Q. 9 D4.0( 4 Votes )

# Prove that:

LHS

= sin 600° cos 390° + cos 480° sin 150°

= sin (90° × 6 + 60°) cos (90° × 4 + 30°) + cos (90° × 5 + 30°) sin (90° × 1 + 60°)

We know that when n is odd, sin cos and cos sin.

= [-sin 60°] cos 30° + [-sin 30°] cos 60°

= -sin 60° cos 30° - sin 30° cos 60°

= -[sin 60° cos 30° + cos 60° sin 30°]

We know that sin A cos B + cos A sin B = sin (A + B)

= -sin (60° + 30°)

= -sin 90°

= -1

= RHS

Hence proved.

Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
Related Videos
Conditional Identities31 mins
Trigonometric Functions - 0568 mins
Trigonometric Functions - 0152 mins
Graphs of trigonometric ratios | Trigonometric Identities39 mins
Quiz on Graph of Trigonometric Ratios40 mins
Trigonometric Functions - 0366 mins
Trigonometric Functions - 0658 mins
Trigonometric Series45 mins
Interactive Quiz on trigonometric ratios & identities73 mins
Trigonometric Functions - 0268 mins
Try our Mini CourseMaster Important Topics in 7 DaysLearn from IITians, NITians, Doctors & Academic Experts
Dedicated counsellor for each student
24X7 Doubt Resolution
Daily Report Card
Detailed Performance Evaluation
view all courses
RELATED QUESTIONS :

Prove that :

RD Sharma - Mathematics

prove that :

RD Sharma - Mathematics

Prove that:

RD Sharma - Mathematics

Prove that:

RD Sharma - Mathematics