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Derive function of 120 degree from the functions of 60 degree and check by using relation between functions of supplementary angles.
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Take θ = 60.
Now, sin 2 θ = 2 sin θ . cos θ⇒sin 120 = 2 sin 60 .
cos 60 = 2×3√2×12 = 3√2Now, sin 120 = sin (180−60) = sin 60 =3√2cos 2θ = 2 cos2θ−1 ⇒cos 120 = 2 cos260 − 1 = 2 × (12)2 − 1 = − 12Now,
cos 120 = cos(180−60) = − cos 60 = − 12Now, tan 2θ = sin 2θcos 2θ = 2 sin θ . cos θ2 cos2θ−1 tan 120 = 2 sin 60 .
cos 60
Now, sin 2 θ = 2 sin θ . cos θ⇒sin 120 = 2 sin 60 .
cos 60 = 2×3√2×12 = 3√2Now, sin 120 = sin (180−60) = sin 60 =3√2cos 2θ = 2 cos2θ−1 ⇒cos 120 = 2 cos260 − 1 = 2 × (12)2 − 1 = − 12Now,
cos 120 = cos(180−60) = − cos 60 = − 12Now, tan 2θ = sin 2θcos 2θ = 2 sin θ . cos θ2 cos2θ−1 tan 120 = 2 sin 60 .
cos 60
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