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Problem :

If $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$, then what is the value of $\cos(\theta -\frac{\pi}{4})$?

My approach :

Solution: $\tan(\pi \cos\theta) =\cot(\pi \sin\theta)$

$\Rightarrow \tan(\pi \cos\theta) = \tan \{ \frac{\pi}{2} - (\pi \sin\theta) \} $

$\Rightarrow \pi \cos\theta = \frac{\pi}{2} - (\pi \sin\theta)$

$\Rightarrow \frac{1}{2} =\frac{1}{\sqrt{2}}[\sin\frac{\pi}{4} \cos\theta + \cos\frac{\pi}{4} \sin\theta] $

$\Rightarrow \frac{1}{\sqrt{2}} = \sin(\frac{\pi}{4} + \theta)$ $\Rightarrow \frac{\pi}{4} = \frac{\pi}{4} + \theta$

$\Rightarrow \theta = 0$

$\therefore \cos(\theta - \frac{\pi}{4})$ = $\frac{1}{\sqrt{2}}$ But this is wrong answer.. please suggest where I am wrong... thanks.

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$\Rightarrow \frac{1}{2} =\sqrt{2}[sin\frac{\pi}{4} cos\theta + cos\frac{\pi}{4} sin\theta]$. – njguliyev Sep 30 '13 at 11:56
that was silly mistake from my side.. thanks a lot... – sultan Sep 30 '13 at 15:28

I would use

$$\frac{\sin{(\pi \cos{\theta})}}{\cos{(\pi \cos{\theta})}} = \frac{\cos{(\pi \sin{\theta})}}{\sin{(\pi \sin{\theta})}}$$

from which I get

$$\cos{[\pi (\cos{\theta}+\sin{\theta})]}=0$$

or, in one case,

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


$$\sqrt{2} \pi \cos{\left ( \theta-\frac{\pi}{4}\right )} = \frac{\pi}{2}$$

You can take it from here.

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$$\tan(\pi\cos\theta)=\cot(\pi\sin\theta)\implies \sin(\pi\cos\theta)\sin(\pi\sin\theta)=\cos(\pi\cos\theta)\cos(\pi\sin\theta)\implies$$


$$\implies\cos\left(\pi(\cos\theta+\sin\theta)\right)=0\iff\cos\theta+\sin\theta=\frac{2n+1}2\;\;,\;\;n\in\Bbb Z$$

But we can only have $\;n\in\{-2,-1,0,1\}\;$ (why?), so

$$\sin\theta+\cos\theta=k\iff \sin x\cos\frac\pi4+\sin\frac\pi4\cos\theta=k\cos\frac\pi4\iff$$

$$\iff\sin\left(\theta+\frac\pi4\right)=\frac k{\sqrt2}\;\ldots$$

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