Given that arccos x = 2 arcsin x, Find the value of arctan x, giving your answer in radians?

Answer 1

suppose that arccos x = 2 arcsin x = θ.

since θ = arccos x, that means x = cos θ. since θ = 2 arcsin x, that means x = sin(θ/2).

so we need to solve for cos θ = sin(θ/2). the half-angle formula says that sin(θ/2) = √((1-cos θ)/2), so we have
cos θ = ±√((1-cos θ)/2).
it's kind of messy, but at least it's an equation where we can solve for cos θ.

square both sides and bring all the terms to one side:
cos²θ + 1/2 cos θ - 1/2 = 0.
now factor:
(cosθ +1)(cos θ - 1/2) = 0
so cosθ = -1 or 1/2.

remember that x = cos θ, so x = -1 or 1/2. now you just find arctan(-1) and arctan(1/2).

Answer 2

Lets think in English not Math for a moment. ArcCos of X means the angle that has a cosine of X, and 2 ArcSin of X means twice the angle that has a sine of X. We'll call that angle A. So converting the English above to math we get

Cos(A) = X
Sin(2A) = X
or
Cos(A) = Sin(2A)
This is true for A = Pi/2 (both are Zero at that point).
So Cos(Pi/2) = X = 0
Sin(2*Pi/2)=Sin(Pi) = X = 0

So, what's the arctan(0) = ? or the angle whose tangent is zero? Clearly 0 radians works as would Pi.

Answer 3

If you take the cosine of both sides of this equation, you get:

x = cos (2arcsinx)

Let arcsinx = A, so we really have
x = cos (2A)
x = 1-2sin²A
Since sin A = x from above, we have
x = 1 - 2x²
2x² + x - 1 = 0
(2x-1)(x+1) = 0
x = 1/2, -1

You are asked to find arctan x:
arctan (1/2) = .4636
arctan (-1) = -π/4

Answer 4

arccos x = pi/2 - arcsin (x) = 2 arcsin (x)

(i) pi/2 - arcsin (x) = 2 arcsin (x)

pi/2 = 3 arcsin(x)

arcsin (x) = pi/6

x = 0.5

arctan x = 0.4636...