Question

In ΔABC, prove that a cos A + b cos B + c cos C = 4R sin A sin B sin C.

Answer 1

Solution :

LHS = acosA + bcosB + ccosC

= 2RsinAcosA

+ 2RsinBcosB

+ 2RsinCcosC

= R[ 2sinAcosA+2sinBcosB+2sinCcosC]

= R[ sin 2A + sin 2B + sin 2C ]

{ In ∆ABC , A + B + C = 180° }


= R{2sin[(2A+2B)/2]cos[(2A-2B)/2]+sin2C}

= R[ 2sin(A+B)cos(A-B)+sin2C ]

= R[2sinCcos(A-B)+2sinCcosC]

=2RsinC[cos(A-B) + cosC ]

= 2RsinC[cos(A-B) - cos(A+B)]

=2RsinC{ 2sin[(A-B+A+B)/2]sin[(A+B-A+B)/2]

= 2RsinC( 2sinAsinB )

= 4RsinAsinBsinC

= RHS

••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

now, acosA + bcosB + ccosC

= 2RsinAcosA + 2RsinBcosB + 2RsinCcosC

=> R[2sinAcosA + 2sinBcosB + 2sinCcosC]

=> R[ sin 2A + sin 2B + sin 2C ]

[ In ∆ABC , A + B + C = 180° ]

=> R{2sin[(2A+2B)/2]cos[(2A-2B)/2]+sin2C}

=> R[ 2sin(A+B)cos(A-B)+sin2C ]

=> R[2sinCcos(A-B)+2sinCcosC]

=> 2RsinC[cos(A-B) + cosC ]

=> 2RsinC[cos(A-B) - cos(A+B)]

=> 2RsinC{ 2sin[(A-B+A+B)/2]sin[(A+B-A+B)/2]

=> 2RsinC( 2sinAsinB )

=> 4RsinAsinBsinC

I HOPE IT'S HELP YOU DEAR,

THANKS