Find the value of $\cos(\theta - \frac{\pi}{4})$

Question

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

Then find the value of the $\cos(\theta - \frac{\pi}{4}) $

Answer 1

$$\tan(x)=\cot\left(\frac\pi2-x\right)$$

$$\implies\tan(\pi\cos(\theta))=\cot\left(\frac\pi2-\pi\cos(\theta)\right)=\cot\left(\pi\sin(\theta)\right)$$

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$${\cot(x)=\cot(y)\implies x-y=n\pi}\tag{*}$$

$$\implies \left(\frac\pi2-\pi\cos(\theta)\right)-\pi\sin(\theta)=n\pi$$

Rearranging,

$$\frac\pi2-n\pi=\pi\cos\left(\theta-\frac\pi4\right)$$

or equivalently,

$$\frac12-n=\cos\left(\theta-\frac\pi4\right)$$

Since $\cos:\Bbb R\to[-1,1]$, the only allowed values are $n=0,1$. Which gives us to possible values, $\frac12$and $-\frac12$

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$(*)$ This is valid because,

$$\cot(x-y)=\frac{1+\cot(x)\cot(y)}{\cot(x)-\cot(y)}$$

Therefore, if $\cot(x)=\cot(y)$, $\cot(x-y)$ is undefined, and so $x-y=n\pi$

Answer 2

As $\displaystyle\cot(\pi \sin \theta)=\tan\left(\frac\pi2-\pi \sin \theta\right) $

we have $\displaystyle\tan(\pi\cos\theta)=\tan\left(\frac\pi2-\pi \sin \theta\right) $

$\displaystyle\implies\pi\cos\theta=m\pi+\frac\pi2-\pi \sin\theta $ where $m$ is any integer

$\displaystyle\implies\cos\theta=m+\dfrac12-\sin\theta\iff\cos\theta+\sin\theta=\dfrac{2m+1}2 $

$\displaystyle\implies\cos\left(\theta-\dfrac\pi4\right)=\cos\frac\pi4\cos\theta+\sin\frac\pi4\sin\theta=\frac{\cos\theta+\sin\theta}{\sqrt2}=\dfrac{2m+1}{2\sqrt2}\ \ \ \ (1)$

Now for real $\displaystyle\theta,-1\le\cos\left(\theta-\dfrac\pi4\right)\le1$

$\displaystyle\implies -1\le\dfrac{2m+1}{2\sqrt2}\le1\ \ \ \ (2)$

$\displaystyle\implies -2\sqrt2\le2m+1\le2\sqrt2\iff -2\sqrt2-1\le2m\le2\sqrt2-1$

$\displaystyle\implies2m\le2\sqrt2-1<2$ as $2\sqrt2<3$ as $(2\sqrt2)^2<3^2$

$\displaystyle\implies m<1$

and $\displaystyle2m+1\ge -2\sqrt2-1>-4$ as $-2\sqrt2>-4+1\iff 3>2\sqrt2 $

$\displaystyle\implies m>-2$

As $m$ is an integer, $\displaystyle-2<m<1\implies m=-1,0$ which needs to be replaced in $(1)$

Another way: $(2)$ also implies $\displaystyle\left(\dfrac{2m+1}{2\sqrt2}\right)^2\le1\iff (2m+1)^2\le(2\sqrt2)^2$ which is $<9$

$\displaystyle\implies (2m+1)^2<3^2\iff (m+2)(m-1)<0\implies -2<m<1$