# Trig identity for sqrt(2)

• Feb 3rd 2011, 09:36 AM
skippets
Trig identity for sqrt(2)
Hi

Upon checking an intergral (that I had worked out by hand) using mathcad(some math software I have) I noticed a simplification of part of my intergral the computer had performed and couldnt work out how it was made.

it is

-sec(6pi/8) = sqrt(2)

I was wondering if anyone knows how one gets from one to the other ?

I have even checked with my my tutor and he isnt sure either !

is there some identity I am not aware of ?
• Feb 3rd 2011, 09:45 AM
Quote:

Originally Posted by skippets
Hi

Upon checking an intergral (that I had worked out by hand) using mathcad(some math software I have) I noticed a simplification of part of my intergral the computer had performed and couldnt work out how it was made.

it is

-sec(6pi/8) = sqrt(2)

I was wondering if anyone knows how one gets from one to the other ?

I have even checked with my my tutor and he isnt sure either !

is there some identity I am not aware of ?

$\displaystyle\ -sec\left(\frac{6{\pi}}{8}\right)=-\frac{1}{cos\left(\frac{6{\pi}}{8}\right)}=\frac{1 }{cos\left(\frac{2{\pi}}{8}\right)}$

since $cos(\pi-A)=-cosA$

giving

$\displaystyle\frac{1}{cos\left(\frac{\pi}{4}\right )}=\frac{1}{\left(\frac{1}{\sqrt{2}}\right)}=\sqrt {2}$
• Feb 3rd 2011, 09:59 AM
skeeter
Quote:

Originally Posted by skippets
...

-sec(6pi/8) = sqrt(2)

I was wondering if anyone knows how one gets from one to the other ?

I have even checked with my my tutor and he isnt sure either !

$\dfrac{6\pi}{8} = \dfrac{3\pi}{4}$ ... a unit circle angle.

since ...

$\cos\left(\dfrac{3\pi}{4}\right) = -\dfrac{\sqrt{2}}{2}$

$\sec\left(\dfrac{3\pi}{4}\right) = -\dfrac{2}{\sqrt{2}} = -\sqrt{2}$

$-\sec\left(\dfrac{3\pi}{4}\right) = \sqrt{2}$

http://hernandiaz.org/wp-content/upl...unitcircle.gif
• Feb 3rd 2011, 10:24 AM
skippets
Thanks very much to both of you!!

(Clapping)(Clapping)