Maximum value of $f(x) = \cos x \left( \sin x + \sqrt {\sin^2x +\sin^2a}\right)$ Can we find maximum value of $$f(x) = \cos x \left(  \sin x + \sqrt
{\sin^2x +\sin^2a}\right)$$
where '$a$' is a given constant.
Using derivatives makes calculation too complicated.
 A: Let $$y=\cos x\left[\sin x+\sqrt{\sin^2 x+\sin^2 a}\right] = \sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}$$
Now Using $\bf{Cauchy\; Schwartz\; Inequality}$
We get $$(\sin^2 x+\cos ^2 x)\cdot \left[\cos^2 x+\sin^2 x+\sin^2 a\right]\geq \left(\sin x\cdot \cos x+\cos x\cdot \sqrt{\sin^2 x+\sin^2 a}\right)^2$$
So we get $$y^2\leq (1+\sin^2 a)\Rightarrow |y| \leq\sqrt{1+\sin^2 a}$$
A: Hint $f(x)\leq \frac{cos^2(x)+(sin(x)+\sqrt{sin^2(x)+sin^2(a)})^2}{2}$ now first expand square to use $cos^2(x)+sin^2(x)=1$ and then one can  easily use derivative to find the minima of the function to get maxima of $f(x)$ by $AM-GM$ inequality. 
A: HINT:
$$y=\cos x(\sin x+\sqrt{\sin^2x+\sin^2A})$$
$$\iff y\sec x-\sin x=\sqrt{\sin^2x+\sin^2A}$$
Squaring we get  $$y^2\tan^2x-2y\tan x+y^2-\sin^2A=0$$
As $\tan x$ is real, the discriminant  $$(2y)^2-4y^2(y^2-\sin^2A)\ge0$$
A: the first derivative is given by $$f'(x)=-\sin \left( x \right)  \left( \sin \left( x \right) +\sqrt { \left( 
\sin \left( x \right)  \right) ^{2}+ \left( \sin \left( a \right) 
 \right) ^{2}} \right) +\cos \left( x \right)  \left( \cos \left( x
 \right) +{\frac {\sin \left( x \right) \cos \left( x \right) }{\sqrt 
{ \left( \sin \left( x \right)  \right) ^{2}+ \left( \sin \left( a
 \right)  \right) ^{2}}}} \right) 
$$ simplifying this you have to solve the equation for $x$:
$$- \left( \sin \left( x \right) +\sqrt { \left( \sin \left( x \right) 
 \right) ^{2}+ \left( \sin \left( a \right)  \right) ^{2}} \right) 
 \left( \sin \left( x \right) \sqrt { \left( \sin \left( x \right) 
 \right) ^{2}+ \left( \sin \left( a \right)  \right) ^{2}}- \left( 
\cos \left( x \right)  \right) ^{2} \right) 
=0$$ 
