Question

1. If B = 30°, prove that 3 sin ß-4 sin3 B = sin 3B.​

Answer 1

Answer:

MATHS

In a triangle ABC sin

3

Acos(B−C)+sin

3

Bcos(C−A)+sin

3

Ccos(A−B) is equal to

December 30, 2019avatar

Nikil Punjabi

SHARE

ANSWER

The value of sin

3

Acos(B−C)=sin

2

Asin(B+C)sin(B−C)

=

2

1

sin

2

A[sin2B+sin2C]=sin

2

A[sinBcosB+sinCcosC]

So the given expression is equal to

sinAsinB(sinAcosB+cosAsinB)+sinBsinC(sinBcosC+cosBsinC)

+sinCsinA(sinAcosC+cosAsinC)

=sinAsinBsin(A+B)+sinBsinCsin(B+C)+sinCsinAsin(C+A)

=3sinAsinBsinC