extremely simple trigonometry question, however I have never taken it and thus I have no clue how to solve it:
Find the angles whereby $\cos(t) = \sin(t)$ where $t$ is between 0 and $4\pi$?
From a calculator, I can get $\arctan(1) = \pi/4$.
However, once I have $\pi/4$ - how do I get all the other angles where it is equal in between $0$ and $4\pi$?
Thanks!
Hint. Since $\sin(t)-\cos(t)=\sqrt{2}\sin(t-\pi/4)$, it follows that the solutions of $\sin(t)-\cos(t)=0$ are $t=\pi/4+k\pi$ with $k\in\mathbb{Z}$. What are the solutions in $[0,4\pi]$?
For a circle of radius $r=1$ one can read the values of $\cos\alpha$ and $\sin\alpha$ as $x$ and $y$ coordinate of the increasing angle.
This gives four possible angles where both have the same length $\lvert x\rvert = \lvert y \rvert$, if one considers only non-negative angles less than 360 degrees (in radians: $2\pi$, not $4\pi$).
And two of them where they have same values.
$$\sin x = \cos x$$ $$\tan x= 1 \implies \tan x = \tan \pi/4, : x\ne (2n+1)\pi/2 $$
$$\therefore x= n\pi+ \pi/4 : n \in \mathbb{Z}$$
$$n=0, x = \pi/4 \\ n=1, x= 5\pi/4\\n=2, x= 9\pi/4 \\ n=3, x=13\pi/4$$
$n=4$ gives $x=4\pi+\pi/4 \ge 4\pi$
so the final answer is $\pi/4,5\pi/4,9\pi/4, 13\pi/4$
