## ChristopherW28 2 years ago Verify that (1-cos^2x)(1+cos^2x)=2sin^2x-sin^4x is a trig identity

1. myininaya

By a certain Pythagorean identity we can say 1-cos^2(x)=?

2. ChristopherW28

Sin^2x?

3. myininaya

yep! :)

4. myininaya

so this means we now have $\sin^2(x)(1+\cos^2(x))=2\sin^2(x)-\sin^4(x)$ now this is an identity we are trying to prove now we have one side in terms of sin and another side in terms of mixture of sin and cos

5. myininaya

We could use that same identity to rewrite cos^2(x)

6. myininaya

$1-\cos^2(x)=\sin^2(x) => \cos^2(x)=$

7. myininaya

that is a blank for you to fill in

8. ChristopherW28

I'm lost

9. myininaya

Ok I want to rewrite cos^2(x) because the other side is just in terms of sin I know an identity that has cos^2(x) and sin^2(x) in it.

10. myininaya

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

11. myininaya

So cos^2(x)=?

12. ChristopherW28

Sin^x+ 1

13. myininaya

1-sin^2(x)

14. ChristopherW28

Sin^2x+1?

15. ChristopherW28

I was close

16. myininaya

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

17. myininaya

So we can back to what we are trying to prove and replace the cos^2(x) with 1-sin^2(x) $\sin^2(x)(1+(1-\sin^2(x))=2\sin^2(x)-\sin^4(x)$

18. myininaya

I replaced mr.cos^2(x) with mrs. (1-sin^2(x))

19. myininaya

now we don't need those parenthesis since there is a + in front of that parenthesis lets drop that extra stuff giving us: $\sin^2(x)(1+1-\sin^2(x))=2\sin^2(x)-\sin^4(x)$

20. myininaya

now 1+1=2 which I know you know (:p) so now you distribute on that left hand side

21. ChristopherW28

I gtg I can figure out what's left. Friend me on Facebook @ Christopher McElhannon. I will need your help again :)

22. myininaya

I will be here most likely. I'm not much of a facebook user. I'm on facebook free diet right now. :p