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Given A = 60° and B = 30°, prove that : cos (A + B) = cos A cos B - sin A sin B - Mathematics

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प्रश्न

Given A = 60° and B = 30°,
prove that : cos (A + B) = cos A cos B - sin A sin B

योग
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उत्तर

Given A = 60° and B = 30°

LHS = cos(A+B)
= cos(60° + 30°)
= cos90°
=0

RHS = cos A cos B – sin A sin B
= cos 60° cos 30° – sin 60° sin 30°

= `(1)/(2) (sqrt3)/(2) – (sqrt3)/(2) (1)/(2)`

=`(sqrt3)/(4) – (sqrt3)/(4)`

= 0
LHS  = RHS

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अध्याय 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (B) [पृष्ठ २९३]

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सेलिना Concise Mathematics [English] Class 9 ICSE
अध्याय 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
Exercise 23 (B) | Q 1.2 | पृष्ठ २९३
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