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

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Discussion Overview

The discussion revolves around expressing the trigonometric functions cos(2 tan^-1(x/4)) and sin(2 tan^-1(x/4)) as algebraic expressions in terms of x. Participants are exploring the relationships between the trigonometric identities and the inverse tangent function, with a focus on deriving these expressions correctly.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents their derived expressions for cos(2 tan^-1(x/4)) and sin(2 tan^-1(x/4)) as cos(theta) = 8*sqrt(x^2+64)/(x^2+64) and sin(theta) = x*sqrt(x^2+64)/(x^2+64).
  • Another participant suggests using the identities cos(2t) = 2cos^2(t) - 1 and sin(2t) = 2sin(t)cos(t) after defining t = tan^-1(x/4), providing expressions for sin(t) and cos(t) in terms of x.
  • A participant questions the interpretation of 2*tan^-1(x/4), suggesting a potential simplification to x/8, which is challenged by another participant who clarifies the distinction between the angle and its double.

Areas of Agreement / Disagreement

Participants express differing interpretations of the relationship between the angles and their tangent functions, leading to a lack of consensus on the correct approach to derive the expressions.

Contextual Notes

Some participants rely on specific trigonometric identities and relationships, which may depend on the correct interpretation of the angles involved. There are unresolved steps in the derivation process, particularly regarding the manipulation of the inverse tangent function.

Elissa89
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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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