If cos(θ) = 3x and 0 <= θ <= π/2 which of the following is equal to 4θ+sin(2θ)?

Question

a) 4arccos(3x) + 2x sqrt (1-9x^2)
b) 3arccos(4x) + 2x sqrt (1+9x^2)
c) 4arccos(3x) + 6 sqrt (x^2-9x^4)
d) 3arccos(4x) + 6x sqrt (1+9x^2)
e) 4arccos(3x) + 2 sqrt (1-9x^2)


How in the world do you know? I get the first part you take the arccos of 3x and then you plug it in for theta 4arccos. But I don't understand the rest. I tried using double angle formulas but got nowhere. Thanks.

Answer 1

I think someone posted this question previously. Here is what I said then:

...For the other part of the sum, remember that:

sin(2theta) = 2sin(theta)cos(theta)

We know that cos(theta) = 3x.

Construct a unit right triangle having rotation angle theta.

The x-length = cos(theta) = 3x.

The hypotenuse = 1.

The y length = sin(theta)

(3x)^2 + y^2 = 1
y = sqrt(1 - 9x^2)

So sin(2theta) = 2sin(theta)cos(theta) = 2*sqrt(1 - 9x^2)*3x

This equals 6x*sqrt(1 - 9x^2).

That takes care of the sin(2theta) part. If I think of anything for the 4theta part, I will get back to you :D


...So then, it must be (d)

Answer 2

cos(θ) = 3x
So, θ = arccos(3x)
Therefore, 4θ = 4arccos(3x)

Use the formula : sin^2(θ) + cos^2(θ) = 1
Rearranging gives :
sin(θ) = √[1 - cos^2(θ) ] = √(1 - 9x^2)

Now use the formula : sin(2θ) = 2sin(θ)cos(θ)

sin(2θ) = 2 * √(1 - 9x^2) * 3x = 6x√(1 - 9x^2)

Thus, 4θ+sin(2θ) = 4arccos(3x) + 6x√(1 - 9x^2)

So if that's correct, then none of a) to e) is correct.