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### Trigonometry

Posted: Thu Aug 25, 2016 11:58 am
1. $x$ বাস্তব সংখ্যা হলে কোনটি বড়? $\sin(\cos x)$ নাকি $\cos(\sin x)$?

2. Prove that $\tan\left(\frac{n+1}{2}a\right) =\frac{\sin a+\sin 2a+\ldots+\sin na}{\cos a+\cos 2a+\ldots+\cos na}$

### Re: Trigonometry

Posted: Sat Sep 24, 2016 12:35 pm
I remember the first problem from Neurone Onuronon. [Ah, nostalgia :") ] Anyway, here are your solutions:

1. Set $y=\sin x$. After some tedious calculations, it becomes apparent that $\cos y>\sin \left(\sqrt{1-y^2}\right)$. In other words, $\cos(\sin x) > \sin(\cos x)$.
[You can also show it by computing the derivatives of both functions. However, that would be, in my opinion, not 'ethical'.]

2. Do telescoping like this:
$\sin ia+\sin(n+1-i)a=2\sin\left(\frac{n+1}{2}a\right)\cos\left(\frac{n+1-2i}2a\right)$

### Re: Trigonometry

Posted: Sat Sep 24, 2016 7:08 pm
$\cos (\sin x)$ and $\sin(\cos x)$ are both continuous functions. So if one isn't always greater than the other, they must intersect at some value $k$ for which$\cos(\sin k) = \sin(\cos k)$

$\sin x$ ranges from $-1$ to $+1$, so $\cos(\sin x)$ is always positive. So $\cos (\sin k) = \sin(\cos k) > 0$

Now for positive values of $\cos x$, $\cos x = \sin y \rightarrow x + y = \frac {\pi}{2}$

So, $\cos k + \sin k = \frac {\pi}{2}$

But $\cos k + \sin k$ has a max value of $\sqrt{2}$. Thus no $k$ exists for which $\cos(\sin k) = \sin(\cos k)$.

Then plugging in any value of x and checking we can conclude that $\cos(\sin x) > \sin(\cos x)$ for all $x$.