Prove this
[ (2cos^2x-1)* cos x/2]- [(4cos^3x - 3cosx)*cos9x/2)]= sin5x sin5x/2
samrat00725100:
Please recheck the question ... The answer will be cos5x cos5x/2
no RHS is sin5xsin5x/2 ,itis a book question
Dear saniarisha why did you deleted my answer.. I made the photo myself using Microsoft Word.... Do you know what that means.. Don't unknowingly delete anyone's answer ...
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Taking RHS
[ (2cos²x-1) cos x/2]- [(4cos³x - 3cosx) cos9x/2)]
= [(cos2x)(cosx/2) - (cos3x)(cos9x/2)]
Multiplying and diving by 2
= 1/2 [2(cos2x)(cosx/2) - 2(cos3x)(cos9x/2)]
Note: 2cosA cosB = cos(A+B) cos(A-B)
= 1/2 [cos(2x+x/2) - cos(3x+9x/2)]
= 1/2 [ cos5x/2 - cos15x/2]
Note: cosD-cosC = -2 sin(D+C/2) sin(D-C/2)
= 1/2 {-2sin([5x/2+15x/2]/2) sin([5x/2-15x/2]/2)}
= {-sin20x/4 (-sin10x/4)}
= {sin5x sin5x/2} ------------ RHS
Therefore LHS = RHS
[ (2cos²x-1) cos x/2]- [(4cos³x - 3cosx) cos9x/2)] = {sin5x sin5x/2}
Hence it is proved..
[ (2cos²x-1) cos x/2]- [(4cos³x - 3cosx) cos9x/2)]
= [(cos2x)(cosx/2) - (cos3x)(cos9x/2)]
Multiplying and diving by 2
= 1/2 [2(cos2x)(cosx/2) - 2(cos3x)(cos9x/2)]
Note: 2cosA cosB = cos(A+B) cos(A-B)
= 1/2 [cos(2x+x/2) - cos(3x+9x/2)]
= 1/2 [ cos5x/2 - cos15x/2]
Note: cosD-cosC = -2 sin(D+C/2) sin(D-C/2)
= 1/2 {-2sin([5x/2+15x/2]/2) sin([5x/2-15x/2]/2)}
= {-sin20x/4 (-sin10x/4)}
= {sin5x sin5x/2} ------------ RHS
Therefore LHS = RHS
[ (2cos²x-1) cos x/2]- [(4cos³x - 3cosx) cos9x/2)] = {sin5x sin5x/2}
Hence it is proved..
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