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

[jimen113]

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

 

[D H]

Why did you change x to N?

 

[rocomath]

\tan{(\cos^{-1}\frac 1 2)}=\tan x

x=\cos^{-1}\frac 1 2

x=\frac{\pi}{3}

Ok, you have x? So your equation just ... \tan x ... solve for tangent!

 

[jimen113]

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)

 

[HallsofIvy]

tan(x)= tan(\pi/3)= \frac{sin(\pi/3)}{cos(\pi/3)}[/itex]<br /> Surely you know sin(\pi/3) and cos(\pi/3)!

 

[dirk_mec1]

[/tex]tan(x)= tan(\pi/3)= \frac{sin(\pi/3)}{cos(\pi/3)}[/itex]
Surely you know sin(\pi/3) and cos(\pi/3)!

The first line of ivy should be:
tan(x)= tan(\pi/3)= \frac{sin(\pi/3)}{cos(\pi/3)}[/itex]

 

[Dick]

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.

 

[jimen113]

Thank you!