# exact value check

• Mar 14th 2009, 04:25 AM
jvignacio
exact value check
find the exact value of $\displaystyle cos(\frac{\pi}{8})$

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

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

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

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

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

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

therefore: we take the positive one because $\displaystyle \frac{\pi}{8}$ 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 $\displaystyle cos(\frac{\pi}{8})$

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

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

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

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

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

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

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

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

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

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

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

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

then

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

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

i split up the $\displaystyle \frac{\pi}{8}$ to $\displaystyle \frac{\pi}{4}$ 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 ?

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

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

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

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

sorry yes..

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

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

$\displaystyle sin^2(x)=\frac{1-cos(2x)}{2}$
$\displaystyle sin(x) = \pm\sqrt{\frac{1-cos(2x)}{2}}$
• Mar 14th 2009, 05:45 AM
running-gag
You should rather take this one

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

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

Originally Posted by running-gag
You should rather take this one

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

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

ahh darn how would i know which one to use ?
• Mar 14th 2009, 05:51 AM
running-gag
Which one among what ?

This one

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

is useful because you know the value of $\displaystyle \cos \left(\frac{\pi}{4}\right)$
• Mar 14th 2009, 06:00 AM
jvignacio
Quote:

Originally Posted by running-gag
Which one among what ?

This one

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

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

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

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

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

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

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

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