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How do you use the angle sum or difference identity to find the exact value of #sin((5pi)/12)#?

1 Answer

Shwetank Mauria

Oct 2, 2016

#sin((5pi)/12)=(1+sqrt3)/(2sqrt2)#

Explanation:

As #sin(A+B)=sinAcosB+cosAsinB#

and #(5pi)/12=(2pi+3pi)/12=(2pi)/12+(3pi)/12=pi/6+pi/4#

#sin((5pi)/12)=sin(pi/6+pi/4)=sin(pi/6)cos(pi/4)+cos(pi/6)sin(pi/4)#

= #1/2xx1/sqrt2+sqrt3/2xx1/sqrt2#

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

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