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anonymous

  • 5 years ago

prove that \[\cos^4θ−\sin^4θ=\cos^2θ−\sin^2θ\]

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  1. darthsid
    • 5 years ago
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    Think about the following two identities: \[a^2 - b^2 = (a+b) (a-b)\] and \[\cos^2\theta + \sin^2\theta = 1\]

  2. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Um Ok, and how would you show that?

  4. anonymous
    • 5 years ago
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    How would you show sin^ + cos^ = 1?

  5. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    Yes. It is a rule. But, the proof has to do with the right triangle and pythagorean theorem.

  7. anonymous
    • 5 years ago
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    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
    • 5 years ago
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    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. darthsid
    • 5 years ago
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    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
    • 5 years ago
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    \[Cos^2\theta-\sin^2\theta(\cos^2\theta+\sin^\theta)? \]or \[(Cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^\theta) ?\]

  11. anonymous
    • 5 years ago
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    latter

  12. anonymous
    • 5 years ago
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    So how (when expanding the bracket) does it simplify to cos^2(theta)-sin^2(theta)?

  13. anonymous
    • 5 years ago
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    Never mind. stupid question

  14. anonymous
    • 5 years ago
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    Thanks for helping! I understand now

  15. darthsid
    • 5 years ago
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    Haha, there are no stupid questions, please feel free to ask.

  16. anonymous
    • 5 years ago
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    Good job!

  17. anonymous
    • 5 years ago
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    I just realized! I would not have thought of it though... need to think more outside the box!

  18. anonymous
    • 5 years ago
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    Sorry again, but again, It wasn't a stupid question... How does it simplify to that answer?

  19. anonymous
    • 5 years ago
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    That is good. Discussing stimulates thought. Also, sometimes not doubting your own good capability helps too.

  20. anonymous
    • 5 years ago
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    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. darthsid
    • 5 years ago
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    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. darthsid
    • 5 years ago
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    er typo... i meant sin^4 theta in the left side

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

  24. darthsid
    • 5 years ago
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    You take the first equation, and put the value of cos^2 + sin^2 from the second equation in it.

  25. anonymous
    • 5 years ago
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    Ohhhhh!

  26. anonymous
    • 5 years ago
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    Thanks. Registration has finally hit!

  27. darthsid
    • 5 years ago
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    That's great! I'm glad I was able to help

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