Simplify sin(arccos2x)?

Question

Please show your work

Answer 1

let y = sin [ (cos^(-1) (2 x) ]
let θ = cos^(-1) (2 x)
θ = sin^(-1)(√(1 - 4 x ²)
sin θ = √ (1 - 4 x ² )
y = sin θ
y = √(1 - 4 x ²)

Answer 2

The answer is: sin(arccos2x) = +/- sqrt (1 - 4x^2), depending on what quadrant the angle arccos2x is in.

[There's a convention --- regrettably not always followed --- that Arccos (with the CAPITAL LETTER) gives you the "Principal Value," lying between 0 and π radians or 180 degrees, while arccos has an unrestricted range. Because of the possibility of ambiguity, it is best to state whether the inverse trig functions are to be interpreted as principal values or not.]

If angle A = arccos2x, that means that cos A = 2x.

But sin^2 A + cos^2 A = 1, so that sin^2 A = 1 - cos^2 A,

giving sin(arccos2x) = +/- sqrt (1 - 4x^2),

depending on what quadrant the angle arccos2x is in.

However, sin (Arccos2x) would be + sqrt (1 - 4x^2); the Arccos function then lying between 0 and π radians or 180 degrees, the sine of any angle in that range is positive.

Live long and prosper.

Answer 3

1 diamond= 1x

1x diamond(1 carat2x) = 1 big bling

Answer 4

arccos2x=A

cosA=2x

sinA=(1-cos^2A)^1/2

=(1-4X^2)^1/2