heya!!Here is a question from class11 maths trigonometry
➡CONTENT QUALITY NEEDED⬅
Q.
Pls solve with explanation
VijayaLaxmiMehra1:
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we know,
cos(90 + ø) = sinø
cos²x = (1 + sin2x)/2
we have to
now,
(1 + cos2x)/2 + [1 + cos(2x + )]/2 + [ 1 + cos(2x - )]/2
1/2 [ 1 + cos2x + 1 + cos(2x + ) + 1 + cos(2x - )]
1/2[3 + cos2x + cos(2x + ) + cos(2x - )]
1/2[ 3 + cos2x + (2cos(2x) × cos ]
1/2[3 + cos2x + 2cos2X × cos(120) ]
1/2[3 + cos2X + 2cos2X × cos(90 + 30) ]
1/2[3 + cos2X + 2cos2X × (-sin30)]
1/2[3 + cos2X + 2cos2X × (-1/2)]
1/2[ 3 + cos2X - cos2x]
3/2
cos(90 + ø) = sinø
cos²x = (1 + sin2x)/2
we have to
now,
(1 + cos2x)/2 + [1 + cos(2x + )]/2 + [ 1 + cos(2x - )]/2
1/2 [ 1 + cos2x + 1 + cos(2x + ) + 1 + cos(2x - )]
1/2[3 + cos2x + cos(2x + ) + cos(2x - )]
1/2[ 3 + cos2x + (2cos(2x) × cos ]
1/2[3 + cos2x + 2cos2X × cos(120) ]
1/2[3 + cos2X + 2cos2X × cos(90 + 30) ]
1/2[3 + cos2X + 2cos2X × (-sin30)]
1/2[3 + cos2X + 2cos2X × (-1/2)]
1/2[ 3 + cos2X - cos2x]
3/2
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