just for fun

[jade smelt]

1. Wait patiently for a helper to come along.
2. Once someone helps you, say thank you and close the thread with:

+close

3. Feel free to nominate the person for helper of the week in #helper-nominations
4. Do not ping the mods, unless someone is breaking the rules.
5. If you're happy with the help you got here, and the server overall, you can contribute financially as well:

[vapid edge]

2cos^2(y)-1
= cos^2(y) -(1-cos^2(y))
= cos^2y - sin^2y
= cosy cosy - sin y sin y
= cos(y+y)
=cos(2y)

you can do the rest

[spark temple]

i appreciate your willingness to help, but instead of solving the problem for OP, it's better to give hints/guide and let him/her solve it on his/her own, and come back when he/she is stuck

[vapid edge]

I just showed OP the trigonometric identity that's related to the problem

[spark temple]

oh yeah u r right my bad

,tex you can use \verb+,align+ to render math

[zealous shale]

vin100

[spark temple]

,align
&2\cos^2(y)-1\
&= \cos^2(y) -(1-\cos^2(y))\
&= \cos^2y - \sin^2y\
&= \cos y \cos y - \sin y \sin y\
&= \cos(y+y)\
&=\cos(2y)

[zealous shale]

vin100

[vapid edge]

I know, I didn't wanna go through the trouble of using latex on something as small

[spark temple]

or you may use '²' instead of "^2"
2cos²(y)-1
= cos²(y) -(1-cos²(y))
= cos²y - sin²y
= cosy cosy - sin y sin y
= cos(y+y)
=cos(2y)

it's Alt+0178

on M$ Win*