Question #d07cf

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Answer 1 · 2016-12-11T09:25:05

#sin(-pi/12)#

#=-sin(pi/12)#

#=-sqrt(1/2(1-cos((2xxpi)/12)#

#=-sqrt(1/2(1-cos(pi/6)#

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

#=-sqrt(1/8(4-2sqrt3)#

#=-sqrt(1/8((sqrt3)^2+1^2-2sqrt3xx1)#

#=-sqrt(1/8((sqrt3-1)^2)#

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

Answer 2 · 2016-12-12T19:07:44

#(sqrt2-sqrt6)/4#

#sin(-pi/12) =sin(pi/6-pi/4)#

Now use the formula #sin(A-B) =sin A cos B-cos A sin B# to evaluate #sin(pi/6-pi/4)#. That is,

#sin(pi/6-pi/4)=sin (pi/6) cos (pi/4) - cos (pi/6) sin (pi/4)#

#=1/2*sqrt2/2 - sqrt3/2*sqrt2/2#

#=sqrt2/4-sqrt6/4#

#:=(sqrt2-sqrt6)/4#