Asked by Kewal
Prove that
cos(A+B)cosC - cos(B+C)cosA = sinBsin(C-A)
cos(A+B)cosC - cos(B+C)cosA = sinBsin(C-A)
Answers
Answered by
Reiny
LS
= (cosAcosB - sinAsinB)cosC - (cosBcosC- sinBsinC)(cosA)
= cosAcosBcosC - sinAsinBcosC - cosAcosBcosC + cosAsinBsinC
= cosAsinBsinC - sinAsinBcosc
RS = sinB(sinCcosA - cosCsinA)
= cosAsinBsinC -sinAsinBcosC
= LS
= (cosAcosB - sinAsinB)cosC - (cosBcosC- sinBsinC)(cosA)
= cosAcosBcosC - sinAsinBcosC - cosAcosBcosC + cosAsinBsinC
= cosAsinBsinC - sinAsinBcosc
RS = sinB(sinCcosA - cosCsinA)
= cosAsinBsinC -sinAsinBcosC
= LS
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