SOLUTION: Hi i have to solve this problem, thanks for help. Sin^4x - Cos^4x = Sin^2x - Cos^2x

Algebra ->  Trigonometry-basics -> SOLUTION: Hi i have to solve this problem, thanks for help. Sin^4x - Cos^4x = Sin^2x - Cos^2x      Log On


   

Question 1042158: Hi i have to solve this problem, thanks for help.
Sin^4x - Cos^4x = Sin^2x - Cos^2x
Answer by ikleyn(52879) About Me  (Show Source):
You can put this solution on YOUR website!
.
Hi i have to solve this problem, thanks for help.
Sin^4x - Cos^4x = Sin^2x - Cos^2x
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 

Are you familiar with this identity: a%5E2+-+b%5E2 = {a+b)*(a-b)?


If not, or if you are not sure, look into the lesson The difference of squares formula in this site.



OK. Now, apply this identity to the left side. You will get


Sin%5E4%28x%29+-+Cos%5E4%28x%29 = %28sin%5E2%28x%29+%2B+cos%5E2%28x%29%29%2A%28sin%5E2%28x%29+-+cos%5E2%28x%29%29.

But the first parentheses,  %28sin%5E2%28x%29+%2B+cos%5E2%28x%29%29 is equal to 1, as everybody knows.

Therefore, 

Sin%5E4%28x%29+-+Cos%5E4%28x%29 = sin%5E2%28x%29+-+cos%5E2%28x%29.

QED.