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Question

Find the value of sin23π12

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Solution

sin(23π12)
=sin(5π3+π4)
=sin(5π3)cos(π4)+cos(5π3)sin(π4)
=(32)cos(π4)+(12)sin(π4)
=6+24
sin(23π12)=6+24
cos(23π12)=1(6+2)216=1662+21216=4+128
cos(23π12)=12sin2(23π24)
sin(23π12)=(14+128)2
sin(23π12)=4+1216=2+38
sin(23π12)=±2+322

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