How do you find the exact value of #sin60°cos30° + sin30°cos60°#?

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Answer 1 · 2015-04-26T02:08:40

#30^o " and " 60^o "are angles of one of the standard triangles"#

#sin(30^o) = 1/2#

#cos(30^o) = sqrt(3)/2#

#sin(60^o) = sqrt(3)/2#

#cos(60^o) = 1/2#

So
#sin(60^o)*cos(30^o) + sin(30^o)*cos(60^o)#

#=(sqrt3/2)(sqrt3/2) +(1/2)(1/2) = 3/4 +1/4#

#=1#