anonymous
  • anonymous
prove that \[\cos^4θ−\sin^4θ=\cos^2θ−\sin^2θ\]
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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darthsid
  • darthsid
Think about the following two identities: \[a^2 - b^2 = (a+b) (a-b)\] and \[\cos^2\theta + \sin^2\theta = 1\]
anonymous
  • 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.
anonymous
  • anonymous
Um Ok, and how would you show that?

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anonymous
  • anonymous
How would you show sin^ + cos^ = 1?
anonymous
  • 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.
anonymous
  • anonymous
Yes. It is a rule. But, the proof has to do with the right triangle and pythagorean theorem.
anonymous
  • 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?
anonymous
  • 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".
darthsid
  • darthsid
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
anonymous
  • anonymous
\[Cos^2\theta-\sin^2\theta(\cos^2\theta+\sin^\theta)? \]or \[(Cos^2\theta-\sin^2\theta)(\cos^2\theta+\sin^\theta) ?\]
anonymous
  • anonymous
latter
anonymous
  • anonymous
So how (when expanding the bracket) does it simplify to cos^2(theta)-sin^2(theta)?
anonymous
  • anonymous
Never mind. stupid question
anonymous
  • anonymous
Thanks for helping! I understand now
darthsid
  • darthsid
Haha, there are no stupid questions, please feel free to ask.
anonymous
  • anonymous
Good job!
anonymous
  • anonymous
I just realized! I would not have thought of it though... need to think more outside the box!
anonymous
  • anonymous
Sorry again, but again, It wasn't a stupid question... How does it simplify to that answer?
anonymous
  • anonymous
That is good. Discussing stimulates thought. Also, sometimes not doubting your own good capability helps too.
anonymous
  • 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?
darthsid
  • darthsid
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 \[\cos^4\theta - \sin^2\theta = (1)(\cos^2\theta-\sin^2\theta)\]
darthsid
  • darthsid
er typo... i meant sin^4 theta in the left side
anonymous
  • anonymous
Not completely no. I realize the first and second identity, but I'm not sure how it links up!? to the last equation?
darthsid
  • darthsid
You take the first equation, and put the value of cos^2 + sin^2 from the second equation in it.
anonymous
  • anonymous
Ohhhhh!
anonymous
  • anonymous
Thanks. Registration has finally hit!
darthsid
  • darthsid
That's great! I'm glad I was able to help

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