Question

Prove that $f\left(x\right)=\frac{4\sin x}{2+\cos x}-x$ is an increasing function for $xϵ(0,\frac{\pi }{2})$ .

Answer 1

$y=\frac{4\sin x}{2+\cos x}-x$

$\frac{dy}{dx}=\frac{4\cos x(2+\cos x)-(-\sin x)\left(4\sin x\right)}{(2+\cos x{)}^2}-1$

$\frac{dy}{dx}=\frac{8\cos x+4(\cos x^2+\sin x^2)}{(2+\cos x{)}^2}-1$

$=\frac{8\cos x+4}{(2+\cos x{)}^2}-1$

$=\frac{8\cos x+4-(2+\cos x{)}^2}{(2+\cos x{)}^2}$

$=\frac{8\cos x-4\cos x+4-4-{\cos }^{2}x}{(2+\cos x{)}^2}$

$\frac{dy}{dx}=\frac{4\cos x-{\cos }^{2}x}{(2+\cos x{)}^2}$

For function to be increa\sin g,

$\frac{dy}{dx}=\frac{\cos x(4-\cos x)}{(2+\cos x{)}^2}x>0$

Therefore, $\cos x(4-\cos x)>0forxϵ[0,\frac{\pi }{2}]$

$0\le \cos x\le 1$...................(i)

Adding 4 both sides,

$3\le 4-\cos x\le 4$

Therefore $4-\cos x$ is positive.

Therefore$\cos x(4-\cos x)>0for\theta ϵ[0,\frac{\pi }{2}]$

Hence, $y=\frac{4\sin x}{2+\cos x}-x$ is increa\sin g function for $xϵ[0,\frac{\pi }{2}]$