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Question Number 23419 by mondodotto@gmail.com last updated on 30/Oct/17

Answered by $@ty@m last updated on 31/Oct/17

Solution (a) LHS=cos A+cos B+cos C =2cos ((A+B)/2)cos ((A−B)/2)+cos C =2cos ((π−C)/2)cos ((A−B)/2)+cos C =2sin (C/2)cos ((A−B)/2)−2sin^2 (C/2)+1 =2sin (C/2)(cos ((A−B)/2)−sin (C/2))+1 =2sin (C/2)(cos ((A−B)/2)−sin ((π−(A+B))/2))+1 =2sin (C/2)(cos ((A−B)/2)−cos ((A+B)/2))+1 =2sin (C/2).(2sin (B/2)sin (A/2))+1 =4sin (A/2)sin (B/2)sin (C/2)+1 =RHS (Pl. recheck question)

Solution(a)LHS=cosA+cosB+cosC=2cosA+B2cosA−B2+cosC=2cosπ−C2cosA−B2+cosC=2sinC2cosA−B2−2sin2C2+1=2sinC2(cosA−B2−sinC2)+1=2sinC2(cosA−B2−sinπ−(A+B)2)+1=2sinC2(cosA−B2−cosA+B2)+1=2sinC2.(2sinB2sinA2)+1=4sinA2sinB2sinC2+1=RHS(Pl.recheckquestion)

Answered by $@ty@m last updated on 31/Oct/17

Solution (b) LHS=cos 2A+cos 2B+cos 2C =2cos (A+B)cos (A−B)+cos 2C =2cos (π−C)cos (A−B)+cos 2C =−2cos Ccos (A−B)+2cos^2 C−1 =−2cos C{cos (A−B)−cos C}−1 =−2cos C(2cosAcos B) −1 =−4cos CcosAcos B −1

Solution(b)LHS=cos2A+cos2B+cos2C=2cos(A+B)cos(A−B)+cos2C=2cos(π−C)cos(A−B)+cos2C=−2cosCcos(A−B)+2cos2C−1=−2cosC{cos(A−B)−cosC}−1=−2cosC(2cosAcosB)−1=−4cosCcosAcosB−1

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