# exact value check

• Mar 14th 2009, 04:25 AM
jvignacio
exact value check
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..
• Mar 14th 2009, 04:41 AM
running-gag
Quote:

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

$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$
• Mar 14th 2009, 04:49 AM
jvignacio
Quote:

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
• Mar 14th 2009, 04:58 AM
jvignacio
Quote:

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}}
$
• Mar 14th 2009, 05:45 AM
running-gag
You should rather take this one

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

with $x=\frac{\pi}{8}$
• Mar 14th 2009, 05:48 AM
jvignacio
Quote:

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 ?
• Mar 14th 2009, 05:51 AM
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 $\cos \left(\frac{\pi}{4}\right)$
• Mar 14th 2009, 06:00 AM
jvignacio
Quote:

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!!
• Mar 14th 2009, 07:55 PM
jvignacio
Quote:

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

?
• Mar 15th 2009, 04:25 AM
running-gag
$\cos (x) = \pm\sqrt{\frac{1+\cos (2x)}{2}}$

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