# Get value of sin(x)

I need help with trigonometry:

How can I calculate $\alpha$: $$\sin(\alpha)=\frac{-\sqrt{2}}{2}$$ ? I can't find value of this angle in any table...

• I just want to be sure the negative is inside the root? – randomgirl Apr 7 '15 at 20:53
• Have you seen sin defined in terms of exponentials? – George Moore Apr 7 '15 at 20:53
• I see you edited it where the negative is outside the root. You should be able to whip out your unit circle and find in the 3rd and 4th quadrant that the sine value is $-\frac{\sqrt{2}}{2}$ for $\alpha=\frac{ 5 \pi}{4} , \frac{ 7 \pi}{4}$ – randomgirl Apr 7 '15 at 21:08
• $\alpha = -45^\circ = -\pi/4$ wont do? – abel Apr 7 '15 at 21:08
• – randomgirl Apr 7 '15 at 21:09

$$\sin \alpha = -\frac{\sqrt{2}}{2} \iff \sin(-\alpha) = \frac{1}{\sqrt{2}}$$ The positive angle $-\alpha$ can be found in the triangle given by the end points $A=(0,0)$, $B=(0,1)$ and $C=(1,0)$. As the angle $-\alpha = \pi/4$. Thus $\alpha = -\pi/4 = 2\pi - \pi/4 = 7\pi/4$.

And from $$\sin(\alpha + \pi) = -\left(-\frac{1}{\sqrt{2}}\right) \Rightarrow \\ \alpha + \pi = \pi/4 \Rightarrow \alpha = \pi/4 - \pi = -3\pi/4 = 2 \pi -3\pi/4 = 5\pi/4$$

Bonus:

$$\sin \alpha = \frac{\sqrt{-2}}{2}$$

That is only possible in the complex domain: $$\sin z = i \frac{1}{\sqrt{2}}$$ and $$\sin(-iu) = \sinh(u) / i$$ this gives $$\sinh u = -\frac{1}{\sqrt{2}} \Rightarrow u = - \sinh^{-1}{\frac{1}{\sqrt{2}}} \Rightarrow z = i \sinh^{-1}{\frac{1}{\sqrt{2}}}$$ where $\sinh^{-1}(1/\sqrt{2}) = 0.658478..$.

Adding the periodicity of $\sin(z) = \sin(z + 2\pi k)$ it should be $$z = 2\pi k + i \sinh^{-1}{\frac{1}{\sqrt{2}}} \quad (k \in \mathbb{Z})$$ And using the symmetry $\sin(z) = - \sin(z + \pi)$ we get $$z = 2\pi k + \pi - i \sinh^{-1}{\frac{1}{\sqrt{2}}} \quad (k \in \mathbb{Z})$$

• wolframalpha.com/input/?i=sin%28x%29%3Dsqrt%28-2%29%2F2 wolfram gives another solution $z=2 \pi k+\pi-i \sinh^{-1}(\frac{1}{\sqrt{2}})$ I guess we can get this solution in a similar way maybe. – randomgirl Apr 7 '15 at 21:22
• There is also a third quadrant solution to the intended problem. – Ian Apr 7 '15 at 21:37
• Thanks Ian, I overlooked a symmetry again it seems. – mvw Apr 7 '15 at 21:42