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

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

or,

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

You can take it from here.

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

$$\cos\left(\pi(\cos\theta-\sin\theta)\right)-\cos\left(\pi(\cos\theta+\sin\theta)\right)=\cos\left(\pi(\cos\theta-\sin\theta)\right)+\cos\left(\pi(\cos\theta+\sin\theta)\right)$$

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

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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It is given that,

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

Therefore,

$\pi \cos\theta + \pi \sin\theta = 2n\pi \pm \frac \pi 2 , [n \in Z]\\ \cos\theta + \sin\theta = 2n \pm \frac 1 2$

Since $\sin\theta + \cos\theta$ always lies between $\sqrt 2$ and $-\sqrt 2$, $n$ can only assume the value $0$. Thus,

$\cos\theta + \sin\theta = \pm \frac 1 2\\ \frac 1 {\sqrt 2} \cos\theta + \frac 1 {\sqrt 2} \sin\theta = \pm \frac 1 {2 \sqrt 2}\\ \cos\frac \pi 4 \cos\theta + \sin\frac \pi 4 \sin\theta = \pm \frac 1 {2 \sqrt2}\\ \cos(\theta - \frac \pi 4) = \pm \frac 1 {2 \sqrt 2}$

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  • $\begingroup$ Use Mathjax for formatting otherwise this is not helpful at all. $\endgroup$ – StubbornAtom Dec 28 '16 at 7:27
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$$\tan (\pi \cos x)=\cot(\pi \sin x)$$

or,

$$\tan (\pi \cos x)=\tan (\frac{\pi}{2} - \pi \sin x)$$ [∵, tan(π/2+Ф)=cotФ]

or, πcosx=π/2-πsinx

or, π(sinx+cosx)=π/2

or, sinx+cosx=1/2

or, (sinx+cosx)1/√2=1/2√2

or, cosx(1/√2)+sinx(1/√2)=1/2√2

or, cosxcosπ/4+sinxsinπ/4=1/2√2

or, cos(x-π/4)=1/2√2 

∴, a)1/2√2 is the right answer

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  • $\begingroup$ Hi, welcome on the MathSE! Do you know Latex? This site supports it, just type in $\tan \pi$ and you get $\tan \pi$. I did it with your first some formulas, the rest would be yours :-) $\endgroup$ – peterh says reinstate Monica Jul 20 '18 at 16:14

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