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anonymous

  • one year ago

does cos^2 - sin^2 = 2cos^2 - 1?

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  1. Valpey
    • one year ago
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    By the Pythagorean Theorum, \(\sin^2 + \cos^2 = 1\). Is there a way you can use that to simplify your equation?

  2. anonymous
    • one year ago
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    i was thinking about that but i wasn't sure and this is just the last step of my equation but yea

  3. Valpey
    • one year ago
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    One way to make it clear might be to say: Let \(X = \cos^2\) Let \(Y = \sin^2\) Now, your question would be: Does \(X - Y = 2X - 1\)? (given that we know \(X+Y = 1\))

  4. anonymous
    • one year ago
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    that sounds helpful

  5. anonymous
    • one year ago
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    \[\cos \left( A+B \right)=\cos A \cos B-\sin A \sin B\] If B=A \[\cos \left( A+A \right)=\cos A \cos A-\sin A \sin A=\cos ^2A-\sin ^2A\] \[\cos ^2A+\sin ^2A=1,\sin ^2A=1-\cos ^2A\] ?

  6. anonymous
    • one year ago
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    thank you for that but i dont think that helps my situation right now ;/

  7. anonymous
    • one year ago
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    \[\cos ^2A-\sin ^2A=\cos ^2A-(1-\cos ^2A)=\cos ^2A-1+\cos ^2A=2 \cos ^2 A-1\]

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