Вашата кошница е празна
Inverse functions (extra-curricular topic)
In many exercises it is necessary to find the number (the radian measure of the corresponding angle) for a given trigonometric functional value to which the function assigns the given value. If there is such a value, then there are infinitely many such values. Acalculator gives one of these; it is our task to determine the rest.
For example when we are solving the equation \latex{\sin x=\frac{1}{2}}, i.e. the question is to which real number x does the sine function assign the value \latex{\frac{1}{2}}, then the answer of the calculator is: \latex{x=0.5236+2k\pi(k\in\Z)} and \latex{x=\pi-0.5236+2n\pi(n\in\Z)} are all good, just like all the corresponding values.
On some calculators, after entering \latex{\frac{1}{2}} pressing the buttons “inv” and “sin” is needed to displayed the corresponding value of x. It means that we calculate the inverse function of the sine at the place \latex{\frac{1}{2}}.
In general, just like any function, the inverse function of the sine assigns a number to a number. According to this the inverse function of the sine, the function “arc sine”, we denote it by arcsin, is defined as follows:
DEFINITION: If \latex{-1\leq a\leq 1}, then arcsin \latex{ a } means the real number from the interval \latex{\left[-\frac{\pi }{2};\frac{\pi }{2} \right]} the sine of which is \latex{ a }.
So the function arcsin is the inverse of the sine function restricted to the interval \latex{\left[-\frac{\pi }{2};\frac{\pi }{2} \right]}.
For example: arcsin \latex{\frac{1}{2} =\frac{\pi }{6}}, arcsin \latex{(-1) =\frac{\pi }{2}}.
The definition implies that sin(arcsin \latex{ a }) \latex{ = a }.
Example 1
Let us plot the graph of the function \latex{\left[-1;1\right] \rightarrow \R,x\mapsto } arcsin \latex{ x }.
Solution
The point (\latex{ x }; arcsin \latex{ x }) lies on the graph of the inverse function if and only if the point (arcsin \latex{ x }; sin(arcsin \latex{ x }) \latex{ = x }) lies on the part of the graph of the sine function that belongs to the interval \latex{\left[-\frac{\pi }{2};\frac{\pi }{2} \right]}.
So the two curves are each other's mirror images about the straight line with the equation y = x (Figure 33).
\latex{-\frac{\pi}{2} }
\latex{\frac{\pi }{2} }
\latex{\frac{\pi }{2} }
\latex{-\frac{\pi}{2} }
\latex{y=\arcsin\; x}
\latex{y=\sin\; x}
\latex{ 1 }
\latex{ -1 }
\latex{ -1 }
\latex{ 1 }
\latex{ y }
\latex{ x }
Figure 33
Example 2
Let us plot the graphs of the following functions:
- \latex{f:\left[-1;1\right] \rightarrow \R, f\left(x\right)=\sin \left(\arcsin x\right)};
- \latex{g:\R\rightarrow \R,g\left(x\right) =\arcsin \left(\sin x\right)}.
\latex{y=f(x)}
\latex{ 1 }
\latex{ x }
\latex{ -1 }
\latex{ -1 }
\latex{ 1 }
\latex{ y }
Figure 34
Solution (a)
Based on the comment we made after the definition the image of the function \latex{ f } can be seen in Figure 34.
Solution (b)
The function \latex{ g } is periodic with the period \latex{2\pi}, so it is enough to examine for example on the interval \latex{\left[-\frac{\pi }{2};\frac{3\pi }{2} \right]}.
If \latex{-\frac{\pi }{2}\leq x\leq \frac{\pi }{2}}, then \latex{\arcsin \left(\sin x\right)=x} based on the definition.
If \latex{\frac{\pi }{2}\leq x\leq \frac{3\pi }{2}}, then \latex{-\frac{\pi }{2}\leq \pi -x\leq \frac{\pi }{2}} and \latex{\sin x=\sin \left(\pi -x\right)}, so again based on the definition \latex{\arcsin \left(\sin x\right) =\pi -x}. The graph of the function \latex{ g } an be seen in Figure 35.
\latex{-2\pi}
\latex{-\frac{3\pi }{2} }
\latex{-\pi}
\latex{\frac{\pi }{2} }
\latex{\pi}
\latex{\frac{3\pi }{2} }
\latex{2\pi}
\latex{\frac{\pi }{2} }
\latex{-\frac{\pi }{2} }
\latex{-\frac{\pi }{2} }
\latex{y=g(x)}
\latex{ y }
\latex{ x }
Figure 35
Let us define the inverse of the tangent function similarly to the inverse of the sine; its name is: arc tangent, its notation is: \latex{\arctan }.
DEFINITION: If \latex{a\in\R}, then \latex{\arctan a} is the real number belonging to the interval \latex{\left]-\frac{\pi }{2};\frac{\pi }{2}\right[ } the tangent of which is \latex{ a }.
In other words the function \latex{\arctan} is the inverse of the tangent function stricted to the interval \latex{\left]-\frac{\pi }{2};\frac{\pi }{2}\right[ }.
For example: \latex{\arctan 1=\frac{\pi }{4},\;\arctan 0=0}.
The definition implies that \latex{\tan \left(\arctan a\right)=a}.
Example 3
Let us plot the graph of the function \latex{\R\rightarrow \R,\;x\mapsto \arctan x}.
Solution
The train of thought we became familiar with in the example \latex{ 1 } can also be applied here, the graph of the arctan function is the mirror image of the graph of the function obtained when restricting the tangent function to the open interval \latex{\left]-\frac{\pi }{2};\frac{\pi }{2}\right[ } about the straight line \latex{ y = x } (Figure 36).
\latex{\frac{\pi }{2} }
\latex{\frac{\pi }{2} }
\latex{-\frac{\pi }{2} }
\latex{-\frac{\pi }{2} }
\latex{y=\arctan x}
\latex{y=\tan x}
\latex{ x }
\latex{ y }
Figure 36
Exercises
{{exercise_number}}. Define the function arccos as the inverse of the cosine function restricted to the interval \latex{\left[0;\pi \right] }, and plot its graph.
{{exercise_number}}. Plot the graphs of the following functions:
- \latex{\R\rightarrow \R,\;x\mapsto \tan \left(\arctan x\right) };
- \latex{\left(\R\setminus \left\{\frac{\pi }{2}+k\pi \mid k\in \Z \right\} \right) \rightarrow \R,\;x\mapsto \arctan \left(\tan x\right) }.
{{exercise_number}}. Define the function arccot as the inverse of the cotangent function restricted to the interval \latex{\left[0;\pi \right] }, and plot its graph.
{{exercise_number}}. Verify the following identities:
- \latex{\arctan x+\arctan \frac{1}{x} =\frac{\pi }{2}\times \text{sgn }x }, if \latex{x\neq0};
- \latex{\arcsin x+\arccos x=\frac{\pi }{2} }, if \latex{-1\leq x\leq 1}.
