Exact Value of Inverse Trig Function

[communitycoll]

Homework Statement

How do I find out the exact value of tan^-1 (1 / sqrt(3))?

Homework Equations

nada

The Attempt at a Solution

I don't know where to start.

 

[Curious3141]

Homework Statement

How do I find out the exact value of tan^-1 (1 / sqrt(3))?

Homework Equations

nada

The Attempt at a Solution

I don't know where to start.

Consider the right triangle with a hypotenuse of 2 and one side equal to √3. What are the angles in this triangle?

If you can't see it immediately, reflect the triangle about the side of length √3 and see what sort of triangle you get when the mirror images are placed next to each other.

 

[communitycoll]

How do I find the angles in the triangle?

 

[Curious3141]

How do I find the angles in the triangle?

What's the third side in the right triangle?

What happens when you reflect it as I suggested? What sort of triangle do you get? What are the angles in that sort of triangle?

(see diagram attached. Figure out the sides marked by the question marks and everything should become clear).

 

[asymptote]

Homework Statement

How do I find out the exact value of tan^-1 (1 / sqrt(3))?

This is one of the 'special' triangles, for which we know exact trigonometric ratios. Look at the 60°-30°-90° triangle for the answer. The point of the problem is to solve the question without using a calculator, only the special triangle.

 

[Millennial]

$\frac{1}{\sqrt{3}}=\tan(x)=\frac{\sin(x)}{\cos(x)}$, so it follows from here that $\cos(x)=\sin(x)\sqrt{3}$, and squaring both sides yields $\cos^2(x)=3\sin^2(x)$. We want to make use of the Pythagorean trigonometric identity, so we replace sine by cosine to get $\cos^2(x)=3-3\cos^2(x)$ which gives $\cos(x)=\frac{\sqrt{3}}{2}$.

Your equation is essentially equivalent to this equation. Can you solve this one for x?

 

[asymptote]

$\frac{1}{\sqrt{3}}=\tan(x)=\frac{\sin(x)}{\cos(x)}$, so it follows from here that $\cos(x)=\sin(x)\sqrt{3}$, and squaring both sides yields $\cos^2(x)=3\sin^2(x)$. We want to make use of the Pythagorean trigonometric identity, so we replace sine by cosine to get $\cos^2(x)=3-3\cos^2(x)$ which gives $\cos(x)=$.

Your equation is essentially equivalent to this equation. Can you solve this one for x?

Wow, that's the very long way around :) Though elegant, you can avoid all of this by simply looking at a 30-60-90 triangle and choosing the angle whose $\tan^{-1} = \frac{1}{sqrt(3)}$
Find the angle where $\frac{opposite}{adjacent} =\frac{1}{sqrt(3)}$