How do you find the exact value of #sin((3pi)/4+(5pi)/6)#?

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Answer 1 · 2017-01-28T12:41:17

#-(sqrt6+sqrt2)/4#

You could apply the formula:

#sin(alpha+beta)=sinalphacosbeta+cosalphasinbeta#

Then:

#sin(3/4pi+5/6pi)=sin(3/4pi)cos(5/6pi)+cos(3/4pi)sin(5/6pi)#

#=sqrt2/2(-1/2)+(-sqrt2)/2xxsqrt3/2#

#=-sqrt(2)/4-sqrt(6)/4#

#-(sqrt6+sqrt2)/4#