exact value check

**jvignacio** · March 14th 2009, 04:25 AM

find the exact value of $cos(\frac{\pi}{8})$

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

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

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

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

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

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

therefore: we take the positive one because $\frac{\pi}{8}$ is in the first quadrant and its positive.

this correct ? much appreciated..

**running-gag** · March 14th 2009, 04:41 AM

Originally Posted by jvignacio

find the exact value of $cos(\frac{\pi}{8})$

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

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

Could you explain the way you go from the first identity to the second one ?

$\cos (2x) = 1-2\sin^2(x)$

$\sin (x) = \pm\sqrt{\frac{1-\cos (2x)}{2}}$

Or $\cos (2x) = 2\cos^2(x)-1$

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

**jvignacio** · March 14th 2009, 04:49 AM

Originally Posted by running-gag

Could you explain the way you go from the first identity to the second one ?

$\cos (2x) = 1-2\sin^2(x)$

$\sin (x) = \pm\sqrt{\frac{1-\cos (2x)}{2}}$

Or $\cos (2x) = 2\cos^2(x)-1$

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

well its $sin^2x+cos^2x = 1$ (fundamental identity)

then

$ sin^2x = 1-cos^2x $
$ sinx = \sqrt{1-cos^2x} $

and in this case x = $\frac{\pi}{4}$

i split up the $\frac{\pi}{8}$ to $\frac{\pi}{4}$ to work with 45 degrees

**jvignacio** · March 14th 2009, 04:58 AM

Originally Posted by running-gag

Could you explain the way you go from the first identity to the second one ?

$\cos (2x) = 1-2\sin^2(x)$

$\sin (x) = \pm\sqrt{\frac{1-\cos (2x)}{2}}$

Or $\cos (2x) = 2\cos^2(x)-1$

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

sorry yes..

$cos2x = 1-2sin^2(x)$

$2sin^2(x)=1-cos(2x)$

$sin^2(x)=\frac{1-cos(2x)}{2}$
$ sin(x) = \pm\sqrt{\frac{1-cos(2x)}{2}} $

**running-gag** · March 14th 2009, 05:45 AM

You should rather take this one

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

with $x=\frac{\pi}{8}$

**jvignacio** · March 14th 2009, 05:48 AM

Originally Posted by running-gag

You should rather take this one

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

with $x=\frac{\pi}{8}$

ahh darn how would i know which one to use ?

**running-gag** · March 14th 2009, 05:51 AM

Which one among what ?

This one

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

is useful because you know the value of $\cos \left(\frac{\pi}{4}\right)$

**jvignacio** · March 14th 2009, 06:00 AM

Originally Posted by running-gag

Which one among what ?

This one

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

is useful because you know the value of $\cos \left(\frac{\pi}{4}\right)$

yeh sorry i was just getting confused but now i understand. thanks!!

**jvignacio** · March 14th 2009, 07:55 PM

Originally Posted by running-gag

Which one among what ?

This one

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

is useful because you know the value of $\cos \left(\frac{\pi}{4}\right)$

so should i do $\cos (x) = \pm\sqrt{\frac{1+\cos (\frac{2\pi}{4})}{2}}$

?

**running-gag** · March 15th 2009, 04:25 AM

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

$\cos \left(\frac{\pi}{8}\right) = \pm\sqrt{\frac{1+\cos \left(\frac{\pi}{4}\right)}{2}}$

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