1) find the value of : 4cos20 - √3 cot20
2) 2cos40 - cos20sin20
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2 Answers
11)
let 20 = θ
we are asked to find cosθ(4sinθ-√3)sinθ
we have 3sinθ-4sin3θ = √32 ....(1)
and 4cos3θ - 3cosθ = 12 .....(2)
from (2), 4sinθ - √3 = 2sinθ(4sin2θ-1)
=> cosθ(4sinθ-√3)sinθ = 2cosθ(4sin2θ-1)1 = 2(3cosθ-4cos3θ)1
= -1
2622) \frac{cos40+(cos40-cos20)}{sin20}
=\frac{cos40-2sin30sin10}{sin20}
=\frac{cos40-cos80}{sin20}
=\frac{2sin60sin20}{sin20}
=\sqrt{3}