Answer
Ans. $sin3θ=3sinθ-4sin^3θ$
$cos3θ=4cos^3θ-3cosθ$
Work Step by Step
Sol:- find out the value of sin3θ and cos3θ
a. $sin3θ=sin(2θ+θ)$
=$sin2θcosθ+cos2θsinθ$
We know that
$Sin2θ=2sinθcosθ$ , $cos2θ=cos^2θ-sin^2θ$
=$2sinθcosθ+(cos^2θ-sin^2θ)sinθ$
=$3sinθcos^2θ-sin^3θ$
Putting $cos^2θ=1-sin^2θ$
=$3sinθ(1-sin^2θ)-sin^3θ$
=$3sinθ-4sin^3θ$
b. $Cos3θ=cos(2θ+θ)$
=$cos2θcosθ-sin2θsinθ$
We know that
$Sin2θ=2sinθcosθ$, $cos2θ=cos^2θ-sin^2θ$
=$(cos^2θ-sin^2θ) cosθ-2sinθcosθsinθ$
=$cos^3θ-3sin^2θcosθ$
putting the value of$sin^2θ=1-cos^2θ$
=$4cos^3θ-3cosθ$
Ans.
