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