How do you simplify $2 sin 35 cos 35$ $2 sin 35 cos 35$ ?

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Answer 1 · 2015-05-12T21:08:42

Use $e^{i θ} = cos θ + i sin θ$ $e^(i theta) = cos theta + i sin theta$

$cos 2 θ + i sin 2 θ$ $cos 2theta + i sin 2theta$

$= e^{2 i θ}$ $= e^(2 i theta)$

$= e^{i θ} e^{i θ}$ $= e^(i theta)e^(i theta)$

$= (cos θ + i sin θ) (cos θ + i sin θ)$ $= (cos theta + i sin theta)(cos theta + i sin theta)$

$= ({cos}^{2} θ - {sin}^{2} θ) + i (2 cos θ sin θ)$ $= (cos^2 theta - sin^2 theta) + i (2 cos theta sin theta)$

Looking at the coefficients of $i$ $i$ , we get

$sin (2 θ) = 2 cos θ sin θ = 2 sin θ cos θ$ $sin (2 theta) = 2 cos theta sin theta = 2 sin theta cos theta$

So $2 {sin 35}^{o} {cos 35}^{o} = sin (2 \cdot 35^{o}) = {sin 70}^{o}$ $2 sin 35^o cos 35^o = sin ( 2*35^o ) = sin 70^o$

How do you simplify 2sin35cos352 sin 35 cos 35?