How to manually calculate the sine? I started studying trigonometry and I'm confused.
How can I manually calculate the sine? For example: $\sin(\frac{1}{8}\pi)$?
I was told to start getting the sum of two values which will result the sine's value. For $\sin(\frac{7}{12}\pi)$, it would be $\sin(\frac{1}{4}\pi + \frac{1}{3}\pi)$. However, I find this way confusing. For example, I don't know which sum will result $\frac{1}{8}$ in the example above.
Is there a better/easier way to do it?
Please, can anyone explain step by step how to do it?
 A: Write \begin{align}\sin(\pi/8) = \sin(\pi/4 - \pi/8) = \sin(\pi/4)\cos(\pi/8) - \cos(\pi/4)\sin(\pi/8) = \frac{\sqrt{2}}{2}(\cos(\pi/8) - \sin(\pi/8)).\end{align}
Now set $t = \sin(\pi/8)$, we thus get an equation:
$$\sqrt{2}t = \sqrt{1 - t^2} - t.$$
Then solve this for $t$.

Details on solving the equation: Squaring on both sides:
\begin{align}
& (\sqrt{2} + 1)^2 t^2 = 1 - t^2 \\
& (3 + 2\sqrt{2} + 1)t^2 = 1 \\
& t^2 = \frac{1}{4 + 2\sqrt{2}} \\
& t = \frac{1}{\sqrt{4 + 2\sqrt{2}}}
\end{align}
Notice that the negative square root is discarded, since $\sin(\pi/8) > 0$.
A: For $\frac{\pi}{8}$ you need to use double angle formula.
Recall that $\sin[2\theta)=2\sin\theta\cos\theta=2\sin\theta\sqrt{1-\sin^2\theta}$
Then let $\theta=\frac{\pi}{8}$:
$$\sin\left(\frac{\pi}{4}\right)=2\sin\frac{\pi}{8}\sqrt{1-\sin^2\frac{\pi}{8}}$$
$$\frac{1}{2}=4\sin^2\frac{\pi}{8}\left(1-\sin^2\frac{\pi}{8}\right)$$
$$\frac{1}{8}=4\sin^2\frac{\pi}{8}-\sin^4\frac{\pi}{8}$$
Solving as a quadratic in $\sin^2\frac{\pi}{8}$ gives:
$$\sin^2\frac{\pi}{8}=\frac{1}{4}\left(2-\sqrt{2}\right)$$
(we can ignore the other solution to the quadratic as we know $\sin\frac{\pi}{8}<\sin\frac{\pi}{3}=\frac{1}{2}$)
$$\sin\frac{\pi}{8}=\frac{1}{2}\sqrt{2-\sqrt{2}}$$
(we can ignore the negative solution as we know $\sin\frac{\pi}{8}>0$)
Note: this is just one way of doing it. You could have started from any of the multiple angle formula.
