If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos\left (\theta -\frac{\pi}{4}\right)$

Question

If $\tan (\pi \cos \theta) =\cot (\pi \sin \theta) $ then find the value of $\cos \left(\theta -\frac{\pi}{4}\right).$

I could not get any idea to solve. However I tried by using $\theta =0^\circ $. But could not get the answer.

Answer 1

Rearrange into sines and cosines. Form a cosine addition identity.

\begin{align} \tan(\pi\cos t)\tan(\pi\sin t) &= 1 \\ \sin(\pi\cos t)\sin(\pi\sin t) &= \cos(\pi\cos t)\cos(\pi\sin t) \\ 0 &= \cos(\pi\cos t)\cos(\pi\sin t) - \sin(\pi\cos t)\sin(\pi\sin t) \\ 0 &= \cos(\pi\cos t + \pi\sin t) \\ 0 &= \cos(\pi[\cos t + \sin t]) \end{align}

The argument of the cosine in the previous line must be an odd integer multiple of $\pi/2$, so we get \begin{align} \cos t + \sin t &= \frac{2n+1}{2} \end{align}

where $n\in\mathbb{Z}$ (EDIT: More accurately, $n \in \{-1,0\}$ to be in the range of $\cos t + \sin t$). Use the cosine subtraction formula to achieve the desired result. \begin{align} \frac{1}{\sqrt2}\cos t + \frac{1}{\sqrt2}\sin t &= \frac{2n+1}{2\sqrt{2}} \\ \cos (t)\cos(\pi/4) + \sin (t)\sin(\pi/4) &= \frac{2n+1}{2\sqrt{2}} \\ \\ \cos(t -\pi/4) &= \frac{2n+1}{2\sqrt{2}} \end{align}

EDIT: Following lab bhattacharjee's comment, the range of cosine restricts the possible solutions to $$\cos(t -\pi/4) = \frac{\pm1}{2\sqrt{2}}.$$

It doesn't appear like the answer is unique to me.

Answer 2

Hint -

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

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

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

$\cos \theta = \frac 12 - \sin \theta$

$\sin \theta + \cos \theta = \frac 12$

Now we have,

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

$= \cos\theta \cos \frac{\pi}4 +\sin\theta \sin \frac{\pi}4$

$= \cos\theta \frac 1{\sqrt{2}} + \sin\theta \frac 1{\sqrt{2}}$

$= \frac1{\sqrt 2} \left( \cos\theta +\sin\theta \right)$

$ = \frac 1{\sqrt2} \left(\frac 12\right)$

$= \frac 1{2\sqrt2}$

Answer 3

Since $$\tan(\pi\cos\theta)=\cot(\pi\sin\theta),$$

we can write $$\tan(\pi\cos\theta)=\cot(90-(\pi\cos\theta)).$$

Thus $$\cot(90-(\pi\cos\theta))=\cot(\pi\sin\theta)$$

$$\therefore\quad 90-(\pi\cos\theta)=\pi\sin\theta$$

$$90=\pi(\sin\theta+\cos\theta)$$

$$\frac{90}{\pi}=\sin\theta+\cos\theta$$

$$\frac{1}{2}=\sin\theta +\cos\theta$$

Now \begin{align*} \cos\left(\theta-\frac{\pi}{4}\right)&=\cos\theta \cos45^{\circ}+\sin\theta \sin45^{\circ}\\ &=\frac{\cos\theta}{\sqrt{2}}+\frac{\sin\theta}{\sqrt{2}}\\ &=\frac{\cos\theta +\sin\theta}{\sqrt{2}} \end{align*}

Since $$\sin\theta + \cos\theta =\frac{1}{2},$$

$$\cos\left(\theta-\frac{\pi}{4}\right)=\frac{1}{2\sqrt{2}}.$$

Answer 4

$$\begin{align} \tan (\pi \cos \theta) &= \cot (\pi \sin \theta) \\ \implies \frac{\sin(\pi \cos \theta)}{\cos(\pi \cos \theta)} &= \frac{\cos(\pi \sin \theta)}{\sin(\pi \sin \theta)} \\ \implies \sin(\pi \cos \theta)\sin(\pi \sin \theta) &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) \\ \end{align}$$ and then $$\begin{align} 0 &= \cos(\pi \sin \theta)\cos(\pi \cos \theta) - \sin(\pi \cos \theta)\sin(\pi \sin \theta) \\ &= \cos(\pi\sin\theta + \pi\cos\theta) \\ &= \cos(\pi(\sin\theta + \cos\theta)) \\ \implies \frac{\pi}{2} &= \pi(\sin\theta + \cos\theta) \\ \implies \frac{1}{2} &= \sin\theta + \cos\theta \\ \end{align}$$

Now we use the fact that $\sin \pi/4 = \cos \pi/4 = \frac{1}{\sqrt{2}}$. $$\begin{align} \sin\theta + \cos\theta &= \sqrt{2}\bigg(\frac{1}{\sqrt{2}}\sin\theta + \frac{1}{\sqrt{2}}\cos\theta\bigg) \\ &= \sqrt{2}(\sin\theta\cos \pi/4 + \cos\theta\sin \pi/4) \\ &= \sqrt{2}\sin(\theta + \pi/4) \\ &= \frac{1}{2} \end{align}$$ Finally, $\cos(\theta - \pi/4) = \sin(\theta + \pi/4) = \frac{1}{2\sqrt{2}}$.