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 $$1+\sin^2(\cos x)-\sin^2(\sin x) = 1+\sin(\cos x+\sin x)\cdot \sin(\cos x-\sin x)$$

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 $$\displaystyle x = \frac{\pi}{2}\;\;,\frac{3\pi}{2}\;\;,\frac{5\pi}{2},.....$$
