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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.

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  • $\begingroup$ Graphing the equation I got that B. is the maximum. $\endgroup$
    – Sigma6RPU
    Commented Aug 29, 2016 at 16:40

4 Answers 4

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$$\max_{\theta\in\mathbb{R}}\{\cos^2(\cos\theta) + \sin^2(\sin\theta)\}\le\max_{-1\le x \le 1}\cos^2 x+\max_{-1\le y \le 1}\sin^2 y=1+\sin^21$$

$$\cos^2 \left( \cos \frac{\pi}{2} \right) + \sin^2 \left( \sin \frac{\pi}{2} \right) = 1 + \sin^2 1$$

So, the maximum is at most $1+\sin^21$, and this value is achieved. Hence the answer is b).

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  • $\begingroup$ +1 Sorry I ran out of votes. :( Your answer is inspirational. $\endgroup$
    – Sigma6RPU
    Commented Aug 29, 2016 at 17:10
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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 $f(\pi/2)=1+\sin^2(1)$.

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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 $(b)$.

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  • $\begingroup$ I like your answer because at every $\pi/2$ the maximum occurs when I graphed it. $\endgroup$
    – Sigma6RPU
    Commented Aug 29, 2016 at 16:44
  • $\begingroup$ Dude you confused the letters. The answer you got is B. and wrote down C. please edit your post. $\endgroup$
    – Sigma6RPU
    Commented Aug 29, 2016 at 17:03
  • $\begingroup$ @Sigma6RPU nice catch. $\endgroup$
    – Alex Ortiz
    Commented Aug 29, 2016 at 17:04
  • $\begingroup$ Your welcome I was like wait right answer wrong letter. $\endgroup$
    – Sigma6RPU
    Commented Aug 29, 2016 at 17:05
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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.

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  • $\begingroup$ I'm not sure about your line where you claim $\sin(\theta + \pi/2) = \sin \pi$. This isn't true if $\theta = 0$, for instance. $\endgroup$
    – Alex Ortiz
    Commented Aug 29, 2016 at 17:17
  • $\begingroup$ @AOrtiz I set $\sin(\theta+\frac{\pi}2)=0$ You are right but just to point out one solution I set it to sine of $\pi$ $\endgroup$
    – Sigma6RPU
    Commented Aug 29, 2016 at 17:28

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