# Solving the equation $\sin t = -\sqrt{2}/2$

Solving the equation $$\sin t = -\frac{\sqrt{2} }{2} .$$

I know the solution is $1.25$ and $1.75$, but I do not know how to get there.

An explanation would be GREATLY appreciated, thanks!

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You have your unit circle values committed to memory, yes? Alternatively, try drawing a right triangle with hypotenuse 1 below (yes, that's a big hint) the horizontal axis. – J. M. Sep 19 '11 at 1:58
In any case, the answer is not $1.25$, but rather $\frac{5}{4}\pi = 1.25\pi$ and $1.75\pi$. Someone seems to have eaten up your pi(e). – Arturo Magidin Sep 19 '11 at 2:01
Do you how to solve $\sin t = \frac{\sqrt{2}}{2}$? – Srivatsan Sep 19 '11 at 2:12

Let us observe trigonometry unit circle shown below.Note that $\sin t$ is represented on $y$-axis,and has positive value between $0$ and $\pi$,and negative value between $\pi$ and $2\pi$.We can show from the right triangle with cathetuses that are sides of the square and hypotenuse which is diagonal of the square that $\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}$. Note that $\sin \frac{3\pi}{4}=\frac{\sqrt{2}}{2}$ also. Now if we look at trigonometry unit circle we can figure out that $\sin t$ has value $\frac{-\sqrt{2}}{2}$ when $t=\frac{5\pi}{4}$ or when $t=\frac{7\pi}{4}$.

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$$\sin(t) = \frac{-1}{\sqrt{2}} = \sin \bigg(\frac{-π}{4}\bigg)$$

$$\implies t=n (π)+(-1)^n \bigg(\frac{-π}{4}\bigg), \space n \in \mathbb{Z}$$

Taking $n=1,2,3,...$, we have the following solutions:

For $n=1$ , $t= π -\big(-\frac{π}{4}\big) = π + \frac{π}{4} = \frac{5π}{4}$ and so on ....

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All you have to know is these formulas: $\sin(x+\pi)=-\sin(x)$ and $sin(2\pi-x)=-sin(x)$ At the end you have to determine which degree is $\frac{\sqrt{2}}{2}$ so that is $\frac{\pi}{4}$. I hope you underestand.

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solving $sin( x) = sin(y)$ gives that $x=y$ or $x=\pi-y$ (modulo $2\pi$ of course). Our equation is $sin(t)=-\sqrt2/2$ so we have to find a value $y$ such that $sin(y)= -\sqrt2/2$. Now since it is known that $sin(\pi/4)=\sqrt2/2$, $y=\pi/4+\pi=5\pi/4$ is a good candidate.so our equation is $sin(x)=sin(5\pi/4)$ which gives that $x=5\pi/4$ or $x=\pi-5\pi/4=-\pi/4$ or if we want a positive representative then we obtain $-\pi/4+2\pi=7\pi/4$ so the set of solutions is $x=5\pi/4+2k\pi$ and $x=7\pi/4+2k\pi$, $k$ being any integer.

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Typeset $\sin$ as "\sin" and it looks better. – Stefan Smith Jun 9 '12 at 14:02

$\sin t = -\frac{\sqrt{2} }{2}$ We have $$\sin t = -\frac{\sqrt{2} }{2} = -\sin\frac{\pi}{4} = \sin\left(-\frac{\pi}{4}\right)$$ $\Leftrightarrow t = -\frac{\pi}{4} + k\cdot2\pi$ or $t = \pi - \left(-\frac{\pi}{4}\right) + k\cdot2\pi = \frac{5\pi}{4}+k\cdot2\pi$ with $k\in\mathbb{Z}$

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HINT: Think the hypotenuse of a $45, 45, 90 \Delta$

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