tan 2A

Find tan θ.

It says π/2 ≤ θ ≤ π.
So draw a terminal side
in quadrant II.

Reference Angle

cos θ = -3/5

Cosine is CAH:
Cosine,
Adjacent side (-3),
Hypotenuse (5).

So draw a right triangle
whose adjacent side is -3
and whose hypotenuse is 5.

See the right triangle.

The sides are -3, (opposite side), and 5.

So this is a [3, 4, 5] right triangle.

So the opposite side is 4.

Pythagorean Triple

Find tan θ.

Tangent is TOA:
Tangent,
Opposite side (4),
Adjacent side (-3).

So tan θ = 4/(-3) = -4/3.

tan θ = -4/3

So tan 2θ = (2⋅[-4/3]) / (1 - [-4/3]²).

2⋅[-4/3] = -8/3

[-4/3]² = 16/9

Multiply 9
to both of the numerator and the denominator.

[-8/3]⋅9 = -8⋅3
(Cancel the denominator 3
and reduce 9 to, 9/3, 3.)

(1 - 16/9)⋅9 = 9 - 16

-8⋅3 = -24

9 - 16 = -7

-24/(-7) = 24/7

So tan 2θ = 24/7.

Tangent means the slope.
And the slope of y = [1/2]x is 1/2.

So tan θ = 1/2.

The central angle of y = mx is
θ + θ = 2θ.

So m = tan 2θ.

tan θ = 1/2

So tan 2θ = (2⋅[1/2]) / (1 - [1/2]²).

2⋅[1/2] = 1

[1/2]² = 1/4

Multiply 4
to both of the numerator and the denominator.

1⋅4 = 4

(1 - 1/4)⋅4 = 4 - 1

4 - 1 = 3

So m = 4/3.