if sin alpha =3/5 and 90°<alpha<180°,then cos3alpha =
Answers
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
Learn More:
Ifsin 990° sin 780° sin 390°+=K(tan 405° – tan 360°) then Kicos 540 ...
brainly.in/question/22239477
Maximum value of sin(cos(tanx)) is(1) sint (2)
brainly.in/question/11761816
evaluate cot[90-theta].sin[90-theta] / sin theta+cot 40/tan 50 - [cos ...
brainly.in/question/2769339
Answer:
`(tan(18 0^(@)-alpha)cos(180^(@)-alpha)tan(90^(@)-alpha))/(sin(9 0^(@)+alpha)cos(9 0^(@)-alpha)tan(9 0^(@)+alpha)`, expression wherever it is defined, is equal to