First slide

Trigonometric transformations

Question

If $α + β - γ = π, then \sin^{2} α + \sin^{2} β - \sin^{2} γ$ is equal to

Moderate

A

$2 \sin αsin βcos γ$
B

$2 \cos αcos βcos γ$
C

$2 \sin αsin βsin γ$
D

None of these

Solution

we have $α + β - γ = π$
Now, $\begin{array}{l} \sin^{2} α + \sin^{2} β - \sin^{2} γ \\ = \sin^{2} α + \sin (β - γ) \sin (β + γ) \\ = \sin^{2} α + \sin (π - α) \sin (β + γ) (∵ α + β - γ = π) \\ = \sin^{2} α + \sin αsin (β + γ) \\ = \sin α [\sin α + \sin (β + γ)] \\ = \sin α [\sin (π - (β - γ)) + \sin (β + γ)] \end{array}$
$\begin{array}{l} = \sin α [\sin (β - γ) + \sin (β + γ)] \\ = \sin α [2 \sin βcos γ] \\ = 2 \sin αsin βcos γ \end{array}$

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