How do you evaluate #sin (pi/4) sin (pi/6)#?

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Answer 1 · 2016-05-03T10:27:14

#sin(pi/4)xxsin(pi/6)=0.35355#

#sin(pi/4)xxsin(pi/6)#

= #1/sqrt2xx1/2=1/(2sqrt2)=1/4xxsqrt2=1.4142/4=0.35355#

Alternatively, #sin(pi/4)xxsin(pi/6)#

= #1/2{cos(pi/4-pi/6)-cos(pi/4+pi/6)#

= #1/2{cos((3pi-2pi)/12)-cos((3pi+2pi)/12)}#

= #1/2{cos(pi/12)-cos((5pi)/12)}#

= #1/2{0.9659-0.2588}=1/2xx0.7071=0.35355#