Question

sin(2cos-1(3x))

Answer 1

oh is it
\[\sin (2 \cos^{-1} (3x))\]

Answer 2

\[\theta = \cos^{-1} (3x)\]
\[\cos \theta = 3x\]
\[\sin \theta = \pm \sqrt{1-\cos^2 \theta} = \pm \sqrt{1-9x^2}\]
\[\sin 2 \theta = 2 \sin \theta \cos \theta\]
\[\sin 2 \theta = \pm 6x \sqrt{1-9x^2}\]

Answer 3

we can use the identity

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

treating cos^-1(3x) as an angle we have

sin(2cos^-1(3x)) = 2 sin(cos^-1(3x))*cos(cos^-1(3x))

now cos(cos^-1(3x)) that cancels since they are inverses, so it equals to 3x

and sin(cos^-1(3x)) , to figure this out you can draw a triangle

Answer 4

so I get
2 * sqrt( 1 - (3x)^2) * 3x

Answer 5

sorry my draw button does not work, i would draw a triangle

Answer 6

wowowowow! Im lost how did you get this

Answer 7

so are you saying that the sin(and whats in the parenthesis) is the double angle formula...? and so I am suppose to enter cos3x into the double angle formula?

Answer 8

thank you! i got it now!