## anonymous 4 years ago prove that $\cos^4θ−\sin^4θ=\cos^2θ−\sin^2θ$

1. anonymous

Think about the following two identities: $a^2 - b^2 = (a+b) (a-b)$ and $\cos^2\theta + \sin^2\theta = 1$

2. anonymous

Difference of squares. So, you have: (cos^2(theta) - sin^2(theta))(cos^2(theta) + sin^2(theta)) Since that latter factor is 1 by trig identity, you got the first factor as the result.

3. anonymous

Um Ok, and how would you show that?

4. anonymous

How would you show sin^ + cos^ = 1?

5. anonymous

It's just a rule, isn;t it? I mean, how would you show the workings? Not sure I completely get this topic yet.

6. anonymous

Yes. It is a rule. But, the proof has to do with the right triangle and pythagorean theorem.

7. anonymous

Yes, that's true. But this question didn't want me to answer with the triangle shown... so In algebraic terms, w/o the diagram, how would the steps be shown?

8. anonymous

It is as I showed: (cos^2(theta) - sin^2(theta))(cos^2(theta) + sin^2(theta)) It is like using (a^2-b^2) = (a-b)(a+b) except that here a and b are "sin^2" and "cos^2".

9. anonymous

What GT said is a good way to go, substitute Sin and Cos with 'a' and 'b' to show the working. If you want to prove the rule, here's how Take a right triangle. Let one of the acute angles be A. Let the base (smaller side of the angle) be 'b' Let the perpendicular (side opposite to the angle) be 'p' and Let the hypotenuse (longest side of the triangle) be 'h' Sin A = p/h Cos A = b/h $\sin^2A = \frac{p^2}{h^2}$ $\cos ^2A = \frac{b^2}{h^2}$ $\sin^2A + \cos^2A= \frac{p^2}{h^2} + \frac{b^2}{h^2}$ $\sin^2A + \cos^2A= \frac{p^2+b^2}{h^2}$ now p^2 + b^2 = h^2 according to pythagoras theorem

10. anonymous

$Cos^2\theta-\sin^2\theta(\cos^2\theta+\sin^\theta)?$or $(Cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^\theta) ?$

11. anonymous

latter

12. anonymous

So how (when expanding the bracket) does it simplify to cos^2(theta)-sin^2(theta)?

13. anonymous

Never mind. stupid question

14. anonymous

Thanks for helping! I understand now

15. anonymous

16. anonymous

Good job!

17. anonymous

I just realized! I would not have thought of it though... need to think more outside the box!

18. anonymous

Sorry again, but again, It wasn't a stupid question... How does it simplify to that answer?

19. anonymous

That is good. Discussing stimulates thought. Also, sometimes not doubting your own good capability helps too.

20. anonymous

I understand the (a-b)(a+b) rule, but how does that simplify it to the second answer... wouldn't it just come back to the first answer?

21. anonymous

So you realize that $\cos^4\theta - \sin^2\theta = (\cos^2\theta + \sin^2\theta)(\cos^2\theta-\sin^2\theta)$ But since $\cos^2\theta + \sin^2\theta = 1$ we get $\cos^4\theta - \sin^2\theta = (1)(\cos^2\theta-\sin^2\theta)$

22. anonymous

er typo... i meant sin^4 theta in the left side

23. anonymous

Not completely no. I realize the first and second identity, but I'm not sure how it links up!? to the last equation?

24. anonymous

You take the first equation, and put the value of cos^2 + sin^2 from the second equation in it.

25. anonymous

Ohhhhh!

26. anonymous

Thanks. Registration has finally hit!

27. anonymous

That's great! I'm glad I was able to help