- Thread starter
- #1

Express cos(2 tan^-1(x/4)) and sin(2tan^-1(x/4) as an algebraic expression in x

I got:

cos(theta)=8*sqrt(x^2+64)/x^2+64

sin(theta)=x*sqrt(x^2+64)/x^2+64

- Thread starter Elissa89
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- Thread starter
- #1

Express cos(2 tan^-1(x/4)) and sin(2tan^-1(x/4) as an algebraic expression in x

I got:

cos(theta)=8*sqrt(x^2+64)/x^2+64

sin(theta)=x*sqrt(x^2+64)/x^2+64

- Mar 1, 2012

- 639

Let $t = \tan^{-1}\left(\dfrac{x}{4}\right) \implies \tan{t} = \dfrac{x}{4}, \, \cos{t} = \dfrac{4}{\sqrt{x^2+16}}, \, \sin{t} = \dfrac{x}{\sqrt{x^2+16}}$

Express cos(2 tan^-1(x/4)) and sin(2tan^-1(x/4) as an algebraic expression in x

I got:

cos(theta)=8*sqrt(x^2+64)/x^2+64

sin(theta)=x*sqrt(x^2+64)/x^2+64

$\cos(2t) = 2\cos^2{t}-1$

$\sin(2t) = 2\sin{t}\cos{t}$

take it from here?

- Thread starter
- #3

but its 2*tan^-1(x/4). Isn't that the same as 2*tan(theta)=x/4. So wouldn't I divide both sides by 2 and get x/8 and go from there?Let $t = \tan^{-1}\left(\dfrac{x}{4}\right) \implies \tan{t} = \dfrac{x}{4}, \, \cos{t} = \dfrac{4}{\sqrt{x^2+16}}, \, \sin{t} = \dfrac{x}{\sqrt{x^2+16}}$

$\cos(2t) = 2\cos^2{t}-1$

$\sin(2t) = 2\sin{t}\cos{t}$

take it from here?

- Mar 1, 2012

- 639

but its 2*tan^-1(x/4).Isn't that the same as 2*tan(theta)=x/4. So wouldn't I divide both sides by 2 and get x/8 and go from there?

note $\tan^{-1}\left(\dfrac{x}{4}\right)$ is an angle and $2 \tan^{-1}\left(\dfrac{x}{4}\right)$ is double that angle