Given $sin60^circ=sqrt3/2$ and $cos60^circ=1/2$ , how do you find $sin30^circ$ ?

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Answer 1 · 2017-03-15T04:37:22

$sin 30^o = 1/2$

by using identity proving,

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

$2 sin 30^o cos 30^o = sqrt 3/2$

$sin 30^o cos 30^o = sqrt 3/4$ --->a

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

$cos^2 30^o - sin^2 30^o = 1/2$

$2 cos^2 30^o - 1 = 1/2$ , where $sin^2 30^o = 1 - cos^2 30^o$

$cos^2 30^o = 3/4$

$cos 30^o = sqrt 3/2$ --->b

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

$sin 30^o = sqrt 3/4 * 2/ sqrt 3 = 1/2$

we also can use as

$sin 30^o = cos (90^o -30^o) = cos 60^o = 1/2$

Given sin60^circ=sqrt3/2sin60^circ=sqrt3/2 and cos60^circ=1/2cos60^circ=1/2, how do you find sin30^circsin30^circ?