Solve tan(cos^-1 0.5): Find Exact Value

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Homework Help Overview

The problem involves finding the exact value of the function tan(cos^-1(0.5)), which relates to trigonometric functions and inverse trigonometric functions.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relationship between the angles and their trigonometric functions, with some attempting to express the problem in terms of known values like π/3. Questions arise regarding the notation used, such as the change from x to N.

Discussion Status

The discussion is ongoing, with various approaches being explored, including direct evaluation of the tangent function and geometric interpretations involving right triangles. Some participants provide guidance on how to proceed with the calculations.

Contextual Notes

There is a mention of using specific trigonometric identities and the need to clarify notation, as well as the potential for multiple methods to arrive at the solution.

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Homework Statement


Find the exact value of the function:
tan(cos^-1 0.5)


Homework Equations



I used this formula: y=sin^-1=sin y=x for the first part, how do I solve the second part of the problem tanx (x)?

The Attempt at a Solution


x=cos^-1 (0.5)
cos(N)=0.5
N= pi/3
Then I tried tanx(x)=
I'm stuck
 
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Why did you change x to N?
 
[tex]\tan{(\cos^{-1}\frac 1 2)}=\tan x[/tex]

[tex]x=\cos^{-1}\frac 1 2[/tex]

[tex]x=\frac{\pi}{3}[/tex]

Ok, you have x? So your equation just ... [tex]\tan x[/tex] ... solve for tangent!
 
D H said:
Why did you change x to N?
Don't know? good question, but I see that I just have to plug in pi/3 into tanx (x)
 
[tex]tan(x)= tan(\pi/3)= \frac{sin(\pi/3)}{cos(\pi/3)}[/itex]<br /> Surely you know [itex]sin(\pi/3)[/itex] and [itex]cos(\pi/3)[/itex]![/tex]
 
Last edited by a moderator:
HallsofIvy said:
[/tex]tan(x)= tan(\pi/3)= \frac{sin(\pi/3)}{cos(\pi/3)}[/itex]
Surely you know [itex]sin(\pi/3)[/itex] and [itex]cos(\pi/3)[/itex]!

The first line of ivy should be:
[tex]tan(x)= tan(\pi/3)= \frac{sin(\pi/3)}{cos(\pi/3)}[/itex][/tex]
 
There's another way to do this. Draw a right triangle that has a angle of cos^(-1)(1/2) in it. E.g. hypotenuse 1 and adjacent side 1/2. Use Pythagoras to find the missing side. Now find tan by opposite/adjacent.
 
Thank you!
 

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