If f-sin 2-sin 2-2 -3-sin 2-4 -3- thenf -15

The correct option is C 3/2 f(θ)=sin^2 θ +sin^2(θ+2 π/3)+sin^2 ( θ+4 π/3 ) =1 cos 2 θ/2+1 cos (2 θ +4 π/3)/2+1 cos(2 θ+8 π/3)/2 =1/2(3 cos 2 θ cos (4 π/3+2 ...

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Byju's Answer

Standard XI

Mathematics

Trigonometric Ratios of Multiples of an Angle

If fθ=sin 2θ+...

Question

If $f (θ) = s i n^{2} θ + s i n^{2} (θ + \frac{2 π}{3}) + s i n^{2} (θ + \frac{4 π}{3}), t h e n f (\frac{π}{15})$

$\frac{2}{3}$

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$\frac{3}{2}$

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$\frac{1}{2}$

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$\frac{1}{3}$

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Solution

The correct option is B
$\frac{3}{2}$

$f (θ) = s i n^{2} θ + s i n^{2} (θ + \frac{2 π}{3}) + s i n^{2} (θ + \frac{4 π}{3}) = \frac{1 - c o s 2 θ}{2} + \frac{1 - c o s (2 θ + \frac{4 π}{3})}{2} + \frac{1 - c o s (2 θ + 8 \frac{π}{3})}{2} = \frac{1}{2} (3 - c o s 2 θ - c o s (\frac{4 π}{3} + 2 θ) - c o s (\frac{8 π}{3} + 2 θ)) = \frac{1}{2} (3 - c o s 2 θ - 2 c o s (2 π + 2 θ) c o s \frac{2 π}{3}) = \frac{1}{2} (3 - c o s 2 θ + c o s 2 θ) f (θ) = \frac{3}{2} ∴ f (\frac{π}{15}) = \frac{3}{2}$

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Similar questions

$f (θ) = s i n^{2} θ + s i n^{2} (θ + \frac{2 π}{3}) + s i n^{2} (θ + \frac{4 π}{3}) t h e n f (\frac{π}{15})$ is equal to

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If f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3),thenf(π15)

The correct option is B 32 f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3)=1−cos2θ2+1−cos(2θ+4π3)2+1−cos(2θ+8π3)2=12(3−cos2θ−cos(4π3+2θ)−cos(8π3+2θ))=12(3−cos2θ−2cos(2π+2θ)cos2π3)=12(3−cos2θ+cos2θ)f(θ)=32∴f(π15)=32

If $f (θ) = s i n^{2} θ + s i n^{2} (θ + \frac{2 π}{3}) + s i n^{2} (θ + \frac{4 π}{3}), t h e n f (\frac{π}{15})$