Trig Question
- Thread starter jsc90
- Start date
Hello, jsc90!
I will assume we're dealing with "principal values" here.
\(\displaystyle \;\;\)The angle will be between -\(\displaystyle 90^o\) and \(\displaystyle 90^o\)
[If they want <u>all</u> the answers, I'll let you generalize the solution.]
\(\displaystyle \;\;\)That is, \(\displaystyle \sin^{-1}\left(-\frac{1}{2}\right)\) is some angle \(\displaystyle \theta\) whose sine is -\(\displaystyle \frac{1}{2}\)
\(\displaystyle \;\;\)In other words: \(\displaystyle \:\sin\theta \,=\,-\frac{1}{2}\)
You're also expected to know that:
\(\displaystyle \;\;\)If \(\displaystyle \sin\theta\,=\,-\frac{1}{2},\,\) then \(\displaystyle \,\theta\,=\,-30^o\)
Therefore: \(\displaystyle \,\cot(-30^o)\:=\:-\sqrt{3}\)
I will assume we're dealing with "principal values" here.
\(\displaystyle \;\;\)The angle will be between -\(\displaystyle 90^o\) and \(\displaystyle 90^o\)
[If they want <u>all</u> the answers, I'll let you generalize the solution.]
You're expected to understand that an inverse trig expression is an <u>angle</u>.
\(\displaystyle \;\;\)That is, \(\displaystyle \sin^{-1}\left(-\frac{1}{2}\right)\) is some angle \(\displaystyle \theta\) whose sine is -\(\displaystyle \frac{1}{2}\)
\(\displaystyle \;\;\)In other words: \(\displaystyle \:\sin\theta \,=\,-\frac{1}{2}\)
You're also expected to know that:
\(\displaystyle \;\;\)If \(\displaystyle \sin\theta\,=\,-\frac{1}{2},\,\) then \(\displaystyle \,\theta\,=\,-30^o\)
Therefore: \(\displaystyle \,\cot(-30^o)\:=\:-\sqrt{3}\)
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:
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.
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.
Firstly:
If we halve an equilateral triangle with side lengths 2, we have the right-angled triangle:
Code:
*
/|
/ |
/\_|
/pi |
/ -- |
/ 6 |
/ |
2 / | _
/ | √3
/ |
/ |
/ |
/ |
/\ pi |
/ | -- |
/ / 3 |
+----------------+
1From 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.
pka
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- Jan 29, 2005
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“I will consider the domain \(\displaystyle [0,2\pi )\)”: the domain of what function?
Unco, what are you saying? You cannot change the commonly understood domains!
The domain of the function arcsine is \(\displaystyle \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]\), 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.
Unco, what are you saying? You cannot change the commonly understood domains!
The domain of the function arcsine is \(\displaystyle \left[ {\frac{{ - \pi }}{2},\frac{\pi }{2}} \right]\), 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.