SOLUTION: Sin^4x = 1÷8 (3-4cos2x+cos4x)

Algebra ->  Trigonometry-basics -> SOLUTION: Sin^4x = 1÷8 (3-4cos2x+cos4x)      Log On


   

Question 1045298: Sin^4x = 1÷8 (3-4cos2x+cos4x)
Found 2 solutions by robertb, advanced_Learner:
Answer by robertb(5830) About Me  (Show Source):
You can put this solution on YOUR website!
sin%5E4%28x%29+=+%28sin%5E2%28x%29%29%5E2
= %281-cos%5E2%28x%29%29%5E2+=+1+-+2cos%5E2%28x%29+%2B+%28cos%5E2%28x%29%29%5E2
= 1+-+%281%2Bcos2x%29+%2B+%28%281%2Bcos2x%29%2F2%29%5E2
= -cos2x+%2B+%281%2B2cos2x+%2Bcos%5E2%282x%29%29%2F4
= %281%2F4%29%281-2cos2x%2Bcos%5E2%282x%29%29
= %281%2F4%29%281-2cos2x%2B+%28cos4x%2B1%29%2F2%29
= %281%2F4%29%281%2F2%29%282-4cos2x+%2B+cos4x+%2B1%29
= %281%2F8%29%283+-+4cos2x+%2B+cos4x%29.

Answer by advanced_Learner(501) About Me  (Show Source):