Hi everyone, I am looking for some help on math. I looked the solution up in the back of the text book, but I am puzzled on how they are getting that answer. They showed their work but I am just curious on if I am interpreting it right and stuff. Here is the first problem: sin^4(x) = 1/8(3 - 4cos2x + cos4x) = (sin²x)² *I understand this = [1/2(1-cos2x)]² *I believe they substituted 1-cos²x for sin²x...and then to get rid of the ² on the cos, they took the square root of the equation, so they put times 1/2 instead. LMK if that is the correct way of thinking. = 1/4(1 - 2cos²x + cos²2x) *I think they foiled out (1-cos2x) (shown down below) FOILED: (1-cos2x) (1-cos2x) = 1 - cos2x - cos2x + cos²2x = 1 - 2cos2x + cos²2x = 1/4[1 - 2cos2x + 1/2(1 + cos4x)] *I don't know what happened here where the bold face is at. = 1/8(2 - 4cos2x + 1 + cos4x) *IDK = 1/8(3 - 4cos2x + cos 4x) *which is what we were trying to prove ----------------------------------------------------------------------------------------- Second Problem: FINISHED TY TO ARMondak! Use half-angle identities to find all solutions in the interval [0, 2π) cos²x = sin²(x/2) answer: π/3, π, 5π/3 *π = pi symbol just to lyk* I do not understand how to do this one, if someone can explain it would be mostly appreciated. Thank you for reading.

Take the right side and substitute ((1-cosx)/2)^1/2 for sin^2 x/2, keeping the square. The square and square root cancel out, leaving you with: cos^2 x = (1 - cos x)/2 Then multiply both sides by 2 2 cos^2 x = 1 - cos x Bring everything to the left 2 cos^2 x + cos x - 1 = 0 Factor (2 cos x - 1) (cos x + 1) Leaving cos x = 1/2, -1 Which will give you x = 60 (n/3) and 240 (5n/3) when cos x = 1/2 and 180 (n) when cos x = -1 Hope this helps.

I am so sorry but the first line lost me : / I understand what you do from there but I am a little confused on the first step and how you did it.

Google and Wikipedia are your best friends. http://en.wikipedia.org/wiki/Proofs_of_trigonometric_identities Double Angle Identities: cos(2x) = 1-2sin^2(x) <-- move that around a little bit for that one decomposition. cos(2x) = 2cos^2(x)-1 <-- double the X term, and you've got the 2nd to last part.