SATHEE: Trigonometrical Ratios and Identities 1 Question 27

Trigonometrical Ratios and Identities 1 Question 27

27. Without using tables, prove that

$(\sin 12^{\circ}) (\sin 48^{\circ}) (\sin 54^{\circ}) = \frac{1}{8}$ .

(1982, 2M)

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Solution:

$\sin 12^{\circ} \sin 48^{\circ} \sin 54^{\circ} = \frac{1}{2} (2 \sin 12^{\circ} \sin 48^{\circ}) \sin 54^{\circ}$

$= \frac{1}{2} [\cos (36^{\circ}) - \cos (60^{\circ})] \sin 54^{\circ}$

$= \frac{1}{2} \cos 36^{\circ} - \frac{1}{2} \sin 54^{\circ}$

$= \frac{1}{4} (2 \cos 36^{\circ} \sin 54^{\circ} - \sin 54^{\circ})$

$= \frac{1}{4} (\sin 90^{\circ} + \sin 18^{\circ} - \sin 54^{\circ})$

$= \frac{1}{4} 1 + \frac{\sqrt{5} - 1}{4} - \frac{\sqrt{5} + 1}{4}$

$= \frac{1}{4} 1 + \frac{\sqrt{5} - 1 - \sqrt{5} - 1}{4}$

$= \frac{1}{4} 1 - \frac{1}{2} = \frac{1}{8}$

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Trigonometrical Ratios and Identities 1 Question 27

27. Without using tables, prove that

Solution:

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