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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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Question

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

Sum

Solution

Given A = 60° and B = 30°

LHS = cos(A – B)

= cos (60° – 30°)

= cos 30°

= `(sqrt3)/(2)`

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)`

= `(sqrt3)/(2)`

LHS = RHS

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Chapter 23: Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios] - Exercise 23 (B) [Page 293]

APPEARS IN

Selina Concise Mathematics [English] Class 9 ICSE
Chapter 23 Trigonometrical Ratios of Standard Angles [Including Evaluation of an Expression Involving Trigonometric Ratios]
Exercise 23 (B) | Q 1.3 | Page 293

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