If sin-sin 2-8-2-sin 2-8-2- k -2 n- - then value of 2 k 2 is

Sinθsin^2(π8+θ2) sin^2(π8 θ2)=k ⇒ nbsp;k=sinθsin(π8+θ2+π8 θ2)·sin(π8+θ2 π8+θ2) ⇒ nbsp;k =sinθsinπ4·sinθ=√(2) ∴ 2k^2=4 ...

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Trigonometric Ratios of Compound Angles

If sinθ/sin 2...

Question

If $\frac{sin θ}{{sin}^{2} (\frac{π}{8} + \frac{θ}{2}) - {sin}^{2} (\frac{π}{8} - \frac{θ}{2})} = k (θ \neq 2 n π)$ , then value of $2 k^{2}$ is

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sin θ + sin (π + θ) + sin (2 π + θ) + sin (3 π + θ) + \dots

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c o s^{2} (\frac{π}{2} + θ) - s i n^{2} (\frac{π}{2} - θ) =

{sin}^{- 1} (sin (\frac{2 π}{3})) = \frac{π}{3}

{sin}^{- 1} (sin θ) = θ;

θ \in [- \frac{π}{2}, \frac{π}{2}]

L = {sin}^{2} (\frac{π}{16}) - {sin}^{2} (\frac{π}{8})

M = {cos}^{2} (\frac{π}{16}) - {sin}^{2} (\frac{π}{8}),

f (x) = {sin}^{2} (\frac{π}{8} + \frac{x}{2}) - {sin}^{2} (\frac{π}{8} - \frac{x}{2})

f

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