Question 1: \sin^2 \frac{2\pi}{5} - \sin^2 \frac{\pi}{3} = \frac{\sqrt{5}-1}{8}

Answer:

LHS = \sin^2 72^o - \sin^2 60^o

= \sin^2 ( 90^o-18^o) - \Big( \frac{\sqrt{3}}{2} \Big)^2

= \cos^2 18^o - \frac{3}{4}

\because \cos 18^o =  \frac{\sqrt{10+2\sqrt{5}}}{4}

= \Big( \frac{\sqrt{10+2\sqrt{5}}}{4} \Big)^2 - \frac{3}{4}

= \frac{10+2\sqrt{5}}{16} - \frac{3}{4}

= \frac{10 + 2 \sqrt{5}-12}{16} = \frac{2\sqrt{5} - 2 }{16} = \frac{\sqrt{5} - 1}{8} = RHS. Hence proved.

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Question 2: \sin^2 24^o - \sin^2 6^o = \frac{\sqrt{5}-1}{8}

Answer:

LHS = \sin^2 24^o - \sin^2 6^o

\because \sin ( A+B) \sin ( A - B) = \sin^2 A - \sin^2 B

= \sin ( 24^o +6^o) \sin (24^o-6^o)

= \sin 30^o \sin 18^o

\because \sin 18^o = \frac{\sqrt{5}-1}{4}

= \frac{1}{2} \Big( \frac{\sqrt{5}-1}{4} \Big)

= \frac{\sqrt{5}-1}{8} = RHS. Hence proved.

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Question 3: \sin^2 42^o - \cos^2 78^o = \frac{\sqrt{5}+1}{8}

Answer:

LHS = \sin^2 42^o - \cos^2 78^o

= \sin^2 (90^o-48^o) - \cos^2 (90^o-12^o)

= \cos^2 48^o - \sin^2 12^o

= \cos ( 48^o + 12^o) \cos ( 48^o - 12^o)

= \cos 60^o \cos 36^o

\because \cos 36^o = \frac{\sqrt{5}+1}{4}

= \frac{1}{2} \times \frac{\sqrt{5}+1}{4}

= \frac{\sqrt{5}+1}{8} = RHS. Hence proved.

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Question 4: \cos 78^o \cos 42^o \cos 36^o = \frac{1}{8}

Answer:

LHS = \cos 78^o \cos 42^o \cos 36^o

= \frac{1}{2} [ 2 \cos 78^o \cos 42^o ] \cos 36^o

= \frac{1}{2} [ \cos 120^o + \cos 36^o ] \cos 36^o

= \frac{1}{2} \Big[ - \frac{1}{2} + \frac{\sqrt{5}+ 1}{4} \Big] \Big( \frac{\sqrt{5}+ 1}{4} \Big)

= \frac{1}{2} \Big[  \frac{\sqrt{5}- 1}{4} \Big] \Big(  \frac{\sqrt{5}+ 1}{4} \Big)

= \frac{1}{2} \times \frac{5-1}{16} = \frac{1}{8} = RHS

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Question 5: \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{16}

Answer:

LHS = \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{2\sin \frac{\pi}{15}} \Big[  2 \sin \frac{\pi}{15} \cos \frac{\pi}{15} \Big]  \cos \frac{2\pi}{15} \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{4\sin \frac{\pi}{15}} \Big[  2 \sin \frac{2\pi}{15} \cos \frac{2\pi}{15} \Big]   \cos \frac{4\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{8\sin \frac{\pi}{15}} \Big[  2 \sin \frac{4\pi}{15} \cos \frac{4\pi}{15} \Big]   \cos \frac{7\pi}{15}

= \frac{1}{16\sin \frac{\pi}{15}} \Big[  2 \sin \frac{8\pi}{15} \cos \frac{7\pi}{15} \Big]

= \frac{1}{16\sin \frac{\pi}{15}} \Big[ \sin \big( \frac{8\pi}{15} + \frac{7\pi}{15} \big) + \sin \big( \frac{8\pi}{15} - \frac{7\pi}{15} \big) \Big]

= \frac{1}{16\sin \frac{\pi}{15}} \Big[   \sin \pi + \sin \frac{\pi}{15} \Big]

= \frac{1}{16\sin \frac{\pi}{15}} \Big[  \sin \frac{\pi}{15} \Big]

= \frac{1}{16} = RHS. Hence proved.

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Question 6: \cos 6^o \cos 42^o \cos 66^o \cos 78^o = \frac{1}{16}

Answer:

LHS = \cos 6^o \cos 42^o \cos 66^o \cos 78^o

= \frac{1}{4} ( 2 \cos 6^o \cos 66^o)( 2 \cos 42^o \cos 78^o)

= \frac{1}{4} ( \cos 72^o + \cos 60^o) ( \cos 120^o + \cos 36^o)

= \frac{1}{4} \big(\sin 18^o + \frac{1}{2} \big) \big( - \frac{1}{2} + \frac{\sqrt{5}+ 1}{4} \big )

= \frac{1}{4} \big ( \frac{\sqrt{5}-1}{4} + \frac{1}{2} \big)  \big(  + \frac{\sqrt{5}+ 1}{4} - \frac{1}{2} \big )

= \frac{1}{4} \big ( \frac{\sqrt{5}+1}{4} \big) \big( \frac{\sqrt{5}-1}{4} \big )

= \frac{1}{4} \times \frac{4}{16} = \frac{1}{16} = RHS. Hence proved.

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Question 7: \sin 6^o \sin 42^o \sin 66^o \sin 78^o = \frac{1}{16}

Answer:

LHS = \sin 6^o \sin 42^o \sin 66^o \sin 78^o

= \frac{1}{4} ( 2 \sin 6^o \sin 66^o)( 2 \sin 42^o \sin 78^o)

= \frac{1}{4} ( \cos 60^o - \cos 72^o) ( \cos 36^o - \cos 120^o)

= \frac{1}{4} \big( \frac{1}{2} - \sin 18^o \big) \big(  \frac{\sqrt{5}+ 1}{4} + \frac{1}{2} \big )

= \frac{1}{4} \big( \frac{1}{2} - \frac{\sqrt{5}- 1}{4} \big) \big(  \frac{\sqrt{5}+ 1}{4} + \frac{1}{2} \big )

= \frac{1}{4} \big ( \frac{3 - \sqrt{5}}{4} \big) \big( \frac{3 + \sqrt{5}}{4} \big )

= \frac{1}{4} \times \frac{4}{16} = \frac{1}{16} = RHS. Hence proved.

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Question 8: \cos 36^o \cos 42^o \cos 60^o \cos 78^o = \frac{1}{16}

Answer:

LHS = \cos 36^o \cos 42^o \cos 60^o \cos 78^o

= \frac{1}{2} \cos 36^o \cos 60^o ( 2 \cos 42^o \cos 78^o)

= \frac{1}{2} \Big(  \frac{\sqrt{5}+1}{4} \Big) \times \frac{1}{2} ( \cos 120^o + \cos 36^o)

= \Big(  \frac{\sqrt{5}+1}{16} \Big) \Big( - \frac{1}{2} + \frac{\sqrt{5}+1}{4} \Big)

= \Big(  \frac{\sqrt{5}+1}{16} \Big) \Big(  \frac{\sqrt{5}-1}{4} \Big)

= \frac{4}{64} = \frac{1}{16} = RHS. Hence proved.

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Question 9: \sin \frac{\pi}{5} \sin \frac{2\pi}{5} \sin \frac{3\pi}{5} \sin \frac{4\pi}{5} = \frac{5}{16}

Answer:

LHS = \sin 36^o \sin 72^o \sin 108^o \sin 144^o

= \sin 36^o \sin 72^o \sin (180^o - 72^o)   \sin ( 180^o - 36^o)

= \sin 36^o \sin 72^o \sin 72^o   \sin 36^o

= ( 2 \sin 36^o \sin 72^o )^2

= ( 2 \sin 36^o \cos 18^o )^2

= \Big( \frac{\sqrt{10 - 2 \sqrt{5}}}{4} \times \frac{\sqrt{10 + 2 \sqrt{5}}}{4} \Big)

= \frac{100 - 20}{16 \times 16} = \frac{80}{16 \times 16} = \frac{5}{16} = RHS. Hence proved.

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Question 10: \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{5\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15} = \frac{1}{128}

Answer:

LHS = \cos \frac{\pi}{15} \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{2 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{\pi}{15}  \cos \frac{\pi}{15}  \Big] \cos \frac{2\pi}{15} \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{4 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{2\pi}{15}  \cos \frac{2\pi}{15}  \Big]  \cos \frac{3\pi}{15} \cos \frac{4\pi}{15} \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{8 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{4\pi}{15}  \cos \frac{4\pi}{15}  \Big]  \cos \frac{3\pi}{15}  \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{16 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ 2 \sin \frac{3\pi}{15} \cos \frac{3\pi}{15} \Big]  \sin \frac{8\pi}{15}  \cos \frac{6\pi}{15} \cos \frac{7\pi}{15}

= \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ 2 \sin \frac{6\pi}{15} \cos \frac{6\pi}{15}  \Big]  \sin \frac{8\pi}{15}   \cos \frac{7\pi}{15}

= \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ \sin \frac{12\pi}{15}    \sin \frac{8\pi}{15}   \cos \frac{7\pi}{15} \Big]

= \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ \sin ( \pi - \frac{3\pi}{15})    \sin (\pi - \frac{7\pi}{15} )  \cos \frac{7\pi}{15}  \Big]

= \frac{1}{32 \sin \frac{\pi}{15} \sin \frac{3\pi}{15}} \Big[ \sin \frac{3\pi}{15}    \sin \frac{7\pi}{15}   \cos \frac{7\pi}{15} \Big]

= \frac{1}{64 \sin \frac{\pi}{15}} \Big[ 2 \sin \frac{7\pi}{15}   \cos \frac{7\pi}{15} \Big]

= \frac{1}{64 \sin \frac{\pi}{15}} \Big[  \sin \frac{14\pi}{15}   \Big]

= \frac{1}{64 \sin \frac{\pi}{15}} \Big[  \sin ( \pi - \frac{\pi}{15} )   \Big]

= \frac{1}{64 \sin \frac{\pi}{15}} \Big[  \sin \frac{\pi}{15}   \Big]

= \frac{1}{64} = RHS. Hence proved.