How do you find the exact value of #sin(135^circ-30^circ)#?

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Answer 1 · 2017-02-03T19:23:34

#sin(135-30)=(sqrt3+1)/(2sqrt2)#

#sin(a-b)=sinacosb-sinbcosa#

#sin30=1/2#

#cos30=sqrt3/2#

#sin(135-30)=sin135cos30-cos135sin30=sqrt(3)/2sin135-1/2cos135#

as #sin135=sin(180-45)=sin45=1/sqrt2#

and #cos135=cos(180-45)=-cos45=-1/sqrt2#

Hence #sin(135-30)=sqrt3/2xx1/sqrt2-1/2xx-(-1/sqrt2)=(sqrt3+1)/(2sqrt2)#