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ChristopherW28

  • one year ago

Verify that (1-cos^2x)(1+cos^2x)=2sin^2x-sin^4x is a trig identity

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  1. myininaya
    • one year ago
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    By a certain Pythagorean identity we can say 1-cos^2(x)=?

  2. ChristopherW28
    • one year ago
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    Sin^2x?

  3. myininaya
    • one year ago
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    yep! :)

  4. myininaya
    • one year ago
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    so this means we now have \[\sin^2(x)(1+\cos^2(x))=2\sin^2(x)-\sin^4(x)\] now this is an identity we are trying to prove now we have one side in terms of sin and another side in terms of mixture of sin and cos

  5. myininaya
    • one year ago
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    We could use that same identity to rewrite cos^2(x)

  6. myininaya
    • one year ago
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    \[1-\cos^2(x)=\sin^2(x) => \cos^2(x)=\]

  7. myininaya
    • one year ago
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    that is a blank for you to fill in

  8. ChristopherW28
    • one year ago
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    I'm lost

  9. myininaya
    • one year ago
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    Ok I want to rewrite cos^2(x) because the other side is just in terms of sin I know an identity that has cos^2(x) and sin^2(x) in it.

  10. myininaya
    • one year ago
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    \[\cos^2(x)+\sin^2(x)=1 \]

  11. myininaya
    • one year ago
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    So cos^2(x)=?

  12. ChristopherW28
    • one year ago
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    Sin^x+ 1

  13. myininaya
    • one year ago
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    1-sin^2(x)

  14. ChristopherW28
    • one year ago
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    Sin^2x+1?

  15. ChristopherW28
    • one year ago
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    I was close

  16. myininaya
    • one year ago
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    1-cos^2(x)=sin^2(x) 1-sin^2(x)=cos^2(x) cos^2(x)+sin^2(x)=1

  17. myininaya
    • one year ago
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    So we can back to what we are trying to prove and replace the cos^2(x) with 1-sin^2(x) \[\sin^2(x)(1+(1-\sin^2(x))=2\sin^2(x)-\sin^4(x)\]

  18. myininaya
    • one year ago
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    I replaced mr.cos^2(x) with mrs. (1-sin^2(x))

  19. myininaya
    • one year ago
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    now we don't need those parenthesis since there is a + in front of that parenthesis lets drop that extra stuff giving us: \[\sin^2(x)(1+1-\sin^2(x))=2\sin^2(x)-\sin^4(x) \]

  20. myininaya
    • one year ago
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    now 1+1=2 which I know you know (:p) so now you distribute on that left hand side

  21. ChristopherW28
    • one year ago
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    I gtg I can figure out what's left. Friend me on Facebook @ Christopher McElhannon. I will need your help again :)

  22. myininaya
    • one year ago
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    I will be here most likely. I'm not much of a facebook user. I'm on facebook free diet right now. :p

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