Sine and Cosine

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 $\displaystyle \min[f(x)] = 2-\sin^2(1)-1 = \cos^2(1)\;\; \forall x\in \frac{(2n+1)\pi}{2}\;,n\in \mathbb{Z}$

So $\max[f(x)]-\min[f(x)] =1+\sin^2(1)-\cos^2(1) = 1-\cos(2)$