Answer :

LHS = cos 570o sin 510o + sin (-330o) cos (-390o)


We know that sin (-x) = -sin (x) and cos (-x) = +cos (x).


= cos 570o sin 510o + [-sin (330o)] cos (390o)


= cos 570o sin 510o - sin (330o) cos (390o)


= cos (90° × 6 + 30°) sin (90° × 5 + 60°) – sin (90° × 3 + 60°) cos (90° × 4 + 30°)


We know that cos is negative at 90° + θ i.e. in Q2 and when n is odd, sin cos and cos sin.


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


= -cos 30° cos 60° + cos 60° cos 30°


= 0


= RHS


Hence proved.


Rate this question :

How useful is this solution?
We strive to provide quality solutions. Please rate us to serve you better.
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
caricature
view all courses
RELATED QUESTIONS :

Prove that :
<
RD Sharma - Mathematics

prove that :
<
RD Sharma - Mathematics

Prove that:
</
RD Sharma - Mathematics

Find the values oRD Sharma - Mathematics

Find the values oRD Sharma - Mathematics

Find the values oRD Sharma - Mathematics

Find the values oRD Sharma - Mathematics

prove that :
<
RD Sharma - Mathematics

Prove that:
</
RD Sharma - Mathematics

Find the values oRD Sharma - Mathematics