Question

cos(2tan-1(x))

Answer 1

its arc-tan, the -1 is suppose to be a ^-1

Answer 2

must use
\[cos(2\theta)=cos^2(\theta)-sin^2(\theta)\] so you need
\[cos(tan^{-1}(x))\] and \[sin(tan^{-1}(x))\]

Answer 3

do you know how to find them?

Answer 4

no

Answer 5

ok do this {
draw a right triangle and label one angle as theta,
theta =\[tan^{-1}(x)\] the angle whose tangent is x

Answer 6

since tangent is \[\frac{opp}{adj}\] label the opposite angle x and the adjacent angle as 1 so you have \[\frac{x}{1}=x\] as the tangent of your angle

Answer 7

by pythagoras the hypotenuse is\[\sqrt{x^2+1}\]

Answer 8

\[sin(tan^{-1}(x))=\frac{opp}{hyp}=\frac{x}{\sqrt{x^2+1}}\]

Answer 9

\[cos(tan^{-1}(x))=\frac{adj}{hyp}=\frac{1}{\sqrt{x^2+1}}\]

Answer 10

now use the "double angle" formula to find \[cos(2tan^{-1}(x))\]

Answer 11

it is the top line i wrote. calling \[tan^{-1}(x)=\theta\] it is
\[cos^2(\theta)-sin^2(\theta)\]

Answer 12

you get \[\frac{1}{x^2+1}-\frac{x^2}{x^2+1}=\frac{1-x^2}{x^2+1}\]

Answer 13

the important part is being able to find \[sin(tan{-1}(x))\]
or
\[sin(tan{-1}(x))\]

Answer 14

i think i made a mistake on the last post where i said
\[cos(tan^{-1}(x))\] was something else. it is \[\frac{1}{\sqrt{x^2+1}}\]