Struggling with two circular functions questions

[dundundundun]

Would really appreciate it if someone could demonstrate how to work these two questions out. Thanks in advance

[wobblywobbly]

Wow! What a tricky little set of questions.

Alright, the first one is:

Let theta = arccos(2x)

Drawing a triangle with theta, the adjacent is 2x, the hypotenuse is 1 (arccos = adjacent/hypotenuse), and the opposite is therefore sqrt(1 - 4x^2)

And since we let theta = arccos(2x), we are left with sin(2 theta), which is equal to 2sin(theta)cos(theta). Working from the triangle above, we get 2 x 2x x  sqrt(1 - 4x^2) = 4x sqrt(1 - 4x^2), so a=4, b=2, C.

Second one is:

Substitute cos^2 = 1 - sin^2 (I'm omitting the theta here for simplicity's sake), so after expanding, you would get

a - a sin^2 + b sin^2 = c ---> sin^2 = (c-a)/(b-a)

Substitute sin^2 = 1 - cos^2, so after expanding, you would get

a cos^2 + b - b cos^2 = c ---> cos^2 = (c-b)/(a-b)

tan^2 = sin^2/cos^2 so

tan^2 = (c-a)/(b-a) x (a-b)/(c-b)
=(c-a)/(b-a) x -(b-a)/(c-b)
=-(c-a)/(c-b)

Distributing the negative on top leads to (a-c)/(c-b) which is E.

[dundundundun]

Wow! What a tricky little set of questions.

Alright, the first one is:

Let theta = arccos(2x)

Drawing a triangle with theta, the adjacent is 2x, the hypotenuse is 1 (arccos = adjacent/hypotenuse), and the opposite is therefore sqrt(1 - 4x^2)

And since we let theta = arccos(2x), we are left with sin(2 theta), which is equal to 2sin(theta)cos(theta). Working from the triangle above, we get 2 x 2x x  sqrt(1 - 4x^2) = 4x sqrt(1 - 4x^2), so a=4, b=2, C.

Second one is:

Substitute cos^2 = 1 - sin^2 (I'm omitting the theta here for simplicity's sake), so after expanding, you would get

a - a sin^2 + b sin^2 = c ---> sin^2 = (c-a)/(b-a)

Substitute sin^2 = 1 - cos^2, so after expanding, you would get

a cos^2 + b - b cos^2 = c ---> cos^2 = (c-b)/(a-b)

tan^2 = sin^2/cos^2 so

tan^2 = (c-a)/(b-a) x (a-b)/(c-b)
=(c-a)/(b-a) x -(b-a)/(c-b)
=-(c-a)/(c-b)

Distributing the negative on top leads to (a-c)/(c-b) which is E.

Thanks so much, that was really helpful I would never have figured them out!

[Special At Specialist]

I got the same answers as wobblywobbly, although I used a slightly different method for the second question:

acos^2(θ) + bsin^2(θ) = c
=> (a - b)cos^2(θ) + bcos^2(θ) + bsin^2(θ) = c
=> (a - b)cos^2(θ) + b = c
=> cos^2(θ) = (c - b) / (a - b)
Since 1 + tan^2(θ) = sec^2(θ) then tan^2(θ) = 1/cos^2(θ) - 1
Thus tan^2(θ) = (a - b) / (c - b) - 1
= (a - b) / (c - b) + (b - c) / (c - b)
= (a - c) / (c - b)
=> Option E

[abcdefg9999]

sorry to bother, do you mind telling me where do you get those 2 questions from?  thanks