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

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Think about the following two identities: \[a^2  b^2 = (a+b) (ab)\] and \[\cos^2\theta + \sin^2\theta = 1\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Difference 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Um Ok, and how would you show that?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0How would you show sin^ + cos^ = 1?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It's just a rule, isn;t it? I mean, how would you show the workings? Not sure I completely get this topic yet.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes. It is a rule. But, the proof has to do with the right triangle and pythagorean theorem.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, 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?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0It is as I showed: (cos^2(theta)  sin^2(theta))(cos^2(theta) + sin^2(theta)) It is like using (a^2b^2) = (ab)(a+b) except that here a and b are "sin^2" and "cos^2".

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[Cos^2\theta\sin^2\theta(\cos^2\theta+\sin^\theta)? \]or \[(Cos^2\theta\sin^2\theta)(\cos^2\theta+\sin^\theta) ?\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So how (when expanding the bracket) does it simplify to cos^2(theta)sin^2(theta)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Never mind. stupid question

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks for helping! I understand now

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Haha, there are no stupid questions, please feel free to ask.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I just realized! I would not have thought of it though... need to think more outside the box!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Sorry again, but again, It wasn't a stupid question... How does it simplify to that answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That is good. Discussing stimulates thought. Also, sometimes not doubting your own good capability helps too.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0I understand the (ab)(a+b) rule, but how does that simplify it to the second answer... wouldn't it just come back to the first answer?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So 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 \[\cos^4\theta  \sin^2\theta = (1)(\cos^2\theta\sin^2\theta)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0er typo... i meant sin^4 theta in the left side

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Not completely no. I realize the first and second identity, but I'm not sure how it links up!? to the last equation?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0You take the first equation, and put the value of cos^2 + sin^2 from the second equation in it.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Thanks. Registration has finally hit!

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0That's great! I'm glad I was able to help
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