JEE Exam  >  JEE Questions  >  Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? Start Learning for Free
Sin3x sin2x-sinx=4sinxcosx/2cos3x/2?
Most Upvoted Answer
Sin3x sin2x-sinx=4sinxcosx/2cos3x/2?
Solution:

Given equation: sin(3x)sin(2x) - sin(x) = 4sin(x)cos(x)/2cos(3x)/2

To solve this equation, we will simplify the expression step by step.

1. Simplifying the right side of the equation:
4sin(x)cos(x)/2cos(3x)/2 can be simplified as 2sin(x)cos(x)/cos(3x).

2. Applying the double-angle formula for sine:
sin(2x) = 2sin(x)cos(x).

Substituting this in the equation, we get:
sin(3x) * 2sin(x)cos(x) - sin(x) = 2sin(x)cos(x)/cos(3x).

3. Multiplying through by 2cos(3x):
2sin(3x)sin(x)cos(x) - 2sin(x)cos(3x) = 2sin(x)cos(x).

4. Rearranging the terms:
2sin(3x)sin(x)cos(x) - 2sin(x)cos(x) = 2sin(x)cos(3x).

5. Factoring out 2sin(x):
2sin(x)[sin(3x)cos(x) - 1] = 2sin(x)cos(3x).

6. Dividing both sides by 2sin(x):
sin(3x)cos(x) - 1 = cos(3x).

7. Moving -1 to the right side:
sin(3x)cos(x) = cos(3x) + 1.

8. Applying the double-angle formula for cosine:
cos(2x) = 2cos^2(x) - 1.

Substituting this in the equation, we get:
sin(3x)cos(x) = 2cos^2(3x) - 1 + 1.

9. Simplifying:
sin(3x)cos(x) = 2cos^2(3x).

10. Applying the identity sin(2A) = 2sin(A)cos(A):
sin(3x)cos(x) = sin(6x).

11. Cancelling out sin(x) on both sides:
cos(x) = sin(6x).

12. Taking the square of both sides:
cos^2(x) = sin^2(6x).

13. Using the identity cos^2(x) = 1 - sin^2(x):
1 - sin^2(x) = sin^2(6x).

14. Rearranging the terms:
sin^2(x) + sin^2(6x) = 1.

15. Applying the Pythagorean identity:
sin^2(x) + (1 - cos^2(6x)) = 1.

16. Simplifying:
sin^2(x) + 1 - cos^2(6x) = 1.

17. Rearranging the terms:
sin^2(x) - cos^2(6x) = 0.

18. Applying the identity cos(2A) = cos^2(A) - sin^2(A):
sin^2(x) - cos^2(3x)
Explore Courses for JEE exam
Sin3x sin2x-sinx=4sinxcosx/2cos3x/2?
Question Description
Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? for JEE 2024 is part of JEE preparation. The Question and answers have been prepared according to the JEE exam syllabus. Information about Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? covers all topics & solutions for JEE 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for Sin3x sin2x-sinx=4sinxcosx/2cos3x/2?.
Solutions for Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? in English & in Hindi are available as part of our courses for JEE. Download more important topics, notes, lectures and mock test series for JEE Exam by signing up for free.
Here you can find the meaning of Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? defined & explained in the simplest way possible. Besides giving the explanation of Sin3x sin2x-sinx=4sinxcosx/2cos3x/2?, a detailed solution for Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? has been provided alongside types of Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? theory, EduRev gives you an ample number of questions to practice Sin3x sin2x-sinx=4sinxcosx/2cos3x/2? tests, examples and also practice JEE tests.
Explore Courses for JEE exam

Top Courses for JEE

Explore Courses
Signup for Free!
Signup to see your scores go up within 7 days! Learn & Practice with 1000+ FREE Notes, Videos & Tests.
10M+ students study on EduRev