If sin 3-sin 3-2 -3-sin 3-4 -3-a sin b - then the value of -b-a- is

Sinθ + sin( θ+2π3)+sin( θ+4π3) =sinθ+2 sin(θ+π)cos( π3) =0 So, sin^3 θ + sin^3( θ+2π3)+sin^3( θ+4π3) =3 sinθ·sin( θ+2π3) ·sin( θ+4π3) nbsp; =3 sinθ·sin(π ...

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Byju's Answer

Standard XII

Mathematics

Sum of Trigonometric Ratios in Terms of Their Product

If sin 3θ+sin...

Question

If ${sin}^{3} θ + {sin}^{3} (θ + \frac{2 π}{3}) + {sin}^{3} (θ + \frac{4 π}{3}) = a sin b θ,$ then the value of $∣ ∣ ∣ \frac{b}{a} ∣ ∣ ∣$ is

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Solution

$sin θ + sin (θ + \frac{2 π}{3}) + sin (θ + \frac{4 π}{3}) = sin θ + 2 sin (θ + π) cos (- \frac{π}{3}) = 0$
So, ${sin}^{3} θ + {sin}^{3} (θ + \frac{2 π}{3}) + {sin}^{3} (θ + \frac{4 π}{3})$
$= 3 sin θ \cdot sin (θ + \frac{2 π}{3}) \cdot sin (θ + \frac{4 π}{3})$
$= 3 sin θ \cdot sin (π - (θ + \frac{2 π}{3})) \cdot sin (π + (θ + \frac{π}{3})) = - 3 sin θ sin (\frac{π}{3} - θ) sin (\frac{π}{3} + θ) = - 3 sin θ (\frac{3}{4} - {sin}^{2} θ) = - 3 (\frac{3 sin θ - 4 {sin}^{3} θ}{4}) = \frac{- 3}{4} sin 3 θ a = \frac{- 3}{4}, b = 3 ∴ ∣ ∣ ∣ \frac{b}{a} ∣ ∣ ∣ = 4$

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Similar questions

\frac{1}{sin 3 θ} [{sin}^{3} θ + {sin}^{3} (\frac{2 π}{3} + θ) + {sin}^{3} (\frac{4 π}{3} + θ)]

equals

Q. Prove that

\frac{c o s^{3} θ - c o s 3 θ}{c o s θ} + \frac{s i n^{3} θ + s i n 3 θ}{s i n θ} = 3

Q. If

x = a {cos}^{3} θ

and y = b

{sin}^{3} θ,

then the value of

{(\frac{x}{a})}^{2 / 3} + {(\frac{y}{b})}^{2 / 3}

Q. Find the value of

\frac{{sin}^{3} θ + {cos}^{3} θ}{sin θ + cos θ} + \frac{({cos}^{3} θ - {sin}^{3} θ)}{cos θ - sin θ}

Q. If

sin 3 θ = sin θ,

how many solutions exist such that

- 2 π < θ < 2 π

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If sin3θ+sin3(θ+2π3)+sin3(θ+4π3)=asinbθ, then the value of ∣∣∣ba∣∣∣ is

If ${sin}^{3} θ + {sin}^{3} (θ + \frac{2 π}{3}) + {sin}^{3} (θ + \frac{4 π}{3}) = a sin b θ,$ then the value of $∣ ∣ ∣ \frac{b}{a} ∣ ∣ ∣$ is