>>7664390
This is just your homework.

>>7664390
cos(x) = cos^2(x/2)-sin^2(x/2)
therefore cos^2(x/2)=cos(x)+sin^2(x/2)

this version is better because the equality is also valid when x=pi

>>7664390

do your own homework ass.

>trig
>fun

>>7664673
It is fun when you use it in Statics

>>7664697
>statics
>fun

>>7664697

That is a rare Pepe of the most exquisite quality.

>>7664390
Kind of off topic, but I don't know where else to figure out the answer. I have this crazy ass Indian professor for Calc III and he always gives us super tedious problems to fuck with us, even on exams. A lot of the shit isn't even calculus material, its just really complex algebra type stuff with several variables. Sorry normally I wouldn't post homework, but I'm completely lost:

"Evaluate the following function for x = 3^(1/2)/(2^(1/3)), y = (3^(1/2)). Your answer should not be in decimal form.

f(x, y) = [-(x-y)^2(2x^2 + 2xy + 2y^2 -3)^2] - [4e^(2y^2 - 2x^2) * (x^3 -2(x^2)(y^2) + y^5 -7(y^3) +3y)^2]
"

Yes this is an actual problem

f(

>>7664747
sorry I mean
f(x, y) = [-(x-y)^2 * (2x^2 + 2xy + 2y^2 -3)^2] - [4e^(2y^2 - 2x^2) * (x^3 -2(x^2)(y^2) + y^5 -7(y^3) +3y)^2]

>>7664750
answer is 4

>>7664757
How the fuck did you find that? I think its negative cause with my calculator I got ~ -0.4463

Here