Does $\tan\left(\frac{5\pi}{4}-\alpha\right)=\tan\left(\frac{\pi}{4}-\alpha\right)$ hold for any angle, or only when $\pi/4-\alpha$ is acute?

Question

I often find this in math problems, but authors don't make it clear if $\alpha$ needs to be an acute angle or not, so I am confused a little bit. I thought we can only do this kind of transformation if we know that $\frac{\pi}{4}-\alpha$ is less than $\frac{\pi}{2}$. So, my question is why is this true for any angle (if it is)? $$\tan\left(\frac{5\pi}{4}-\alpha\right)=\tan\left(\frac{\pi}{4}-\alpha\right)$$

Answer 1

You can write $\dfrac{5\pi}{4}$ as $\pi+\dfrac\pi4$.

Thus, $\tan(\dfrac{5\pi}{4}-\alpha) = \tan(\pi+\dfrac\pi4-\alpha) = \tan(\dfrac\pi4-\alpha)$

Because for any $\alpha$ ; $\quad$$\tan(\pi+\alpha ) = \tan(\alpha)$ .

This is also true for $\cot(\pi+\alpha)$ as both.

EDIT : When you have $\frac\pi2\pm\alpha $ or $\frac{3\pi}2\pm \alpha$

$\cos \to \sin$

$\tan \to \cot$

$\sec \to \csc$

and vice versa.ie if you have $\cos(\frac\pi2- \frac\pi4) $ it becomes $\sin(\frac\pi4)$

You might find it useful to learn this.

Depending on where your angle lies you should change the sign.

Edit: A full list of this type of identity for the non-hyperbolic trigonometric functions can be found here.

Answer 2

The formula for tangent of sum is

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

We know that $\tan \pi =0$

Let $x=\pi$ and $y=\pi /4 - \alpha $

You get $$ \tan(\dfrac{5\pi}{4}-\alpha) = \tan(\dfrac\pi4-\alpha)$$