Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
any help much appreciated
Hi, jvignacio. To find the exact value of sin(pi/8), you need to know two things: The exact value of cos(pi/4) and the half-angle identity for sin(x).
$\displaystyle \sin\frac{\pi}{8} = \sin\left({\frac{1}{2}*\frac{\pi}{4}}\right)$
$\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{1 - \cos\frac{\pi}{4}}{2}}$
$\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{1 -\frac{\sqrt{2}}{2}}{2}}$
$\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}$
$\displaystyle \sin\frac{\pi}{8} = \pm\sqrt{\frac{2 - \sqrt{2}}{4}}$
$\displaystyle \sin\frac{\pi}{8} = \pm\frac{\sqrt{2 - \sqrt{2}}}{2}$
and since pi/8 is in Quadrant I, you need the positive value. Hope that helps!
A slightly alternatively approach is to note that $\displaystyle \cos (2A) = 1 - 2 \sin^2 (A)$.
Let $\displaystyle A = \frac{\pi}{8}$: $\displaystyle \cos \frac{\pi}{4} = 1 - 2 \sin^2 \frac{\pi}{8} \Rightarrow \frac{\sqrt{2}}{2} = 1 - 2 \sin^2 \frac{\pi}{8}$.
Re-arrange and solve for $\displaystyle \sin \frac{\pi}{8}$ (the negative root solution is rejected - why?).