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Sin(A+B)Sin(A-B)

cos 2 θ + ϕ +...

Question

$cos 2 (θ + ϕ) + 4 cos (θ + ϕ) sin θ sin ϕ + 2 {sin}^{2} ϕ =$

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Solution

$\begin{matrix} cos 2 (θ + ϕ) + 4 cos (θ + ϕ) sin θ sin ϕ + 2 {sin}^{2} ϕ = cos 2 (θ + ϕ) + 2 cos (θ + ϕ) [2 sin θ sin ϕ] + 2 {sin}^{2} ϕ = cos 2 (θ + ϕ) + 2 cos (θ + ϕ) [cos (θ - ϕ) - cos (θ + ϕ)] + 2 {sin}^{2} ϕ = cos 2 (θ + ϕ) + 2 cos (θ + ϕ) cos (θ - ϕ) - {cos}^{2} (θ + ϕ) + 2 {sin}^{2} ϕ = [2 cos 2^{2} (θ + ϕ) - 1] + cos (θ + ϕ + θ - ϕ) + cos (θ + ϕ - θ + ϕ) - 2 {cos}^{2} (θ + ϕ) + 2 {sin}^{2} ϕ = - 1 + cos 2 θ + cos 2 ϕ + (1 - cos 2 θ) = cos 2 ϕ \end{matrix}$

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Q. cos 2θ cos 2ϕ + sin²(θ – ϕ) – sin²(θ + ϕ) is equal to
(a) sin 2 (θ + ϕ)
(b) cos 2 (θ + ϕ)
(c) sin 2 (θ – ϕ)
(d) cos 2 (θ – ϕ)

3 {sin}^{2} θ + 2 {sin}^{2} ϕ = 1

3 sin 2 θ = 2 sin 2 ϕ

0 < θ < \frac{π}{2}

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, then the value of

θ + 2 ϕ

Q. Prove that :

cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos 2 (θ + ϕ)

cos 2 θ cos 2 ϕ + {sin}^{2} (θ - ϕ) - {sin}^{2} (θ + ϕ) = cos (2 θ + 2 ϕ)

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cos 2 θ + 2 {sin}^{2} ϕ =

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Sin(A+B)Sin(A-B)

Standard XII Mathematics

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