Calc- Trig Identity Help! I have a few questions on trig/calc stuff I am having trouble with, for some reason I am just not getting the concept. 1: What happens if you take $B=2\pi$ in the addition formula? Do the results agree with something you already know? so... $\sin A \cos B + \cos A \sin B = \sin A \cos 2\pi + \cos A \sin 2\pi$, without $A$, we cannot get further? 2: Find the function values. $\displaystyle \sin^2(\frac{3\pi}{8})$ Thank you, I would really appreciate it if someone could help me figure these out, and learn the concept! Thanks

1. You can get further without $A$, since you know that $\cos2\pi = 1$ and $\sin2\pi = 0$. (These are really fundamental properties of $\cos$ and $\sin$.) Therefore you have $\sin(A+2\pi) = \sin(A)$, as expected.

2. We know that $\sin^2(\frac{3\pi}8) = (\sin(\frac{3\pi}8))^2$. We can simply plug in $\sin(\frac{3\pi}8)$ into a calculator and square it. If we want to find that out by hand, we can use an identity and find $\sin(\frac{3\pi}8) = \pm \sqrt\frac{1-\cos(\frac{3\pi}4)}2$. We know $\cos(\frac{3\pi}4)=-\frac{\sqrt2}2$. Anyway, plug in everything and you end up with $\sin^2(\frac{3\pi}8)=\frac{2+\sqrt2}4$.