Chapter 7: Q. 111 (page 488)

Show that $\sin^{3} θ + \sin^{3} (θ + 120 °) + \sin^{3} (θ + 240 °) = - \frac{3}{4} \sin (3 θ)$

Short Answer

Expert verified

To show that $\sin^{3} θ + \sin^{3} (θ + 120 °) + \sin^{3} (θ + 240 °) = - \frac{3}{4} \sin (3 θ)$ , take the LHS and simplify using the identity formulas.

Step by step solution

Step 1. Given information.

Consider the given question,

$\sin^{3} θ + \sin^{3} (θ + 120 °) + \sin^{3} (θ + 240 °) = - \frac{3}{4} \sin (3 θ)$

Take the LHS,

localid="1646423145211" $= \sin^{3} θ + \sin^{3} (θ + 120 °) + \sin^{3} (θ + 240 °) = \sin^{3} θ + {(\sin θ \cos 120 ° + \cos θ \sin 120 °)}^{3} + {(\sin θ \cos 240 ° + \cos θ \sin 240 °)}^{3} = \sin^{3} θ + {(- \frac{1}{2} \sin θ + \frac{\sqrt{3}}{2} \cos θ)}^{3} + {(- \frac{1}{2} \sin θ - \frac{\sqrt{3}}{2} \cos θ)}^{3} = \sin^{3} θ + (- \sin θ) (\frac{1}{4} \sin^{2} θ + \frac{3}{4} \cos^{2} θ - \frac{1}{4} \sin^{2} θ + \frac{3}{4} \cos^{2} θ)$

Step 2. Use the identity formulas and simplify the expression.

Continuing the above expression,

$= \frac{3}{4} \sin^{3} θ - \frac{9}{4} \cos^{2} θ \sin θ = \sin^{3} θ + (- \frac{\sin^{3} θ}{4} - \frac{9}{4} \cos^{2} θ \sin θ)$

Use the identity $\cos^{2} θ = 1 - \sin^{2} θ$ ,

= \frac{3}{4} \{\sin^{3} θ - 3 \sin θ (1 - \sin^{2} θ)\} = \frac{3}{4} (4 \sin^{3} θ - 3 \sin θ) = - \frac{3}{4} \sin (3 θ)

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Show that $\sin^{3} θ + \sin^{3} (θ + 120 °) + \sin^{3} (θ + 240 °) = - \frac{3}{4} \sin (3 θ)$

Short Answer

Step by step solution

Step 1. Given information.

Step 2. Use the identity formulas and simplify the expression.

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Short Answer

Step by step solution

Step 1. Given information.

Step 2. Use the identity formulas and simplify the expression.

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Show that $\sin^{3} θ + \sin^{3} (θ + 120 °) + \sin^{3} (θ + 240 °) = - \frac{3}{4} \sin (3 θ)$