In the previous section, we completed a discussion on relations and functions. In this chapter, we will see inverse trigonometric functions.
Some basics can be written in 8 steps:
1. Consider the sine function:
f(x) = sin x
• The input x can be any real number. So the domain is R.
• The output lies in the interval [-1,1]. So the codomain is [-1,1]
• We saw the above details in class 11. We saw a neat pictorial representation of the above details in the graph of the sine function. It is shown again in fig.18.1 below:
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| Fig.18.1 |
2. Let us check whether the sine function is one-one.
• Let input x = $\frac{\pi}{2}$.
♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\sin \left(\frac{\pi}{2} \right)~=~1 $
• Let input x = $\frac{5 \pi}{2}$.
♦ Then the output will be $f \left(\frac{5 \pi}{2} \right)~=~\sin \left(\frac{5 \pi}{2} \right)~=~1 $
• Let input x = $\frac{-3 \pi}{2}$.
♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\sin \left(\frac{-3 \pi}{2} \right)~=~1 $
(We can cross check with the graph and confirm that the above inputs and outputs are correct)
• We see that, more than one input values from the domain can give the same output. So the sine function is not a one-one function.
3. Suppose that, we restrict the input values.
• That is, we take input values only from the set $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.2 below:
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| Fig.18.2 |
• In the green portion, no two inputs will give the same output. So the green portion represents a function which is one-one.
4. Next, we have to prove that, the green portion is onto.
• For that, we can consider any y value from the codomain [-1,1].
• There will be always a x value in $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, which will satisfy the equation y = sin x.
• So the green portion is onto.
5. We see that, the green portion is both one-one and onto.
• We can represent this function in the mathematical way:
$\text{sine}:~ \left[\frac{-\pi}{2}, \frac{\pi}{2} \right]~\to~[-1,1]$, defined as f(x) = sin x.
6. If a function is both one-one and onto, the codomain is same as range.
• So we can write:
For this function, the domain is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ and range is [-1,1]
7. We know that, if a function is one-one and onto, it will be invertible. We have seen the properties of inverse functions. Let us apply those properties to our present case. It can be written in 4 steps:
(i) If y = f(x) = sin x is invertible, then there exists a function g such that:
g(y) = x
(ii) The function g will also be one-one and onto.
(iii) The domain of f will be the range of g. So the range of g is $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$.
(iv) The range of f will be the domain of g. So the domain of g is [-1,1]
(iv) The inverse of sine function is denoted as sin-1. So we can define the inverse function as:
$\sin^{-1}:~ [-1,1]~\to~\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$, defined as x = g(y) = sin-1 y.
8. Let us see an example:
• Suppose that, for the inverse function, the input y is $\frac{1}{2}$
• Then we get an equation: $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\sin^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\sin^{-1} \left(y \right)$
(ii) This is an inverse trigonometric function, where input y = $\frac{1}{2}$.
•
Based on the inverse trigonometric function, we can write the original trigonometric function:
$y ~=~ f(x) ~=~\sin x$
•
In our present case, it is:
$y~=~\frac{1}{2}~=~\sin x$
(iii) So we have a trigonometric equation:
$\sin x~=~\frac{1}{2}$
•
When we solve this equation, we get x.
(iv) We have seen the method for solving trigonometric equations in class 11.
•
In the present case, we do not need to write many steps. We already
know that, $\sin \left(\frac{\pi}{6} \right)~=~\frac{1}{2}$
•
So we can write:
$x ~=~\frac{\pi}{6}$
The above 8 steps help us to understand the basics about inverse trigonometric functions. Now we will see a few more details. It can be written in 6 steps:
1. We saw that, the sine function is not a one-one function. But to make it one-one, we restricted the domain to $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$.
2. There are other possible “restricted domains” available.
• $\left[\frac{-3 \pi}{2}, \frac{- \pi}{2} \right]$ is shown in magenta color in fig.18.3 below:
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| Fig.18.3 |
• $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ is shown in cyan color in fig,18.3 above.
3. There are infinite number of such restricted domains possible.
• We say that:
♦ Each restricted domain gives a corresponding branch of the sin-1 function.
♦ The restricted domain $\left[\frac{- \pi}{2}, \frac{\pi}{2} \right]$ gives the principal branch of the sin-1 function.
4. In class 11, we plotted the sine function. Now we will plot the inverse. It can be done in 4 steps:
(i) Write the set for the original function f. It must contain a convenient number of ordered pairs.
• $\left(\frac{\pi}{6} , \frac{1}{2} \right)$ is an example of the ordered pairs in f.
(To get a smooth curve, we must write a large number of ordered pairs)
(ii) Based on set f, we can write set g.
This is done by picking each ordered pair from f and interchanging the positions.
• For example, the point $\left(\frac{\pi}{6} , \frac{1}{2} \right)$ in the set f will become $\left(\frac{1}{2} , \frac{\pi}{6} \right)$ in set g.
• Thus we will get the required number of ordered pairs in g.
(iii) Mark each ordered pair of g on the graph paper.
• The first coordinate should be marked along the x-axis.
• The second coordinate should be marked along the y-axis.
(iv) Once all the ordered pairs are marked, draw a smooth curve connecting all the marks.
• The smooth curve is the required graph. It is shown in fig.18.4 below:
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| Fi.18.4 |
(v) We see that, the graph is smaller in width but larger in height.
• The graph is smaller in width because:
♦ Values to the left of -1 cannot be used as input values.
♦ Also, values to the right of 1 cannot be used as input values.
• The graph is larger in height because:
♦ Depending upon the branch, values upto +∞ or –∞ can be obtained as output values.
• The green curve is related to the $\left[\frac{-\pi}{2}, \frac{\pi}{2} \right]$ principal branch.
• The cyan curve is related to the $\left[\frac{\pi}{2}, \frac{3 \pi}{2} \right]$ branch.
• The magenta curve is related to the $\left[\frac{-3 \pi}{2}, \frac{\pi}{2} \right]$ branch.
5. Consider any function f for which the inverse g exists.
• The graph of g will be the mirror image of the graph of f.
♦ The line with equation y=x will be the mirror line.
• We will prove this in the case of sine function. It can be done in 5 steps:
(i) Consider the grid of the graph in fig.18.4 above.
• The x-axis is divided into equal parts, with each part equal to 1 unit.
• The y-axis is also divided into equal parts, with each part equal to π/2 units.
• π/2 = 3.14/2 = 1.57, which is larger than 1. This is the reason why we see larger divisions along the y-axis.
• For the next graph in fig.18.5 below, we will use π/2 for both the axes.
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| Fig.18.5 |
(ii) In fig.18.5 above,
♦ The sine function is drawn in red color.
♦ The sin-1 function is drawn in green color.
♦ The mirror line f(x) = x is drawn in white color.
(iii) Any two convenient points P and Q are marked on the sine function.
• We need to show that, their mirror images P' and Q' lie on the sin-1 function.
(iv) First we consider point P.
• From P, drop a perpendicular to the mirror line. Let P1 be the foot of the perpendicular.
• Extend PP1 so as to intersect with the green curve. Let P' be the point of intersection.
• We can measure and see that, PP1 = P1P'.
• Here we see two facts:
♦ PP' is perpendicular to the mirror line.
♦ PP1 = P1P'.
So P' is the mirror image of P.
• We also see that, the coordinates of P' is the interchanged version of P.
(v) Next we consider point Q.
• From Q, drop a perpendicular to the mirror line. Let Q1 be the foot of the perpendicular.
• Extend QQ1 so as to intersect with the green curve. Let Q' be the point of intersection.
• We can measure and see that, QQ1 = Q1Q'.
• Here we see two facts:
♦ QQ' is perpendicular to the mirror line.
♦ QQ1 = Q1Q'.
So Q' is the mirror image of Q.
• We also see that, the coordinates of Q' is the interchanged version of Q.
• In this way, we can prove that, any point on sin-1 function is the mirror image of a corresponding point in the sine function.
6. The sin-1 function is also known as arcsine function.
We have completed a basic discussion on the sin-1 function. In the next section, we will see cos-1 function.
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