Prove that `sin^4 x - cos^4 x = 2*sin^2x - 1`

2 Answers | Add Yours

lfryerda's profile pic

lfryerda | High School Teacher | (Level 2) Educator

Posted on

Another way of solving identities with fourth powers of `sinx` and `cos x`  is to use the identities `sin^2x = 1-cos^2 x` and `cos^2 x = 1-sin^2 x`.

You usually want to start a trig identity from the side that looks more complicated, in this case the LHS.

`LS=sin^4 x-cos^4 x`

`=sin^2 x( 1-cos^2 x) - cos^2 x(1-sin^2 x)`

`=sin^2x - sin^2x cos^2 x -cos^2 x + sin^2xcos^2x`   some terms cancel out!

`=sin^2 x - cos^2x`    now get rid of the cosine

`=sin^2x - (1-sin^2x)`

`=sin^2x-1+sin^2x`    collect like terms

`=2sin^2x - 1`

`=RS` 

justaguide's profile pic

justaguide | College Teacher | (Level 2) Distinguished Educator

Posted on

The identity `sin^4 x - cos^4 x = 2*sin^2x - 1` has to be proved

`sin^4x - cos^4x`

=> `(sin^2x - cos^2x)(sin^2x + cos^2x)`

=> `(sin^2x - (1 - sin^2x))*1`

=> `2*sin^2x - 1`

This proves that `sin^4 x - cos^4 x = 2*sin^2x - 1`

We’ve answered 318,915 questions. We can answer yours, too.

Ask a question