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

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The discussion focuses on expressing cos(2 tan^-1(x/4)) and sin(2 tan^-1(x/4)) as algebraic expressions in terms of x. Participants provide their attempts, with one user suggesting expressions involving square roots and fractions. Another user clarifies the relationship between the angle and its tangent, emphasizing that 2 tan^-1(x/4) represents double the angle rather than a simple division of x. The conversation highlights the need for careful manipulation of trigonometric identities to derive the correct expressions. Overall, the thread illustrates the complexities of transforming inverse tangent functions into algebraic forms.
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So my professor gave us a study guide for the final but no there is no answer key. Could someone check my answers please?

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
 
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Elissa89 said:
So my professor gave us a study guide for the final but no there is no answer key. Could someone check my answers please?

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

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?
 
skeeter said:
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?

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?
 
Elissa89 said:
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?

no.

$\theta = 2\tan^{-1}\left(\dfrac{x}{4}\right) \implies \dfrac{\theta}{2} = \tan^{-1}\left(\dfrac{x}{4}\right) \implies \dfrac{x}{4} = \tan\left(\dfrac{\theta}{2}\right)$

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