A Level (C3) maths question help needed please?

Question

Given that sinθ = 4sin(θ -60°), show that

(2√3)cosθ = sinθ

Hence find the value of θ such that 0° < θ < 180°

The answer is 74 if that helps

Thanks alot

Answer 1

sinθ = 4 sin (θ - 60°)
Use the identity sin (A - B) = sinAcosB - cosAsinB.
sinθ = 4(sinθcos60 - cosθsin60)
Recall: cos 60° = 1/2, sin 60° = √3/2
sinθ = 4(1/2sinθ-cosθ(√3/2))
sinθ = 2sinθ - (2√3)cosθ
sinθ - 2sinθ = -(2√3)cosθ
-sinθ = -(2√3)cosθ
Hence,
(2√3)cosθ = sinθ.

To find the value of theta, divide both sides by cosθ.
tanθ = 2√3
θ = arctan(2√3)
θ = 73.9 degrees (1 dp)
Rounded off, this is 74, which is the answer you got.

Answer 2

Try expanding the right hand side.
4 sin(θ-60) = 4( sin θ cos 60 -cos θ sin 60)
= 4( ½ sin θ - ½ √3 cos θ)= sin θ
Rearranging gives
2√3 cos θ = sin θ
as required.
Incidentally, that says tan θ = 2√3.
I don't see how to do part b, other than by calculator
Or perhaps by using the Taylor series for arctan x.
My calculator gives 73.897 for the answer.

Answer 3

sinθ = 4sin(θ -60°),

knowing that:
sin (a - b) = sin a cos b - sin b cos a

4sin(θ -60°) = 4(sin θ cos 60 - sin cos θ)=sinθ
4( .5 sin θ - √3/2 cos θ =sinθ
2 sin θ - 2√3 cos θ=sinθ
sinθ = 2√3 cos θ


to find the value of θ:
sinθ = 2√3 cos θ
sinθ / cos θ = 2√3
tan θ = 2√3
θ =Arc tan 2√3
θ = 73.89

Answer 4

Using the Trig. Identity:-
sin(x - 60) = sinxcos60 - cosxsin60
sin(x - 60) = 1/2sinx - 0.8660cosx
So
sinx = 4/2sinx - 4sqrt3/2cosx
sinx = 2sinx - 2sqrt3cosx
2sqrt3cosx = 2sinx - sinx
2sqrt3cosx = sinx (As required).
2sqrt3 = sinx/cosx
2sqrt3 = tanx
x = tan^-1(2sqrt3)
x = 73.89788625 degrees (74 degrees to nearest degree).

Hope that helps !!!!