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anonymous

  • one year ago

Please help*** Proving Trigonometric Identities

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  1. anonymous
    • one year ago
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    |dw:1436290113123:dw|

  2. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 1-2\sin^2\theta=2\cos^2\theta-1 }\)

  3. SolomonZelman
    • one year ago
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    \(\large\color{black}{ \displaystyle 1-2\sin^2\theta=2\cos^2\theta-1 }\) knowing that \(\large\color{blue}{ \displaystyle 1=\sin^2\theta+\cos^2\theta }\) \(\large\color{black}{ \displaystyle (\sin^2\theta+\cos^2\theta)-2\sin^2\theta=2\cos^2\theta-(\sin^2\theta+\cos^2\theta) }\) \(\large\color{black}{ \displaystyle \sin^2\theta+\cos^2\theta-2\sin^2\theta=2\cos^2\theta-\sin^2\theta-\cos^2\theta }\) now, finish and simplify.

  4. anonymous
    • one year ago
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    I'm still lost @SolomonZelman

  5. SolomonZelman
    • one year ago
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    you can't add/subtract like terms?

  6. anonymous
    • one year ago
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    would it be sin^2 (theta) + cos^2(theta) = cos^2(theta) - sin^2 (theta) @SolomonZelman

  7. SolomonZelman
    • one year ago
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    x-2x=?

  8. anonymous
    • one year ago
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    -x

  9. SolomonZelman
    • one year ago
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    good

  10. SolomonZelman
    • one year ago
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    sin²θ - 2sin²θ = ?

  11. anonymous
    • one year ago
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    -sin^2(theta)

  12. SolomonZelman
    • one year ago
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    yes

  13. anonymous
    • one year ago
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    so is it

  14. SolomonZelman
    • one year ago
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    so, your left side is \(\large\color{black}{ \displaystyle -\sin^2\theta+\cos^2\theta }\) and the right side is \(\large\color{black}{ \displaystyle \cos^2\theta-\sin^2\theta }\) (the right side is same as left side, so yes, it is an identity)

  15. anonymous
    • one year ago
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    Thanks

  16. SolomonZelman
    • one year ago
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    yw

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