Question

In any triangle ABC, prove that a sin (B – C) + b sin (C – A) + c sin (A – B) = 0.

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Answer 1

Step-by-step explanation:

$\mathbb\red{ \tiny A \scriptsize \: N \small \:S \large \: W \Large \:E \huge \: R}$

$\begin{aligned} \rm L.H.S.& \rm=a \cos A+b \cos B+c \cos C \\ \\ & \rm=K \sin A \cos A+K \sin B \cos B+K \sin C \cos C \\ \\ & \rm=\frac{K}{2}[\sin 2 A+\sin 2 B+\sin 2 C] \\ \\ & \rm =\frac{K}{2}[2 \sin (A+B) \cos (A-B)+\sin 2 C] \\ \\ & \rm=\frac{K}{2}[2 \sin C \cos (A-B)+2 \sin C \cos C] \\ \\ & \rm=K \sin C[\cos (A-B)+\cos C] \\ \\ & \rm=K \sin C\left[\cos (A-B)+\cos \left(180^{\circ}-(A+B)\right]\right.\\ \\ & \rm=K \sin C[\cos (A-B)-\cos (A+B)] \\ \\ & \rm=K \sin C[-2 \sin A \sin B] \\ \\ & \rm=2 K \sin A \sin B \sin C \\ \\ & \boxed{\color{olive} \rm=2 a \sin B \sin C } \rm\qquad\qquad (\because \: \: K \sin A = a\:) \\ \\ & \rm=R . H . S . \end{aligned}$

Answer 2

Answer:

L.H.S.

=acosA+bcosB+ccosC

=KsinAcosA+KsinBcosB+KsinCcosC

=

2

K

[sin2A+sin2B+sin2C]

=

2

K

[2sin(A+B)cos(A−B)+sin2C]

=

2

K

[2sinCcos(A−B)+2sinCcosC]

=KsinC[cos(A−B)+cosC]

=KsinC[cos(A−B)+cos(180

∘

−(A+B)]

=KsinC[cos(A−B)−cos(A+B)]

=KsinC[−2sinAsinB]

=2KsinAsinBsinC

=2asinBsinC

(∵KsinA=a)

=R.H.S.

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