2(\frac{1}{2}\cos x+\frac{\sqrt{3}}{2}\sin x)$

We know, $\cos {60}^{\circ }=\frac{1}{2}$ and $\sin {60}^{\circ }=\frac{\sqrt{3}}{2}$

Substituting the values, $2(\cos {60}^{\circ }\cos x+\sin {60}^{\circ }\sin x)$

Using the identity, $\cos (A-B)=\cos A\cos B+\sin A\sin B$, the above expression becomes,

$2(\cos {60}^{\circ }\cos x+\sin {60}^{\circ }\sin x)$ = $2\cos (x-{60}^{\circ })$

$LHS=RHS$. Hence proved.

Since, the cosine function has a range of $[-1,1]$, the maximum value of $2\cos (x-{60}^{\circ })$ is equal to $2$.

$max\left(2\cos \right(x-{60}^{\circ }\left)\right)$ $=2\cos (x-{60}^{\circ }){|}_{x={60}^{\circ }}$ $=2\times 1=2$