Question

For any triangle abc, prove that acosa + bcosb + ccosc = 2asinbsinc

Answer 1

Here is your answer buddy,

aCosA+bCosB+cCosC = 2aSinBCosC

aCosA+bCosB+cCosC 

=2R sinA cosA +2R SinB CosB + 2R Sin C CosC (From Sine rule)

=R(2 sinA cosA +2 SinB CosB + 2 Sin C CosC)

=R (Sin2A+Sin2B+Sin2C)

=R(2Sin (A+B)Cos(A-B)+2SinCCosC)

=R [ 2 SinC Cos(A-B) + 2SinC CosC]

=2RSinC[Cos(A-B)+CosC]

=2RSinC[Cos(A-B)-Cos(A+B)]

=2R SinC [ 2 SinA Sin B]

=(2R SinA) (2SinB SinC)

=a (2 SinB SinC)

=2a Sin B SinC

Hope this helps.
Be Brainly.

Answer 2

$\textbf{\underline{\underline{According\:to\:the\:Question}}}$

Using rule of sine

$\tt{\rightarrow\dfrac{a}{sinA}=\dfrac{b}{sinB}=\dfrac{c}{sinC}=p}$

a = psinA

b = psinB

c = psinC

Now,

LHS

= acosA + bcosB + ccosC

= p sinA cos A + psinB cosB + psinC cos C

$\tt{\rightarrow\dfrac{1}{2}p(sin2A+sin2B+sin2C}$

$\tt{\rightarrow\dfrac{1}{2}p(2sin(A+B)cos(A-B)+2sinCcosC}$

= p{sin(A + B) cos(A - B) + sinC cosC}

= p{sin(π - C) cos(A - B) + sinC cosC}

= psin C{cos(A - B) + cos C}

= p sin C[cos (A - B) + cos{π - (A + B)}]

= psin C{cos(A - B) - cos(A + B)}

= psin C × 2sinAsinB

= 2(psinA)sinBsinC

= 2asinBsinC

Hence proved

$\Large{\boxed{\bigstar{{LHS = RHS}}}}$