Question

Prove that -
cos 8º cos 10° cos 12°
-sin 8° sin 10° cos 12° - sin 18° sin 12° =​

Answer 1

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Answer 2

Answer:

Value of cos 8⁰ cos 10⁰ cos 12⁰ - sin 8⁰ sin 10⁰ cos 12⁰ - sin 18⁰ sin 12⁰ is √3/2.

Step-by-step explanation:

• Cosine of sum of two angles a and b is given by:

$\cos( a + b) = \cos a \cos b - \sin a \sin b$

• Given that:

$\cos{8}^{0} \cos {10}^{0} \cos {12}^{0} - \sin {8}^{0} \sin { {10}^{0} } \cos{12}^{0} \: - \sin {18}^{0} \sin {12}^{0}$

• Taking common cos 12⁰ from first two terms;

$( \cos {8}^{0} \cos {10}^{0} - \sin {8}^{0} \sin {10}^{0} ) \cos{12}^{0} - \sin {18}^{0} \sin{12}^{0}$

$or \: \cos( {8}^{0} + {10}^{0} ) \cos {12}^{0} - \sin {18}^{0} \sin {12}^{0} \\ or \: \cos {18}^{0} \cos {12}^{0} - \sin {18}^{0} \sin {12}^{0}$

$or \: \cos( {18}^{0} + {12}^{0} ) \\ or \: \cos{30}^{0} = \frac{ \sqrt{3} }{2}$