ARC functions
- Thread starter mathfun
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while i was doing some more trig... i stumble across another question...
so do arcfunctions not give you the answer?? but arcfunctions only tell you if the answer is suppose to be NEGATIVE or POSTIVE??
because somebody told me that...(i think) and the certain things u do with the answer to get it in the right quadrant...... like for example, if you want it to be in the IV quadrant.... u use 2pi and minus the answer...
So, could any body pleaseeeeeeeeeeeeeeeeeee clear this up for me? Thanks!
so do arcfunctions not give you the answer?? but arcfunctions only tell you if the answer is suppose to be NEGATIVE or POSTIVE??
because somebody told me that...(i think) and the certain things u do with the answer to get it in the right quadrant...... like for example, if you want it to be in the IV quadrant.... u use 2pi and minus the answer...
So, could any body pleaseeeeeeeeeeeeeeeeeee clear this up for me? Thanks!
I originally interpreted that you were given the triangle as part of the question. If that was the case, the positive sides mean \(\displaystyle \theta\) is restricted to \(\displaystyle 0 \leq \theta < \frac{\pi}{2}\) and posts which followed giving one solution are correct.
However, if you were simply asked to find \(\displaystyle \L \cos{\left(\arcsin{\left(\frac{3}{4}\right)}\right)} = ?\), then we can give two answers.
You have already let
\(\displaystyle \L \theta = \arcsin{\left(\frac{3}{4}\right)}\) and drawn a right-angled triangle to obtain \(\displaystyle \L \cos{\theta} = \frac{4}{5}\).
If we take the sine of both sides:
. .\(\displaystyle \L \theta = \arcsin{\left(\frac{3}{4}\right)}\)
. . . \(\displaystyle \L \Rightarrow \sin{(\theta)} = \frac{3}{4}\)
You should now understand that there are two solutions for \(\displaystyle \theta\) in the domain \(\displaystyle 0 \leq x < 2\pi\) (remember, this domain will give us all the unique solutions for \(\displaystyle \theta\), ie. angles that aren't 2pi apart).
One is quadrant 1 ('all') and the other is in quadrant 2 ('sin').
\(\displaystyle \L \cos{(\theta)}\) is positive in quadrant 1, and negative in quadrant 2. Therefore \(\displaystyle \L \cos{\theta} \, = \, \frac{4}{5} \, \text{or} \, -\frac{4}{5}\).
Our quadrants look like:
Notice that \(\displaystyle \L \sin{\theta} \, = \, \frac{3}{5}\) for both triangles (both quadrants).
Also notice that \(\displaystyle \L \cos{\theta} \, = \, \frac{+4}{5}\) in quadrant 1 but \(\displaystyle \L \cos{\theta} \, = \, \frac{-4}{5}\) in the quadrant 2.
Conclusion:
. . If \(\displaystyle \L \sin{\theta} \, = \, \frac{3}{5}\), \(\displaystyle \L \cos{\theta{\) can be either \(\displaystyle \L -\frac{4}{5}\) or \(\displaystyle -\frac{4}{5}\).
Alternatively:
. . If \(\displaystyle \L \theta = \arcsin{\left(\frac{3}{5}\right)}\),
. . . \(\displaystyle \L \cos{\theta} = \cos{\left( \arcsin{\frac{3}{5}}\right)} = \frac{4}{5} \, \text{or} \, -\frac{4}{5}\).
However, if you were simply asked to find \(\displaystyle \L \cos{\left(\arcsin{\left(\frac{3}{4}\right)}\right)} = ?\), then we can give two answers.
You have already let
\(\displaystyle \L \theta = \arcsin{\left(\frac{3}{4}\right)}\) and drawn a right-angled triangle to obtain \(\displaystyle \L \cos{\theta} = \frac{4}{5}\).
If we take the sine of both sides:
. .\(\displaystyle \L \theta = \arcsin{\left(\frac{3}{4}\right)}\)
. . . \(\displaystyle \L \Rightarrow \sin{(\theta)} = \frac{3}{4}\)
You should now understand that there are two solutions for \(\displaystyle \theta\) in the domain \(\displaystyle 0 \leq x < 2\pi\) (remember, this domain will give us all the unique solutions for \(\displaystyle \theta\), ie. angles that aren't 2pi apart).
One is quadrant 1 ('all') and the other is in quadrant 2 ('sin').
\(\displaystyle \L \cos{(\theta)}\) is positive in quadrant 1, and negative in quadrant 2. Therefore \(\displaystyle \L \cos{\theta} \, = \, \frac{4}{5} \, \text{or} \, -\frac{4}{5}\).
Our quadrants look like:
Code:
/|\ y
Quad 2 | Quad 1
|
:\ 3 /:
: \ | / :
3 : \5 | 5 / : 3 Q = theta
: \ | / :
: Q \|/ Q :
-------+---------+---------+-------->
-4 | +4 x
|
(Note: although Q is equal in both triangles, we measure the angle of trigonometric
functions from the positive x-axis, so the angle in the second quadrant is
given by pi - Q when measured from the positive x-axis.)Also notice that \(\displaystyle \L \cos{\theta} \, = \, \frac{+4}{5}\) in quadrant 1 but \(\displaystyle \L \cos{\theta} \, = \, \frac{-4}{5}\) in the quadrant 2.
Conclusion:
. . If \(\displaystyle \L \sin{\theta} \, = \, \frac{3}{5}\), \(\displaystyle \L \cos{\theta{\) can be either \(\displaystyle \L -\frac{4}{5}\) or \(\displaystyle -\frac{4}{5}\).
Alternatively:
. . If \(\displaystyle \L \theta = \arcsin{\left(\frac{3}{5}\right)}\),
. . . \(\displaystyle \L \cos{\theta} = \cos{\left( \arcsin{\frac{3}{5}}\right)} = \frac{4}{5} \, \text{or} \, -\frac{4}{5}\).
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mathfun said:
stapel said:
No, I'm afraid I don't.mathfun said:
- "i stumble across another question": Does this mean you're working on another exercise that confused you? If so, what is the text of that exercise? Or do you mean something else? If so, what?
[/*:m:aa9xvh0x] - "do arcfunctions not give you the answer?": The answer to what? What do you mean by "giving" you the answer? In what context?
[/*:m:aa9xvh0x] - "arcfunctions only tell you if the answer is suppose to be NEGATIVE or POSTIVE?": What answer? To what question? On what basis do you make this claim? What are you talking about?
[/*:m:aa9xvh0x] - "somebody told me that...(i think)": Who told you what about what? When? Where? Why? What did you want us to do with it?
[/*:m:aa9xvh0x] - "the certain things u do with the answer to get it in the right quadrant": What things who does to which answer to "get it" where?
[/*:m:aa9xvh0x] - "like for example, if you want it to be in the IV quadrant": If who wants what to be in the fourth quadrant? Why? In what context?
[/*:m:aa9xvh0x] - "u use 2pi and minus the answer": Why? When? Where? What answer?[/*:m:aa9xvh0x]
Eliz.
pka
Elite Member
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What I am about to write is not popular with some who contribute here.
There has been had some disagreement with it.
However, it is standard in most current textbooks.
There are three principle inverse trig functions.
The arccosine function has its domain \(\displaystyle - 1 \le x \le 1\) and its range or output \(\displaystyle 0 \le \arccos (x) \le \pi\).
The arcsine function has its domain \(\displaystyle - 1 \le x \le 1\) and its range or output \(\displaystyle \frac{{ - \pi }}{2} \le \arcsin (x) \le \frac{\pi }{2}\).
The arctangent function has its domain \(\displaystyle - \infty < x < \infty\) and its range or output \(\displaystyle \frac{{ - \pi }}{2} < \arctan (x) < \frac{\pi }{2}\).
These must just be learn!
Knowing the range of these functions is really the most important.
Thus, \(\displaystyle \sin \left( {\arccos \left( {\frac{{ - \sqrt 2 }}{2}} \right)} \right) = \left( {\frac{{\sqrt 2 }}{2}} \right)\) because the output of this arccos is in II.
But \(\displaystyle \left( {\arcsin \left( {\frac{{ - \sqrt 2 }}{2}} \right)} \right) = \left( {\frac{{ - \pi }}{4}} \right)\) because of the range of the arcsine function.
While I on the subject: this is really unpopular.
The trend in teaching this material is to stay away from degrees altogether!
Also, the trend is to use only radians in the domain \(\displaystyle - \pi < x \le \pi\) as ‘angles’ associated polar forms.
There has been had some disagreement with it.
However, it is standard in most current textbooks.
There are three principle inverse trig functions.
The arccosine function has its domain \(\displaystyle - 1 \le x \le 1\) and its range or output \(\displaystyle 0 \le \arccos (x) \le \pi\).
The arcsine function has its domain \(\displaystyle - 1 \le x \le 1\) and its range or output \(\displaystyle \frac{{ - \pi }}{2} \le \arcsin (x) \le \frac{\pi }{2}\).
The arctangent function has its domain \(\displaystyle - \infty < x < \infty\) and its range or output \(\displaystyle \frac{{ - \pi }}{2} < \arctan (x) < \frac{\pi }{2}\).
These must just be learn!
Knowing the range of these functions is really the most important.
Thus, \(\displaystyle \sin \left( {\arccos \left( {\frac{{ - \sqrt 2 }}{2}} \right)} \right) = \left( {\frac{{\sqrt 2 }}{2}} \right)\) because the output of this arccos is in II.
But \(\displaystyle \left( {\arcsin \left( {\frac{{ - \sqrt 2 }}{2}} \right)} \right) = \left( {\frac{{ - \pi }}{4}} \right)\) because of the range of the arcsine function.
While I on the subject: this is really unpopular.
The trend in teaching this material is to stay away from degrees altogether!
Also, the trend is to use only radians in the domain \(\displaystyle - \pi < x \le \pi\) as ‘angles’ associated polar forms.
pka
Elite Member
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I may not have been at this site. Several months ago I posted a similar comment about the range of the arccosine. At two engineers and/or physics types took strong exception to making it a one-to-one function. I was told that in physics domains do not matter. Maybe that is what I was thinking about.
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Inverse Relations are one thing. If you want Inverse FUNCTIONS, you must restrict them. PKA has demonstrated probably the most common restricted definitions. Learn these.
It is VERY UNLIKELY at this level, but if you need some other definition, then define it, but be consistent.
A Principle Inverse Trig FUNCTION will give you a REFERENCE Angle. It is the wise users responsibility to figure out where it belongs (what Quadrant). You must put Right Triangles and Unit Circles deep in your brain. See them. Feel them. Dream about them. Eat them. Carve them into your forehead. (hyperbole!! - don't actually do that!)
It is VERY UNLIKELY at this level, but if you need some other definition, then define it, but be consistent.
A Principle Inverse Trig FUNCTION will give you a REFERENCE Angle. It is the wise users responsibility to figure out where it belongs (what Quadrant). You must put Right Triangles and Unit Circles deep in your brain. See them. Feel them. Dream about them. Eat them. Carve them into your forehead. (hyperbole!! - don't actually do that!)