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Prove that: cos 78 ° cos 42 ° cos 36 ° = 1 8

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प्रश्न

Prove that:  \[\cos 78°  \cos 42°  \cos 36° = \frac{1}{8}\]

संख्यात्मक
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उत्तर

\[LHS = \cos78° \cos42° \cos36° \]
\[ = \frac{\left( 2\cos78°  \cos42° \right)}{2} \cos36° \]
\[ = \frac{\cos\left( 78°  + 42°  \right) + \cos\left( 78°  - 42° \right)}{2} \times \cos36° \]
\[ \left[ 2\text{ cos } A\text{ cos } B = \cos\left( A + B \right) + \cos\left( A - B \right) \right]\]
\[ = \frac{1}{2}\left( \cos120°  + \cos36°  \right) \cos36° \]

\[= \frac{1}{2}\left( - \cos\left( 180°  - 120° \right) + \cos36°  \right) \cos36° \]
\[ = \frac{1}{2}\left( - \cos60°  + \cos36°  \right) \cos36° \]
\[ = \frac{1}{2}\left( - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right)\frac{\sqrt{5} + 1}{4}\]
\[ = \frac{1}{2} \times \frac{\sqrt{5} - 1}{4} \times \frac{\sqrt{5} + 1}{4}\]
\[ = \frac{1}{8}\]
\[ = RHS\]
\[\text{ Hence proved  } .\]

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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]

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आर.डी. शर्मा Mathematics [English] Class 11
अध्याय 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 4 | पृष्ठ ४२

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