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?

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.
