Silly question about trig past pi/2

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In summary, the conversation discusses the difficulties of finding the correct value for arctan(-1) within an interval of pi. The solution is to consider the function as mapping to a particular interval of angles, such as (-pi/2, pi/2). The conversation also mentions that arctan is an odd function and that there are an infinite number of solutions for y to the equation tan(y) = x.
  • #1
motornoob101
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I know all the well know angles from 0 to [tex] \pi/2[/tex]. However, past pi/2 I am rather clueless and have no idea what angle corresponds to what value.

For example, in trying to figure out [tex] tan^{-1}(-1)[/tex] How does one know it is 5pi/4 and not 7pi/8? I was debating between measuring pi/4 from the negative x-axis (which would give me 7pi/8) and measuring pi/4 from the y axis,(which would give me 5pi/4) but as you can see, I choose the wrong value. Is there a trick to it? Thanks.
 
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  • #2
well

[tex]arctan(-1)=-arctan(1)=-\frac{\pi}{4}[/tex] so it means that [tex] \frac{\pi}{4} [/tex] lies on the fourth quadrant of the unit circle. So you can get that value by subtracting it from [tex]2\pi[/tex] that is

[tex]2\pi-\frac{\pi}{4}=\frac{8\pi-\pi}{4}=\frac{7\pi}{4}[/tex] because -pi/4 is the angle counted clockwise direction.
YOu know that it does not lie on the 3rd quadrant because tan(x) there is positive. There are other tricks too.
 
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  • #3
That's the "problem" with inverse trig functions--since, for example, there are an infinite number of solutions for y to the equation [itex]\tan{y} = x[/itex], in order for [itex]f(x) = \arctan{x}[/itex] to be a well-defined function, you have to consider it as mapping to a particular interval of angles. Without qualification, arctan typically returns values between [itex]\frac{-\pi}{2}[/itex] and [itex]\frac{\pi}{2}[/itex], but any interval of length [itex]\pi[/itex] is valid as long as your usage is consistent. There are similar considerations for arcsin and arccos.

Edit: Because I'm not sure if I made it clear, let me say this: [itex]\arctan{x}[/itex] is usually defined as the function from the real numbers to the interval [itex](\frac{-\pi}{2} , \frac{\pi}{2})[/itex] such that [itex]\tan{(\arctan{x})} = x[/itex] for any real x.
 
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  • #4
Oh I forget that arctan is an odd function. Thanks. But what if the question asks for a value for arctan(-1) within pi? I would have tried the 7pi/4 approach if I wasn't restricted to 0 to pi. Thanks.
 
  • #5
motornoob101 said:
Oh I forget that arctan is an odd function. Thanks. But what if the question asks for a value for arctan(-1) within pi? I would have tried the 7pi/4 approach if I wasn't restricted to 0 to pi. Thanks.
If you have to find the value within an interval of a length [tex]\pi[/tex] that doent mean that your interval must be [tex][0,\pi][/tex] , moreover the common interval for this problem is chosen like JohnDuck stated [tex][-\frac{\pi}{2},\frac{\pi}{2}][/tex].

This means that also [tex]
\frac{7\pi}{2}[/tex] is within that interval.
 
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  • #6
sutupidmath said:
One more thing [tex]arctan(-1)[/tex] has no value that lies on the interval [tex][0,\pi][/tex].
[tex]\frac{3\pi}{4}[/tex]?
 
  • #7
JohnDuck said:
[tex]\frac{3\pi}{4}[/tex]?


Blahhhhhh! Yup.
 

1. What is the maximum value of sine and cosine when the angle is greater than pi/2?

The maximum value of sine and cosine when the angle is greater than pi/2 is 1. This is because the values of sine and cosine are equal to the y and x coordinates, respectively, of a point on the unit circle. As the angle increases beyond pi/2, these coordinates approach the point (1,0) on the circle, resulting in a maximum value of 1.

2. Can the values of tangent and cotangent be greater than 1 when the angle is greater than pi/2?

Yes, the values of tangent and cotangent can be greater than 1 when the angle is greater than pi/2. This is because tangent and cotangent are equal to the y and x coordinates, respectively, of a point on the unit circle divided by the x and y coordinates, respectively. As the angle increases, the y coordinate approaches 1 while the x coordinate approaches 0, resulting in a value greater than 1.

3. Why do we use radians instead of degrees to measure angles in trigonometry?

We use radians instead of degrees in trigonometry because they provide a more natural and consistent way of measuring angles. Radians are based on the unit circle, where the length of the arc of a circle is equal to the radius of the circle. This allows for easy conversion between the measures of angles and the corresponding trigonometric ratios.

4. Is there a maximum angle that can be used in trigonometric functions?

No, there is no maximum angle that can be used in trigonometric functions. These functions can be used for any angle, including those greater than pi/2. However, as the angle increases, the values of the trigonometric functions may become very large or undefined.

5. Can we use trigonometric functions to solve real-world problems that involve angles greater than pi/2?

Yes, trigonometric functions can be used to solve real-world problems that involve angles greater than pi/2. In fact, many real-world scenarios, such as calculating the height of a tree or the distance between two points, involve angles greater than pi/2 and can be solved using trigonometry. The use of radians allows for accurate calculations in these scenarios.

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