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Factorization Method Form to Remove Indeterminate Form

The value of ...

Question

The value of $\sin \frac{π}{10} \sin \frac{13 π}{10}$ is
(a) $\frac{1}{2}$

(b) $- \frac{1}{2}$

(c) $- \frac{1}{4}$

(d) 1

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Solution

$\sin \frac{π}{10} \sin \frac{13 π}{10} = \sin \frac{π}{10} \sin (π + \frac{3 π}{10}) = \sin \frac{π}{10} (- \sin \frac{3 π}{10}) (∵ \sin (π + θ) = - sinθ) ∴ \sin \frac{π}{10} \sin \frac{13 π}{10} = - \sin \frac{π}{10} \sin \frac{3 π}{10} . . . (1) Let θ = \frac{π}{10} i . e . 5 θ = \frac{π}{2} i . e . 3 θ + 2 θ = \frac{π}{2} i . e . 3 θ = \frac{π}{2} - 2 θ i . e . \cos 3 θ = \cos (\frac{π}{2} - 2 θ) = \sin 2 θ i . e . 4 \cos^{3} θ - 3 \cos θ = 2 \sin θ \cos θ (using identity of \cos 3 θ and \sin 2 θ)$

$i . e . 4 \cos^{2} θ - 3 = 2 \sin θ i . e . 4 (1 - \sin^{2} θ) - 3 = 2 \sin θ i . e . 4 \sin^{2} θ + 2 \sin θ - 1 = 0 (i . e . Let \sin θ = t 4 t^{2} + 2 t - 1 roots are = \frac{- 2 \pm \sqrt{20}}{2 \times 4} = \pm (\frac{\sqrt{5} - 1}{4})) i . e . \sin θ = \frac{\sqrt{5} - 1}{4} i . e . \sin \frac{π}{10} = \frac{\sqrt{5} - 1}{4}$

$\begin{array}{rcl} \sin \frac{3 π}{10} & = & 3 \sin \frac{π}{10} - 4 \sin^{3} \frac{π}{10} \\ = & 3 (\frac{\sqrt{5} - 1}{4}) - 4 {(\frac{\sqrt{5} - 1}{4})}^{3} \\ = & \frac{3}{4} (\sqrt{5} - 1) - \frac{1}{4^{2}} {(\sqrt{5} - 1)}^{3} \\ = & \frac{3}{4} (\sqrt{5} - 1) - \frac{1}{16} [{(\sqrt{5})}^{3} - {(1)}^{3} - 3 \sqrt{5} (\sqrt{5} - 1)] \\ = & \frac{3}{4} \sqrt{5} - \frac{3}{4} - \frac{1}{16} [5 \sqrt{5} - 1 - 3 \times 5 + 3 \sqrt{5}] \\ = & \frac{3}{4} \sqrt{5} - \frac{3}{4} - \frac{1}{16} [5 \sqrt{5} - 1 - 15 + 3 \sqrt{5}] \\ = & \frac{3}{4} \sqrt{5} - \frac{3}{4} - \frac{1}{16} [8 \sqrt{5} - 16] \\ = & \frac{3}{4} \sqrt{5} - \frac{3}{4} - \frac{1}{2} \sqrt{5} + 1 \\ = & \frac{3}{4} \sqrt{5} - \frac{1}{2} \sqrt{5} + 1 - \frac{3}{4} \\ = & \frac{3 \sqrt{5} - 2 \sqrt{5}}{4} + \frac{4 - 3}{4} \\ = & \frac{\sqrt{5}}{4} + \frac{1}{4} \end{array}$

$\sin \frac{3 π}{10} = \frac{1}{4} (\sqrt{5} + 1) from (1) ∴ \sin \frac{π}{10} \sin \frac{13 π}{10} = - \sin \frac{π}{10} \sin \frac{3 π}{10} = - (\frac{\sqrt{5} - 1}{4}) (\frac{\sqrt{5} + 1}{4}) = - (\frac{5 - 1}{4^{2}}) = \frac{- 4}{16} = \frac{- 1}{4} Hence \sin \frac{13 π}{10} \sin \frac{π}{10} = \frac{- 1}{4}$

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