Trig Question

jsc90

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Jan 6, 2006
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4
Hi, you someone guide me step by step as to solving this equation

cotan[sin^-1 (-1/2)]

It says to determine the exact values.
 
Hello, jsc90!

I will assume we're dealing with "principal values" here.
    \displaystyle \;\;The angle will be between -90o\displaystyle 90^o and 90o\displaystyle 90^o
[If they want <u>all</u> the answers, I'll let you generalize the solution.]

Evaluate: cot[sin1(12)]\displaystyle \cot\left[\sin^{-1}\left(-\frac{1}{2}\right)\right]
You're expected to understand that an inverse trig expression is an <u>angle</u>.

    \displaystyle \;\;That is, sin1(12)\displaystyle \sin^{-1}\left(-\frac{1}{2}\right) is some angle θ\displaystyle \theta whose sine is -12\displaystyle \frac{1}{2}

    \displaystyle \;\;In other words: sinθ=12\displaystyle \:\sin\theta \,=\,-\frac{1}{2}


You're also expected to know that:
    \displaystyle \;\;If sinθ=12,\displaystyle \sin\theta\,=\,-\frac{1}{2},\, then θ=30o\displaystyle \,\theta\,=\,-30^o

Therefore: cot(30o)=3\displaystyle \,\cot(-30^o)\:=\:-\sqrt{3}
 
I will consider the domain \(\displaystyle \L [0, 2\pi)\).

Firstly:

If we halve an equilateral triangle with side lengths 2, we have the right-angled triangle:

Code:
                 *
                /|
               / |
              /\_|
             /pi |
            / -- |
           /  6  | 
          /      |   
     2   /       |   _
        /        |  √3
       /         |
      /          |
     /           |
    /            |
   /\  pi        |
  /  | --        |
 /  /  3         |
+----------------+
        1

From this we have
. . \(\displaystyle \L \sin{\left(\frac{\pi}{6}\right)} \, = \, \frac{opp}{hyp} \, = \, \frac{1}{2}\)

That is, \(\displaystyle \L \sin^{-1}{\left(\frac{1}{2}\right)} \, = \, \frac{\pi}{6}\)

Also,
. . \(\displaystyle \L \cot{\left(\frac{\pi}{6}\right)} \, = \, \frac{adj}{opp} \, = \, \frac{\sqrt{3}}{1} \, = \, \sqrt{3}\)

When considering \(\displaystyle \L \cot{\left(\sin^{-1}{\left(-\frac{1}{2}\right)}\right)\), we can draw a similar triangle in each of quadrants 3 and 4.

Code:
                           /|\ y
                            |
                            |
                            |
                            |
                            |
                 _          |           _
               -√3          |          √3
            ----+-----------+-----------+----->
                :   7pi/6  /|\ 11pi/6   :     x
                :        /  |  \        : 
             -1 :      /    |    \      : -1
                :    / 2    |   2  \    :
                :  /        |        \  :
                :/        -1*          \:
                            |
                            |

   where the angles 7pi/6 and 11pi/6 are both measured counter-clockwise
     from the positive x-axis (pi + pi/6 = 7pi/6 and 2pi - pi/6 = 11pi/6).

Notice that the triangle in quad 3 gives

. . \(\displaystyle \L \sin{\left(\frac{7\pi}{6}\right)} \, = \, -\frac{1}{2}\)

and the triangle in quad 4 gives
. . \(\displaystyle \L \sin{\left(\frac{11\pi}{6}\right)} \, = \, -\frac{1}{2}\)

Whereas from the respective triangles

. . \(\displaystyle \L \cot{\left(\frac{7\pi}{6}\right)} \, = \, \frac{-\sqrt{3}}{-1} \, = \, \sqrt{3}\)

and

. . \(\displaystyle \L \cot{\left(\frac{11\pi}{6}\right)} \, = \, \frac{\sqrt{3}}{-1} \, = \, -\sqrt{3}\)

We conclude:

. If \(\displaystyle \L \sin{\theta} \, = \, -\frac{1}{2}\) ,

. . \(\displaystyle \L \cot{\theta} \, = \, \sqrt{3} \, \text{or} \, -\sqrt{3} \, \, \Leftrightarrow \, \cot{\left(\sin^{-1}{\left(-\frac{1}{2}\right)}\right)} \, = \, \sqrt{3} \, \text{or} \, -\sqrt{3}\).

A similar thread occured recently: http://www.freemathhelp.com/forum/viewtopic.php?t=10356&postdays=0&postorder=asc&start=0. TK and Pka's posts towards the end particularly are a good read.
 
“I will consider the domain [0,2π)\displaystyle [0,2\pi )”: the domain of what function?

Unco, what are you saying? You cannot change the commonly understood domains!
The domain of the function arcsine is [π2,π2]\displaystyle \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right], for you to say otherwise is to be ignorant of mathematics.
BUT, I seem to remember that you are in physics or chemistry.
If you want to move a notch in the intelligent scale learn the basics.
 
Sorry to hear that!
Physicians are just millionaire mechanics today.
That is too bad. You seem to really have a good brain!
 
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