SATHEE: Trigonometrical Ratios and Identities 1 Question 28

Trigonometrical Ratios and Identities 1 Question 28

28. Prove that $\sin^{2} α + \sin^{2} β - \sin^{2} γ = 2 \sin α \sin β \sin γ$ , where $α + β + γ = π$ .

$(1978, 4 M)$

Integer Answer Type Question

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Solution:

LHS $= \sin^{2} α + \sin^{2} β - \sin^{2} γ$

$\begin{aligned} = \sin^{2} α + (\sin^{2} β - \sin^{2} γ) \\ = \sin^{2} α + \sin (β + γ) \sin (β - γ) \\ = \sin^{2} α + \sin (π - α) \sin (β - γ) [∵ α + β + γ = π] \\ = \sin^{2} α + \sin α \sin (β - γ) \\ = \sin α [\sin α + \sin (β - γ)] \\ = \sin α [\sin (π - (β + γ)) + \sin (β - γ)] \\ = \sin α [\sin (β + γ) + \sin (β - γ)] \\ = \sin α [2 \sin β \cos γ] \\ = 2 \sin α \sin β \cos γ = RHS \end{aligned}$

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Trigonometrical Ratios and Identities 1 Question 28

28. Prove that sin2⁡α+sin2⁡β−sin2⁡γ=2sin⁡αsin⁡βsin⁡γ, where α+β+γ=π.

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28. Prove that $\sin^{2} α + \sin^{2} β - \sin^{2} γ = 2 \sin α \sin β \sin γ$ , where $α + β + γ = π$ .