Math, asked by baidikkoirala005, 16 days ago

sinA=3sin(A/3)-4sin^3(A/3).Please solve this eqn step by step.​

Answers

Answered by bathwar23
2

Answer:

sin3A=sin3(30

o

)=sin90

o

=1

3sinA−4sin

3

A=3sin30

o

−4sin

3

30

o

=3(

2

1

)−4(

2

1

)

3

=

2

3

2

1

=1

Step-by-step explanation:

We have to prove sin3A = 3sinA – 4sin3A

Proof

sin 3A can be expressed assin (2A + A)

sin(2A + A)

= sin2A. cosA + cos2A. sinA

= 2sinA.cosA.cosA + (cos2A – sin2A) sinA

= 2sinA.cosA.cosA + (1 – 2 sin2A) sinA

sin2A = 2sinA. cosA

cos2A = cos2A – sin2A = 1 – 2sin2A = 2cos2A – 1

= 2sinA. cos2A + sinA – 2sin3A

= 2sinA(1- sin2A) + sinA – 2sin3 A

= 2 sinA – 2 sin3A + sinA – 2sin3A

= 3sin – 4sin3A

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