## anonymous one year ago Trig / Pre Cal / identities These things are killing me. Please help guide me. Thank you. $(\sin^4 x-\cos^4 x) = 2\sin^2 x -1$ $(\sin^2 x+\cos^2 x)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1$ Am I on the right path?

1. Nnesha

apply special identity $\huge\rm sin^2 x + \cos^2 =1$ solve for sin^2

2. phi

yes that is a good start

3. anonymous

so we have $(1)(\sin^2 x-\cos^2 x)= 2\sin^2 x -1$

4. phi

I would next replace the cos^2

5. Nnesha

sin^2 x + cos ^2 = 1 solve this identity for cos^2

6. anonymous

$(1)(\sin^2 x + 1-\sin^2 x )= 2\sin^2 x -1$

7. Nnesha

$\rm (\sin^2 x-(\color{reD}{{1-sin^2x}}))$ write that in the parentheses and distribute by negative one

8. phi

close, but you lost a minus sign -(1 - sin^2)

9. anonymous

$(1)(\sin^2 x - (1-\sin^2 x))= 2\sin^2 x -1$ Now what?

10. phi

-(a +b) means -1*(a+b) = -1*a + -1*b

11. anonymous

Ah so $(1)(\sin^2 x - 1+\sin^2 x )= 2\sin^2 x -1$ I see it now.

12. anonymous

Thank you both!!!

13. Nnesha

:-)