Given ${sin 60}^{\circ} = \frac{\sqrt{3}}{2}$ $sin60^circ=sqrt3/2$ and ${cos 60}^{\circ} = \frac{1}{2}$ $cos60^circ=1/2$ , how do you find ${sin 30}^{\circ}$ $sin30^circ$ ?

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Answer 1 · 2017-03-15T04:37:22

${sin 30}^{o} = \frac{1}{2}$ $sin 30^o = 1/2$

by using identity proving,

${sin 60}^{o} = sin (30^{o} + 30^{o}) = 2 {sin 30}^{o} {cos 30}^{o}$ $sin 60^o = sin (30^o + 30^o) = 2 sin 30^o cos 30^o$

$2 {sin 30}^{o} {cos 30}^{o} = \frac{\sqrt{3}}{2}$ $2 sin 30^o cos 30^o = sqrt 3/2$

${sin 30}^{o} {cos 30}^{o} = \frac{\sqrt{3}}{4}$ $sin 30^o cos 30^o = sqrt 3/4$ --->a

${cos 60}^{o} = cos (30^{o} + 30^{o}) = {cos}^{2} 30^{o} - {sin}^{2} 30^{o}$ $cos 60^o = cos (30^o + 30^o) = cos^2 30^o - sin^2 30^o$

${cos}^{2} 30^{o} - {sin}^{2} 30^{o} = \frac{1}{2}$ $cos^2 30^o - sin^2 30^o = 1/2$

$2 {cos}^{2} 30^{o} - 1 = \frac{1}{2}$ $2 cos^2 30^o - 1 = 1/2$ , where ${sin}^{2} 30^{o} = 1 - {cos}^{2} 30^{o}$ $sin^2 30^o = 1 - cos^2 30^o$

${cos}^{2} 30^{o} = \frac{3}{4}$ $cos^2 30^o = 3/4$

${cos 30}^{o} = \frac{\sqrt{3}}{2}$ $cos 30^o = sqrt 3/2$ --->b

replace $b$ $b$ in $a$ $a$
${sin 30}^{o} (\frac{\sqrt{3}}{2}) = \frac{\sqrt{3}}{4}$ $sin 30^o (sqrt 3/2) = sqrt 3/4$

${sin 30}^{o} = \frac{\sqrt{3}}{4} \cdot \frac{2}{\sqrt{3}} = \frac{1}{2}$ $sin 30^o = sqrt 3/4 * 2/ sqrt 3 = 1/2$

we also can use as

${sin 30}^{o} = cos (90^{o} - 30^{o}) = {cos 60}^{o} = \frac{1}{2}$ $sin 30^o = cos (90^o -30^o) = cos 60^o = 1/2$

Given sin60∘=√32sin60^circ=sqrt3/2 and cos60∘=12cos60^circ=1/2, how do you find sin30∘sin30^circ?