sin alpha!=3/5 90<alpha<180 then cos 3alpha=?
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Given : Sinα = 3/5
90° < α < 180°
To Find : cos3α
Solution:
90° < α < 180° Hence α lies in 2nd Quadrant
in 2nd Quadrant Cos is negative
Sinα = 3/5
cos²α = 1 - Sin²α
=>cos²α = 1 - (3/5)²
=> cos²α = (4/5)²
=> cosα = ± 4/5
in 2nd Quadrant Cos is negative
=> cosα = - 4/5
cos3α = 4cos³α - 3cosα
=> cos3α = 4(-4/5)³ - 3(-4/5)
=> cos3α = -256/125 + 12/5
=> cos3α = -256/125 + 300/125
=> cos3α = 44/125
cos3α = 44/125 = 0.352
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