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anonymous

  • 4 years ago

(cos^4x−sin^4x)/(1−tan^4x)=cos^4x trig identity?

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  1. anonymous
    • 4 years ago
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    \[(\cos^4x−\sin^4x)/(1−\tan^4x)=\cos^4x\]

  2. anonymous
    • 4 years ago
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    how do you prove this? :/

  3. anonymous
    • 4 years ago
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    i would start by factoring the numerator, and then see that one of the factors is 1. that should make the rest easier

  4. myininaya
    • 4 years ago
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    you can factor Mr. Denominator as well

  5. anonymous
    • 4 years ago
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    i've tried that. i've went \[(\cos^2x+\sin^2x)(\cos^2x-\sin^2x)/(1-(\sin^4x/\cos^4x))\]

  6. myininaya
    • 4 years ago
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    cos^2(x)+sin^2(x)=1

  7. TuringTest
    • 4 years ago
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    you don't need to get it all into sin and cos

  8. myininaya
    • 4 years ago
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    cos^2(x)-sin^2(x)=cos(2x)

  9. anonymous
    • 4 years ago
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    hmm.. when i try, numerator part and denominator part keep gets messed up :/

  10. myininaya
    • 4 years ago
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    \[\frac{\cos(2x)}{1-\frac{\sin^4(x)}{\cos^4(x)}} \cdot \frac{\cos^4(x)}{\cos^4(x)}\] \[\frac{\cos^4(x) \cos(2x)}{\cos^4(x)-\sin^4(x)}=\frac{\cos^4(x)\cos(2x)}{\cos(2x)}=\cos^4(x)\]

  11. myininaya
    • 4 years ago
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    since we already showed \[\cos^4(x)-\sin^4(x)=\cos(2x) \] earlier

  12. anonymous
    • 4 years ago
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    great! thanks a lot.

  13. myininaya
    • 4 years ago
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    you got it? :)

  14. anonymous
    • 4 years ago
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    yup i've tried it out myself! thanks for the help.

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