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Explain why -sin[SUP]2[/SUP]x - cos[SUP]2[SUP]x = -1

  • Thread starter tjohn101
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  • #1
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Homework Statement


I should know this, but I'm using it as a step in one of my proofs and having trouble explaining how it works. Can someone help?

-sin2x - cos2x = -1

Homework Equations





The Attempt at a Solution


Not sure how to prove and explain it.

And sorry about the title gents I'm using a blackberry so I just copied and pasted the equation from my question : /
 

Answers and Replies

  • #2
HallsofIvy
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Derived from what? If you know that [math]sin^2(x)+ cos^2(x)= 1[/math], then just multiply both sides of that equation by -1.

If you cannot use that, how you prove it depends upon the precise definition of sine and cosine. What definitions are you using?
 
  • #3
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Not sure. I'm proving d/dx [cotx] = -csc2x and I have the above equation in there as a step towards my proof. I've proven it, but my professor wants me to explain why the above equation = -1.
 
  • #4
ideasrule
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It's crucial to know what you're allowed to assume. If you can assume sin^2x + cos^2x = 1, then just multiply both sides by -1 and you get that equation. If you can't assume that, you can draw out a triangle and prove that sin^2x + cos^2x = 1 from the definitions of sine and cosine.
 
  • #5
it's the pythagorean theorem. Assume a right triangle formed by a vector and the x,y axes with sides A, B, C. Assume you're looking at angle "a": sin(a)=A/C cos(a)=B/C. Plug those in and multiply both sides by C^2 to get A^2+B^2=C^2
 

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