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pathosdebater

  • 3 years ago

Please help! Use the half-angle identities to express sin 3C in terms of function of 6C.

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  1. hamidic
    • 3 years ago
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    sin(6C) = 2 sin(3C) cos(3C) ;then by the sin inverse function 6C=sin-1(2s in(3C) cos (3C)

  2. pathosdebater
    • 3 years ago
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    (2s in(3c)? Does that mean 2sin^-1(3c)?

  3. hamidic
    • 3 years ago
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    the answer is 6C = sin-1(2 sin(3C) cos(3C) )

  4. pathosdebater
    • 3 years ago
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    It's the inverse of sin, so it's sin^-1, right?

  5. hamidic
    • 3 years ago
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    yes its the inverse of sin

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