Maximum value of trigonometric equation $\cos^2(\cos\theta) + \sin^2(\sin\theta)$ For any real $\theta$ the maximum value of $$\cos^2(\cos\theta) + \sin^2(\sin\theta)$$
A. $1$
B. $1 + \sin^21$
C. $1 + \cos^21$
D. does not exist
I tried it by converting the whole expression into $\sin$ but getting nowhere with that. 
$$1-\sin^2(\cos\theta) + \sin^2(\sin\theta)$$
Now since 1 is constant therefore, $$[\sin^2(\cos\theta) + \sin^2(\sin\theta)]$$ should be minimum but I don't know how to minimize it. 
Also is there a way to think about it's solution graph. 
I have to solve this without using calculus.
Kindly help.
 A: $$\max_{\theta\in\mathbb{R}}\{\cos^2(\cos\theta) + \sin^2(\sin\theta)\}\le\max_{-1\le x \le 1}\cos^2 x+\max_{-1\le y \le 1}\sin^2 y=1+\sin^21$$
$$\cos^2 \left( \cos \frac{\pi}{2} \right) + \sin^2 \left( \sin \frac{\pi}{2} \right) = 1 + \sin^2 1$$
So, the maximum is at most $1+\sin^21$, and this value is achieved. Hence the answer is b).
A: Hint: 
1) $f(x):=\cos^2(\cos\theta) + \sin^2(\sin\theta)$ is a periodic continuous function of period $\pi$.
2) $f(x)=f(\pi-x)$.
3) $f$ is increasing in $[0,\pi/2]$ because $\cos^2(\cos\theta)$ and $\sin^2(\sin\theta)$ are increasing in $[0,\pi/2]$.
4) $f$ attains its maximum at $\pi/2$ and $f(\pi/2)=1+\sin^2(1)$.
A: The idea is that we can maximize the argument of $\sin^2$ while also maximizing $\cos^2\theta$. To do so, we know that $\cos\theta$ (and $\cos^2\theta$) is maximized when $\color{red}{\theta = 0}$ and $\sin\theta$ is maximized when $\color{blue}{\theta = \pi/2}$, hence our result. We only need to notice that $\sin^2$ is increasing on $[0, 1]$ (remembering the graph of $\sin$ helps here) to conclude we have the max:
$$
\cos^2(\color{red}{\cos \pi/2}) + \sin^2(\sin \color{blue}{\pi/2}) = \cos^2(0) + \sin^2 1 = 1 + \sin^2 1.
$$
So the correct answer is $(b)$.
A: I thought about the problem like this when I saw the graph.

$$\cos^2(\cos\theta) + \sin^2(\sin\theta)$$
  $$\cos^2(\sin(\frac{\pi}{2}-\theta)) + \sin^2(\sin\theta)$$
  $$\sin(\theta+\frac{\pi}{2})=0$$
  $$\sin(\theta+\frac{\pi}{2})=\sin(\pi)$$
  $$\theta+\frac{\pi}{2}=\pi$$
  $$\theta=\frac{\pi}{2}$$

I used the co-function identity between $\sin(\theta)$, and $\cos(\theta)$. In case anybody wonders where I got the $\pi$ that is the period of the $\sin^2(x)$, and $\cos^2(x)$.  $\pi$ maximizes cosine and minimizes sine.
