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

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प्रश्न

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

 
संख्यात्मक
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उत्तर

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

\[= \frac{2\sin\frac{\pi}{15}\cos\frac{\pi}{15}}{2\sin\frac{\pi}{15}}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{7\pi}{15}\]
\[ \left[ \text{ On dividing and multiplying by } 2\sin\frac{\pi}{15} \right]\]

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

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

 

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Values of Trigonometric Functions at Multiples and Submultiples of an Angle
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 9: Values of Trigonometric function at multiples and submultiples of an angle - Exercise 9.3 [पृष्ठ ४२]

APPEARS IN

आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 5 | पृष्ठ ४२

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