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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)
Great Answer, thanks!
2 Answers
- No MythologyLv 78 years agoFavorite Answer
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.
- ?Lv 45 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.