# Thread: Trig identity for sqrt(2)

1. ## 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 ?

2. 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 ?

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

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

giving

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

3. 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 !

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

since ...

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

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

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

4. Thanks very much to both of you!!