# Solve trigonometric equation

I am interested in finding all $t$ in $[0,2\pi]$ with $\cos t = - \sin t$. Is there a way to do this without using a calculator? I know $- \sin t = \sin (-t)$ but it doesn't seem to help. Thank you.

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You could make tangent appear, if it makes it easier for you. –  julien Feb 26 '13 at 16:11
I have found answer on my own. By drawing the curves. It is not as difficult to draw as it might seem. Please delete this question. –  colollary Feb 26 '13 at 16:17
How can I delete this question? I have answered it. –  colollary Feb 26 '13 at 16:19
Like many questions you will be asked, this one is easier without a calculator. –  André Nicolas Feb 26 '13 at 18:38

You can use $\cos t + \sin t=\sqrt 2 \sin (t + \frac \pi 4)$ and set it to zero to find $t=\frac {3 \pi}4, \frac {7 \pi}4$

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Thanks, it is good to see different way of getting to same solution. After posting question I found solution by drawing the curves. It works also. –  colollary Feb 26 '13 at 16:29

Hint: $\cos t = \sin t \;\implies \;\;t_1 = \pi /4, \;\;t_2 = 5\pi/4\;\;$ for $0 \leq t \leq 2\pi$

Added: Now what does this mean if $\cos t = -\sin t$?

It means that the solutions for $\;\;0 \leq t \leq 2\pi\;\;$ are the solutions $\;\;t_1 + \pi/2\;\;\;\text{and}\;\;\;t_2 + \pi/2$

Alternatively, $$\cos t = - \sin t \;\; \iff \;\; 1 = -\frac{\sin t}{\cos t}$$ $$\iff\;\; 1 = -\tan t \;\;\iff \tan t = -1 \;; \iff t = \tan^{-1}(-1)$$

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Yes I know how to use a calculator. I don't see how your hint is answering my question. Meanwhile I have found answer using picture. One correct solution is $t= 3 \pi / 4$ –  colollary Feb 26 '13 at 16:16
My hint was about the value of t when $\cos t = \sin t$ (angle in first and third quadrant)...so, I was trying to hint that if $\cos t = \bf{-}\sin t$, we have a rotation of $\pi/2$ from the values I listed: angles in the second and fourth quadrant: which yes in the second quadrant $t = \pi/4 + \pi/2 = 3\pi/4$ and the other being $t = 5\pi/4 + \pi/2 = 7\pi/4$ which is in the fourth quadrant. I did not mean to be demeaning, if that's the way you took my post. –  amWhy Feb 26 '13 at 16:20
@Ross, please see my comment above. I KNOW that there is a minus sign on $\sin$ in the given equation. I was trying to HINT that if you know the values of $t$ for which $\cos t = \sin t$, one can easily deduce the values for $t$ when $\cos t = - \sin t$ –  amWhy Feb 26 '13 at 16:24
@colollary I did not intend to be demeaning. I hope you read by comment above. I wouldn't have answered unless I was trying to be helpful. –  amWhy Feb 26 '13 at 16:33
No, not demeaning. But my question is how to compute this without calculator. Your answer is beside the point and therefore useless: you don't explain how to compute it you post a solution. As for "Alternatively...": how do I calculate $\tan^{-1}$ without calculator? I am not a calculator. –  colollary Feb 26 '13 at 16:35

$\cos t=-\sin t\implies \tan t=-1=-\tan\frac\pi4=\tan \left(-\frac\pi4\right)$ as $\tan(-x)=-\tan x$

$\implies t=n\pi-\frac\pi4$ where $n$ is any integer.

Now, we need $0\le n\pi-\frac\pi4\le 2\pi\implies 0\le4n-1\le 8$

So, $4n-1\ge 0\implies n\ge \frac14\implies n\ge1$ as $n$ is any integer

and $4n-1\le 8\implies n\le\frac94\implies n\le2$ as $n$ is any integer.

So, $n=1,2\implies t=\pi-\frac\pi4=\frac{3\pi}4$ or $t=2\pi-\frac\pi4=\frac{7\pi}4$

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