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richyw

  • 3 years ago

could someone give me a refresher on what to do with somthing like this \[\sin\left( 2\sin^{-1}\left( \frac{1}{2}\right)\right)\]

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  1. Chlorophyll
    • 3 years ago
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    Start from inside out: 1/2 = sin 30° -> 2arcsin1/2 = 2 * 30 = ... => sin (.. ) = ...

  2. richyw
    • 3 years ago
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    thanks, I made a typo though. I meant to say \[\sin\left( 2\sin^{-1}\left( \frac{x}{2}\right)\right)\]

  3. richyw
    • 3 years ago
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    \[y=\sin\left( 2\sin^{-1}\left( \frac{x}{2}\right)\right)\]\[\sin^{-1}y=2\sin^{-1}\left(\frac{x}{2}\right)\]

  4. richyw
    • 3 years ago
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    nope. super lost!

  5. Algebraic!
    • 3 years ago
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    sin2(theta) = sin(theta)*cos(theta)

  6. Chlorophyll
    • 3 years ago
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    Use the half angle formula: sin2a = 2 sina cosa with a = arcsin (x/2)

  7. Algebraic!
    • 3 years ago
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    |dw:1354928836731:dw|

  8. Algebraic!
    • 3 years ago
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    so x/2 * (sqrt(4-x^2))/2

  9. richyw
    • 3 years ago
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    thanks! totally forgot how useful those triangle diagrams are!

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