how to solve sin (22 1/2)degrees
09991506401:
Use the Half-Angle Formula for Sine which follows directly from the Double Angle Formula for Cosine which follows directly from the Sum Formula for Cosine. See Double Angle and Half Angle Formulas for a complete discussion.
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Answered by
76
We know that
Cos 2θ=Cos^2θ - sin^2θ
Cos2θ = 1-sin^θ -sin^2 θ
Cos 2θ = 1- 2sin^2 θ
2sin^2 θ = 1- cos 2θ
Therefore
Sin ^2 θ = (1-cos 2θ)/2----(1)
Let θ = 22 1/2
Put θ value in (1)
Sin ^2 221/2= (1- cos 2×221/2)/2
=(1-cos 45)/2
=(1 - 1/root2)/2
=(root2- 1)/(2root2)
Therefore
Sin 221/2= sqrt[(sqrt2 - 1)/2sqrt2]
Cos 2θ=Cos^2θ - sin^2θ
Cos2θ = 1-sin^θ -sin^2 θ
Cos 2θ = 1- 2sin^2 θ
2sin^2 θ = 1- cos 2θ
Therefore
Sin ^2 θ = (1-cos 2θ)/2----(1)
Let θ = 22 1/2
Put θ value in (1)
Sin ^2 221/2= (1- cos 2×221/2)/2
=(1-cos 45)/2
=(1 - 1/root2)/2
=(root2- 1)/(2root2)
Therefore
Sin 221/2= sqrt[(sqrt2 - 1)/2sqrt2]
Answered by
47
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