Question

In triangle ABC is sin cube A + sin cube b + sin cube c equal to 3 sin a.sin b.sin c then triangle is​

Answer 1

Answer:

MATHS

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Asked on December 20, 2019 by

Prakrati Sah

In any triangle ABC,

sin

3

Acos(B−C)+sin

3

Bcos(C−A)+sin

3

Ccos(A−B)=msinAsinBsinC. Find m

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ANSWER

Given:

A+B+C=π

∴A=π−(B+C)

LHS=sin

3

Acos(B−C)+sin

3

Bcos(C−A)+sin

3

Ccos(A−B)

=sin

2

Asin(B+C)cos(B−C)+sin

2

Bsin(C+A)cos(C−A)+sin

2

Csin(A+B)cos(A−B)

=

2

1

sin

2

A(sin2B+sin2C)+

2

1

sin

2

B(sin2C+sin2A)+

2

1

sin

2

C(sin2A+sin2B)

=sin

2

A(sinBcosB+sinCcosC)+sin

2

B(sinCcosC+sinAcosA)+sin

2

C(sinAcosA+sinBcosB)

=sinAsinB(sinAcosB+sinAcosB)+sinBsinC(sinBcosC+sinBcosC)+

sinCsinA(sinCcosA+sinCcosA)

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

=3sinAsinBsinC= RHS

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