heya!!Here is a question from class11 maths trigonometry
➡CONTENT QUALITY NEEDED⬅
Q.

Pls solve with explanation
VijayaLaxmiMehra1:
Now check my answer
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Answered by
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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 +
1/2 [ 1 + cos2x + 1 + cos(2x +
1/2[3 + cos2x + 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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