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

Question

2960 views around the world

You can reuse this answer
Creative Commons License

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#