Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$

any help much appreciated

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- Mar 10th 2009, 06:23 PMjvignaciofinding the exact value
Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$

any help much appreciated - Mar 10th 2009, 07:12 PMChris L T521
Note that $\displaystyle \sin\left(\frac{\pi}{8}\right)=\sin\left(\frac{1}{ 2}\left(\frac{\pi}{4}\right)\right)=\pm\sqrt{\frac {1-\cos\left(\frac{\pi}{4}\right)}{2}}=\pm\sqrt{\frac {1-\frac{\sqrt{2}}{2}}{2}}=\pm\frac{\sqrt{2-\sqrt{2}}}{2}$.

Does this make sense? - Mar 10th 2009, 07:22 PMjvignacio
- Mar 10th 2009, 07:25 PMChris L T521
- Mar 10th 2009, 07:33 PMsinewave85
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! - Mar 10th 2009, 07:35 PMsinewave85
- Mar 10th 2009, 09:11 PMmr fantastic
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?).