Use app×
Join Bloom Tuition
One on One Online Tuition
JEE MAIN 2025 Foundation Course
NEET 2025 Foundation Course
CLASS 12 FOUNDATION COURSE
CLASS 10 FOUNDATION COURSE
CLASS 9 FOUNDATION COURSE
CLASS 8 FOUNDATION COURSE
+1 vote
4.1k views
in Trigonometry by (47.4k points)
closed by

Prove that cosθ cos θ/2 – cos3 9θ/2 = sin 7θ sin 8θ.

[Hint: Express L.H.S. = 1/2 [2cosθ cos θ/2 – 2 cos3θ cos 9θ/2]

1 Answer

+1 vote
by (49.0k points)
selected by
 
Best answer

To prove: 

As equation on RHS is a simplified expression, so we must opt Left side equation and simplify it further so that we can get

LHS = RHS.

And thus we will be able to prove it.

By seeing the expression we can think that the problem can be solved using transformation formula:

By transformation formula, we have:

2 cos A cos B = cos(A + B) + cos (A – B)

-2 sin A sin B = cos(A + B) - cos (A – B)

But as LHS expression does not contain ‘2’ in its term. So we multiply and divide the expression by 2.

Again, applying the transformation formula:

Welcome to Sarthaks eConnect: A unique platform where students can interact with teachers/experts/students to get solutions to their queries. Students (upto class 10+2) preparing for All Government Exams, CBSE Board Exam, ICSE Board Exam, State Board Exam, JEE (Mains+Advance) and NEET can ask questions from any subject and get quick answers by subject teachers/ experts/mentors/students.

Categories

...