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The value of \( \Large \sin 36 ^{\circ} \sin 72 ^{\circ} \sin 108 ^{\circ} \sin 144 ^{\circ} \) is equal to:

A) \( \Large \frac{1}{4} \)
B) \( \Large \frac{1}{16} \)
C) \( \Large \frac{3}{4} \)
D) \( \Large \frac{5}{16} \)

Correct Answer:




D) \( \Large \frac{5}{16} \)

Description for Correct answer:
\( \Large \sin 36 ^{\circ} \sin 72 ^{\circ} \sin 108 ^{\circ} \sin 144 ^{\circ} \)

= \( \Large \sin^{2}36 ^{\circ} \sin^{2}72 ^{\circ} = \frac{1}{4} \left(2 \sin^{2}36 ^{\circ} \right) \left(2 \sin^{2}72 ^{\circ} \right) \)

= \( \Large \frac{1}{4} \left(1-\cos 72 ^{\circ} \right) \left(1-\cos 144 ^{\circ} \right) \)

= \( \Large \frac{1}{4} \left(1-\sin 18 ^{\circ} \right) \left(1+\cos 36 ^{\circ} \right) \)

= \( \Large \frac{1}{4}\left[ \left(1-\frac{\sqrt{5}-1}{4}\right) \left(1+\frac{\sqrt{5}+1}{4}\right) \right] \)

= \( \Large \frac{1}{4}\left[ 1 + \left(\frac{\sqrt{5}+1}{4}\right)- \left(\frac{\sqrt{5}-1}{4}\right)- \left(\frac{4}{16}\right) \right] \)

= \( \Large \frac{1}{4} \left[ 1+\frac{1}{2}-\frac{1}{4} \right] = \frac{5}{16} \)

Part of solved Trigonometric ratio questions and answers : >> Elementary Mathematics >> Trigonometric ratio

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Similar Questions

1). If \( \Large \cos \left( \theta - \alpha \right),\ \cos \theta\ and\ \cos \left( \theta + \alpha \right) \) are in HP, the: \( \Large \cos \theta \sec \frac{ \alpha }{2} \) is equal to:

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D). none of these

2). If \( \Large \cos x + \cos y + \cos \alpha = 0 \) and \( \Large \sin x + \sin y + \sin \alpha = 0,\) then \( \Large \cot \left(\frac{x+y}{2}\right) \) is equal to:

A). \( \Large \sin \alpha \)

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6). If \( \Large \cos \theta = \cos \alpha \cos \beta .\ then\ \tan \frac{ \theta + \alpha }{2} \tan \frac{ \theta - \alpha }{2} \) is equal to:

A). \( \Large \tan^{2}\frac{ \alpha }{2} \)

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A). 0

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8). A solution \( \Large \left(x, y\right)\ of\ x^{2}+2x \sin xy + 1 = 0 \) is:

A). (1,0)

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10). The distance of the point (-2, 3) from the line x -y = 5 is:

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B). \( \Large 2\sqrt{5} \)

C). \( \Large 3\sqrt{5} \)

D). \( \Large 5\sqrt{3} \)