How do you use the angle sum identity to find the exact value of #sin255#?

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Answer 1 · 2016-07-23T06:18:11

#sin255^o=-(1+sqrt3)/(2sqrt2)=-0.9659#

#sin255^o=sin(135^o +120^o)#

= #sin135^ocos120^o +cos135^osin120^o#

Now as #sin135^o=sin45^o=1/sqrt2#, #cos135^o=-cos45^o=-1/sqrt2#

#sin120^o=sin60^o=sqrt3/2# and #cos120^o=-cos60^o=-1/2#

#sin255^o=1/sqrt2xx(-1/2)+(-1/sqrt2)xxsqrt3/2#

= #-1/(2sqrt2)-sqrt3/(2sqrt2)#

= #-(1+sqrt3)/(2sqrt2)=-(sqrt2(1+sqrt3))/4=-(sqrt2+sqrt6)/4#

= #-(1.4142+2.4495)/4=-0.9659#