Sin 2n- - sin 2n - 1- - sin 2- where n is constant and n 0-1

LHS=sin ^ 2 nθ #160; sin ^ 2 (n 1)θ #160; =sin [n (n 1)]θ #160; .sin [n+(n 1)]θ #160; =sinθ #160; .sin [2n 1]θ #160; ∴sin ...

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Principal Solution of Trigonometric Equation

sin 2nθ - sin...

Question

${sin}^{2} n θ - {sin}^{2} (n - 1) θ = {sin}^{2} θ$ , where $n$ is constant and $n \neq 0, 1$

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{sin}^{2} n θ - {sin}^{2} (n - 1) θ = {sin}^{2} θ

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