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Question

Prove that:
(i) 2 sin 5π12 sinπ12=12
(ii) 2 cos 5π12 cosπ12
(iii) 2 sin 5π12 cos 5π12=3+22

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Solution

(i) 2 sin 5π12 sinπ12=122 sin A sin B=cos(AB)cos(A+B)2 sin 5π12 sin π12=cos(5π12π12)cos(5π12+π12)=cos(4π12)cos(6π12)=cos(π3)cos(π2)=120=12 RHS

(ii) 2 cos 5π12 cosπ12=122 cos A cos B=cos(A+B)+cos(AB)2 sin 5π12 sin π12=cos(5π12+π12)+cos(5π12π12)=cos(π2)+cos(π3)=0+12=12=RHS

(iii) 2 sin 5π12 cos π12=3+222 sin A cos B=sin(A+B)+sin (AB)=(5π12+π12)+sin(5π12π12)=sinπ2+π3=12+32=2+32=RHS
(Taking LCM)


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