# finding the exact value

• Mar 10th 2009, 07:23 PM
jvignacio
finding the exact value
Hi guys, trying to find the exact value of $sin\frac{\pi}{8}$
any help much appreciated
• Mar 10th 2009, 08:12 PM
Chris L T521
Quote:

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

Note that $\sin\left(\frac{\pi}{8}\right)=\sin\left(\frac{1}{ 2}\left(\frac{\pi}{4}\right)\right)=\pm\sqrt{\frac {1-\cos\left(\frac{\pi}{4}\right)}{2}}=\pm\sqrt{\frac {1-\frac{\sqrt{2}}{2}}{2}}=\pm\frac{\sqrt{2-\sqrt{2}}}{2}$.

Does this make sense?
• Mar 10th 2009, 08:22 PM
jvignacio
Quote:

Originally Posted by Chris L T521
Note that $\sin\left(\frac{\pi}{8}\right)=\sin\left(\frac{1}{ 2}\left(\frac{\pi}{4}\right)\right)=\pm\sqrt{\frac {1-\cos\left(\frac{\pi}{4}\right)}{2}}=\pm\sqrt{\frac {1-\frac{\sqrt{2}}{2}}{2}}=\pm\frac{\sqrt{2-\sqrt{2}}}{2}$.

Does this make sense?

okay you lost me on the 3rd step..
• Mar 10th 2009, 08:25 PM
Chris L T521
Quote:

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

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

Does this clarify things?
• Mar 10th 2009, 08:33 PM
sinewave85
Quote:

Originally Posted by jvignacio
Hi guys, trying to find the exact value of $sin\frac{\pi}{8}$
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).

$\sin\frac{\pi}{8} = \sin\left({\frac{1}{2}*\frac{\pi}{4}}\right)$

$\sin\frac{\pi}{8} = \pm\sqrt{\frac{1 - \cos\frac{\pi}{4}}{2}}$

$\sin\frac{\pi}{8} = \pm\sqrt{\frac{1 -\frac{\sqrt{2}}{2}}{2}}$

$\sin\frac{\pi}{8} = \pm\sqrt{\frac{\frac{2 - \sqrt{2}}{2}}{2}}$

$\sin\frac{\pi}{8} = \pm\sqrt{\frac{2 - \sqrt{2}}{4}}$

$\sin\frac{\pi}{8} = \pm\frac{\sqrt{2 - \sqrt{2}}}{2}$

and since pi/8 is in Quadrant I, you need the positive value. Hope that helps!
• Mar 10th 2009, 08:35 PM
sinewave85
Quote:

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

Does this clarify things?

You got two posts typed out while I unawares was wading through the jungle of Latex on one. Ah, redundancies.
• Mar 10th 2009, 10:11 PM
mr fantastic
Quote:

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

A slightly alternatively approach is to note that $\cos (2A) = 1 - 2 \sin^2 (A)$.

Let $A = \frac{\pi}{8}$: $\cos \frac{\pi}{4} = 1 - 2 \sin^2 \frac{\pi}{8} \Rightarrow \frac{\sqrt{2}}{2} = 1 - 2 \sin^2 \frac{\pi}{8}$.

Re-arrange and solve for $\sin \frac{\pi}{8}$ (the negative root solution is rejected - why?).