F -sin 2-sin 2-2 -3-sin 2-4 -3 then f -15 is equal toA- 1B- 2 - 3C- 1 - 3D- 3 - 2

The correct option is D 3/2 f(θ)=sin^2θ+sin^2(θ+2π 3)+sin^2(θ+4π 3) =1 2[3 cos2θ cos(4π 3+2θ) cosθ(8π 3+2θ)] =1 2(3 cos2θ 2cos(2π+2θ)(cos2π 3)) =1 2(3 cos 2θ+c ...

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Standard XII

Mathematics

Definition of Functions

f θ=sin 2θ+si...

Question

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

2/3

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Solution

The correct option is B
3/2

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

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f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3) then f(π15) is equal to

The correct option is B 3/2 f(θ)=sin2θ+sin2(θ+2π3)+sin2(θ+4π3) =12[3−cos2θ−cos(4π3+2θ)−cosθ(8π3+2θ)] =12(3−cos2θ−2cos(2π+2θ)(cos2π3)) =12(3−cos 2θ+cos 2θ)=32 f(π15)=32

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