Prove that: \[\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15} = \frac{1}{16}\]

\[LHS = \cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}\] \[= \frac{2\sin\frac{\pi}{15}\cos\frac{\pi}{15}}{2\sin\frac{\pi}{15}}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}\]\[ \left[ \text{ On dividing and multiplying by } 2\sin\frac{\pi}{15} \right]\] \[= \frac{2\sin\frac{2\pi}{15} \times \cos\frac{2\pi}{15}}{2 \times 2\sin\frac{\pi}{15}}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}\]\[ = \frac{2\sin\frac{4\pi}{15} \times \cos\frac{4\pi}{15}}{2 \times 2 \times 2\sin\frac{\pi}{15}}\cos\frac{7\pi}{15}\]\[ = \frac{\sin\frac{8\pi}{15}}{2 \times 2 \times 2\sin\frac{\pi}{15}}\cos\frac{7\pi}{15}\] \[= \frac{2\sin\frac{8\pi}{15}\cos\frac{7\pi}{15}}{2 \times 2 \times 2 \times 2\sin\frac{\pi}{15}}\]\[ = \frac{2\sin\frac{8\pi}{15}\cos\frac{7\pi}{15}}{16\sin\frac{\pi}{15}}\]\[ = \frac{\sin\left( \frac{8\pi}{15} + \frac{7\pi}{15} \right) + \sin\left( \frac{8\pi}{15} - \frac{7\pi}{15} \right)}{16\sin\frac{\pi}{15}} \left[ \because 2\text{ sin } A\text{ cos } B = \sin\left( A + B \right) + \sin\left( A - B \right) \right]\] \[= \frac{sin\pi + \sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}\]\[ = \frac{0 + \sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}\]\[ = \frac{\sin\frac{\pi}{15}}{16\sin\frac{\pi}{15}}\]\[ = \frac{1}{16}\]\[ = RHS\] Hence proved.

Prove That: Cos π 15 Cos 2 π 15 Cos 4 π 15 Cos 7 π 15 = 1 16 - Mathematics

Question

Prove that: $\cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15} = \frac{1}{16}$

Numerical

Solution

L H S = \cos \frac{π}{15} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15}

$= \frac{2 \sin \frac{π}{15} \cos \frac{π}{15}}{2 \sin \frac{π}{15}} \cos \frac{2 π}{15} \cos \frac{4 π}{15} \cos \frac{7 π}{15}$
$[On dividing and multiplying by 2 \sin \frac{π}{15}]$

$= \frac{2 \sin \frac{2 π}{15} \times \cos \frac{2 π}{15}}{2 \times 2 \sin \frac{π}{15}} \cos \frac{4 π}{15} \cos \frac{7 π}{15}$
$= \frac{2 \sin \frac{4 π}{15} \times \cos \frac{4 π}{15}}{2 \times 2 \times 2 \sin \frac{π}{15}} \cos \frac{7 π}{15}$
$= \frac{\sin \frac{8 π}{15}}{2 \times 2 \times 2 \sin \frac{π}{15}} \cos \frac{7 π}{15}$

= \frac{2 \sin \frac{8 π}{15} \cos \frac{7 π}{15}}{2 \times 2 \times 2 \times 2 \sin \frac{π}{15}}

= \frac{2 \sin \frac{8 π}{15} \cos \frac{7 π}{15}}{16 \sin \frac{π}{15}}

= \frac{\sin (\frac{8 π}{15} + \frac{7 π}{15}) + \sin (\frac{8 π}{15} - \frac{7 π}{15})}{16 \sin \frac{π}{15}} [∵ 2 sin A cos B = \sin (A + B) + \sin (A - B)]

= \frac{s i n π + \sin \frac{π}{15}}{16 \sin \frac{π}{15}}

= \frac{0 + \sin \frac{π}{15}}{16 \sin \frac{π}{15}}

= \frac{\sin \frac{π}{15}}{16 \sin \frac{π}{15}}

= \frac{1}{16}

= R H S

Hence proved.

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle

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Q 5 Q 4 Q 7

Chapter 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.3 [Page 42]

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RD Sharma Mathematics [English] Class 11

Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 5 | Page 42

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Prove That: Cos π 15 Cos 2 π 15 Cos 4 π 15 Cos 7 π 15 = 1 16 - Mathematics

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