Please help with Trig Identity Proof?

Please help with the trig identity proof, thanks.

Prove the identity cos²(3Θ) - cos²(4Θ) = sin(Θ) sin(7Θ)

Update:

Great Answer, thanks!

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  • 8 years ago
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    I'd start with the right side. Note that Θ = 4Θ - 3Θ and 7Θ = 4Θ + 3Θ.

    sinΘsin(7Θ) = sin(4Θ - 3Θ)sin(4Θ + 3Θ)

    = [sin(4Θ)cos(3Θ) - sin(3Θ)cos(4Θ)][sin(4Θ)cos(3Θ) + sin(3Θ)cos(4Θ)]

    (difference of squares)

    = sin²(4Θ)cos²(3Θ) - sin²(3Θ)cos²(4Θ)

    = (1 - cos²(4Θ))cos²(3Θ) - (1 - cos²(3Θ))cos²(4Θ)

    = cos²(3Θ) - cos²(4Θ)cos²(3Θ) - cos²(4Θ) + cos²(3Θ)cos²(4Θ)

    = cos²(3Θ) - cos²(4Θ)

    as required.

  • 4 years ago

    sin(40 5° - ?/2) * cos(40 5° + ?/2) = cos(40 5° + ?/2) * cos(40 5° + ?/2), considering cos(ninety° - x) = sin(x). = ½(a million + cos(ninety° + ?)), considering cos(2x) = 2 cos²(x) - a million. = ½(sin ninety° - sin ?), considering sin ninety° = a million and cos(ninety° + x) = - sin x.

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