# Thread: finding the exact value

1. ## finding the exact value

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

2. Originally Posted by jvignacio
Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
any help much appreciated
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?

3. Originally Posted by Chris 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?
okay you lost me on the 3rd step..

4. Originally Posted by jvignacio
okay you lost me on the 3rd step..
From the second step, you apply the half angle formula $\displaystyle \sin\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos\alpha}{2}}$ where in our case, $\displaystyle \alpha=\frac{\pi}{4}$.

Does this clarify things?

5. Originally Posted by jvignacio
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!

6. Originally Posted by Chris L T521
From the second step, you apply the half angle formula $\displaystyle \sin\left(\frac{\alpha}{2}\right)=\pm\sqrt{\frac{1-\cos\alpha}{2}}$ where in our case, $\displaystyle \alpha=\frac{\pi}{4}$.

Does this clarify things?
You got two posts typed out while I unawares was wading through the jungle of Latex on one. Ah, redundancies.

7. Originally Posted by jvignacio
Hi guys, trying to find the exact value of $\displaystyle sin\frac{\pi}{8}$
any help much appreciated
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?).