Problem related to maximum and minimum value of a trigonometric function I have used the following technique to calculate the maximum value of the function...
but I couldn't proceed with next step..
can anyone guide me please?
Find the maximum and minimum value of the function $$f(x) = \sin^2(\cos x)+\cos^2(\sin x)$$
I have written the following function as $$1+\sin^2(\cos x)-\sin^2(\sin x) = 1+\sin(\cos x+\sin x)\cdot \sin(\cos x-\sin x)$$
 A: Without Using Calculus
Let $$f(x) = \sin^2(\cos x)+\cos^2(\sin x) = 1-\cos^2(\cos x)+1-\sin^2(\sin x)\;$$
So we get $$f(x) = 2-\left[\cos^2(\cos x)+\sin^2(\sin x)\right]$$
Now $$-1\leq \cos x\leq 1\Rightarrow \cos^2(1)\leq \cos^2(\cos x)\leq 1$$ And equality on left and right hold for $\displaystyle x = k\pi$ and $\displaystyle x= \frac{(2n+1)\pi}{2}$ 
Now $$-1\leq \sin x \leq 1\Rightarrow 0\leq \sin^2(\sin x)\leq \sin^2(1)$$
And equality on left and right hold for $\displaystyle x = k\pi$ and $\displaystyle x= \frac{(2n+1)\pi}{2}$
Hence $$\cos^2(1)\leq \cos^2(\cos x)+\sin^2(\sin x)\leq \sin^2(1)+1$$  And equality on left and right hold for $\displaystyle x = k\pi$ and $\displaystyle x= \frac{(2n+1)\pi}{2}$
So we get $$\max[f(x)] = 2-\cos^2(1) = 1+\sin^2(1)\;\; \forall x\in k\pi\;,k\in \mathbb{Z}$$ 
So we get $$\displaystyle \min[f(x)] = 2-\sin^2(1)-1 = \cos^2(1)\;\; \forall x\in \frac{(2n+1)\pi}{2}\;,n\in \mathbb{Z}$$ 
A: If I am not wrong:
1) You can takes $x\in [-\pi,+\pi]$.
2) You have $f(-x)=f(x)$. You can take $x\in [0,\pi]$.
3) Put $t=\cos(x)$. You have $\sin(x)=\sqrt{1-t^2}$, with $t\in [-1,+1]$. Your function is now $g(t)=\sin(t)^2+\cos(\sqrt{1-t^2})^2$
4) You have $g(-t)=g(t)$. You can take $t\in [0,1]$.
5) Now on $[0,1]$, $\sin(t)$ is increasing and positive. Hence $(\sin(t))^2$ is increasing. $\cos(u)$ is decreasing, but $\sqrt{1-t^2}$ also. So $\cos(\sqrt{1-t^2})$ is increasing, and positive and its square is increasing. Hence $g(t)$ is increasing on $[0,1]$, and it is easy to finish.
A: Using Calculus
Let $$f(x) = \sin^2(\cos x)+\cos^2(\sin x) = 1-\cos^2(\cos x)+1-\sin^2(\sin x)\;$$
So we get $$\displaystyle f(x) = 2-\left[\cos^2(\cos x)+\sin^2(\sin x)\right] = 2-\frac{1}{2}\left[2\cos^2(\cos x)+2\sin^2(\sin x)\right]$$
So we get $$\displaystyle f(x) = 2-\frac{1}{2}\left[1+\cos(2\cos x)+1-\cos(2\sin x)\right] = 1+\frac{1}{2}\left[\cos(2\sin x)+\sin(2\cos x)\right]$$
Now Using First Derivative test.
So $$\displaystyle f'(x) = \frac{1}{2}\left[-\sin(2\sin x)\cdot 2\cos x+\cos(2\cos x)\cdot -2\sin x\right]$$
So $$\displaystyle f'(x) = -\left[\sin(2\sin x)\cdot \cos x+\cos(2\sin x)\cdot \sin x\right]$$
Now $f'(x) =0$ at $$\displaystyle x=0\;\;,\frac{\pi}{2}\;\;,\pi\;\;,\frac{3\pi}{2}\;\;,2\pi,......$$
So $$\displaystyle \max[f(x)] = \sin^2(1)+1\;\;$$ at $$\displaystyle x = 0\;\;,\pi\;\;,2\pi,.....$$
And So $$\displaystyle \min[f(x)] = \cos^2(1)\;\;$$ at $$\displaystyle x = \frac{\pi}{2}\;\;,\frac{3\pi}{2}\;\;,\frac{5\pi}{2},.....$$
