prove that : cos28°-sin58°=sin2°
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Hi friend!!
cos28°-sin58°
=cos 28°-sin(90-32)
since, sin(90-A)=cosA
=cos28°-cos32°
By using the identity,
cosA-CosB=-2sin(A+B/2)sin(A-B/2)
we get,
=-2sin(60°/2)sin(-4°/2)
=-2sin30°sin(-2°)
since, sin30°=1/2
=-2×1/2×sin(-2°)
=-sin(-2°)=sin2°
Hence proved.
I hope this will help u ;)
cos28°-sin58°
=cos 28°-sin(90-32)
since, sin(90-A)=cosA
=cos28°-cos32°
By using the identity,
cosA-CosB=-2sin(A+B/2)sin(A-B/2)
we get,
=-2sin(60°/2)sin(-4°/2)
=-2sin30°sin(-2°)
since, sin30°=1/2
=-2×1/2×sin(-2°)
=-sin(-2°)=sin2°
Hence proved.
I hope this will help u ;)
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