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Prove that: cos 6 ° cos 42 ° cos 66 ° cos 78 ° = 1 16 - Mathematics

Numerical

Prove that: \[\cos 6° \cos 42°   \cos 66°    \cos 78° = \frac{1}{16}\]

 
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Solution

\[LHS = \cos6°  \cos42°   \cos66°   \cos78° \]
\[ = \frac{1}{4}\left( 2\cos6°  \cos66°   \right)\left( 2\cos42°  \cos78°   \right) \]
\[ = \frac{1}{4}\left( \cos72°  + \cos60°  \right)\left( \cos120°  + \cos36°  \right) \left[ \because 2\text{ cos }A\text{ cos } B = \cos\left( A + B \right) + \cos\left( A - B \right) \right] \]
\[ = \frac{1}{4}\left\{ \cos\left( 90°  - 72°  \right) + \frac{1}{2} \right\}\left\{ - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right\}\]

\[= \frac{1}{4}\left( \sin18°   + \frac{1}{2} \right)\left( - \frac{1}{2} + \frac{\sqrt{5} + 1}{4} \right)\]
\[ = \frac{1}{4}\left( \frac{\sqrt{5} - 1}{4} + \frac{1}{2} \right)\left( \frac{\sqrt{5} + 1}{4} - \frac{1}{2} \right)\]
\[ = \frac{1}{4}\left( \frac{\sqrt{5} - 1 + 2}{4} \right)\left( \frac{\sqrt{5} + 1 - 2}{4} \right)\]
\[ = \frac{1}{64}\left( \sqrt{5} + 1 \right)\left( \sqrt{5} - 1 \right)\]
\[ = \frac{1}{64}\left( 5 - 1 \right)\]
\[ = \frac{1}{16} = RHS\]
\[\text{ Hence proved }  .\]

Concept: Values of Trigonometric Functions at Multiples and Submultiples of an Angle
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APPEARS IN

RD Sharma Class 11 Mathematics Textbook
Chapter 9 Values of Trigonometric function at multiples and submultiples of an angle
Exercise 9.3 | Q 7 | Page 42
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