In the previous section, we saw sin-1 function. In this section, we will see cos-1 function.
Some basics can be written in 8 steps:
1. Consider the cosine function:
f(x) = cos 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 cosine function. It is shown again in fig.18.6 below:
 |
| Fig.18.6 |
• Let input x = $\frac{\pi}{2}$.
♦ Then the output will be $f \left(\frac{\pi}{2} \right)~=~\cos \left(\frac{\pi}{2} \right)~=~0 $
• Let input x = $\frac{3 \pi}{2}$.
♦ Then the output will be $f \left(\frac{3 \pi}{2} \right)~=~\cos \left(\frac{3 \pi}{2} \right)~=~0 $
• Let input x = $\frac{-3 \pi}{2}$.
♦ Then the output will be $f \left(\frac{-3 \pi}{2} \right)~=~\cos \left(\frac{-3 \pi}{2} \right)~=~0 $
(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 cosine 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 $\left[0, \pi \right]$.
• The codomain is the usual [-1,1]
• Then we will get the green curve shown in fig.18.7 below:
 |
| Fig.18.7 |
• 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 $\left[0, \pi \right]$, which will satisfy the equation y = cos 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:
$\text{cosine}:~ \left[0, \pi \right]~\to~[-1,1]$, defined as f(x) = cos 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 $\left[0, \pi \right]$ 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) = cos 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 $\left[0, \pi \right]$.
(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 cos-1. So we can define the inverse function as:
$\cos^{-1}:~ [-1,1]~\to~\left[0, \pi \right]$, defined as x = g(y) = cos-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: $x ~=~\cos^{-1} \left(\frac{1}{2} \right)$
• Our aim is to find x. It can be done in 4 steps:
(i) $x ~=~\cos^{-1} \left(\frac{1}{2} \right)$ is an equation of the form $x ~=~f(y)~=~\cos^{-1} \left(y \right)$
(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:
$y ~=~ f(x) ~=~\cos x$
•
In our present case, it is:
$y~=~\frac{1}{2}~=~\cos x$
(iii) So we have a trigonometric equation:
$\cos x~=~\frac{1}{2}$
•
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, $\cos \left(\frac{\pi}{3} \right)~=~\frac{1}{2}$
•
So we can write:
$x ~=~\frac{\pi}{3}$
The
above 8 steps help us to understand the basics about cos-1 function. Now we will see a few more details. It can be
written in 5 steps:
1. We saw that, the cos function is not a one-one
function. But to make it one-one, we restricted the domain to
$\left[0, \pi \right]$.
2. There are other possible “restricted domains” available.
• $\left[- \pi, 0 \right]$ is shown in magenta color in fig.18.8 below:
 |
| Fig.18.8 |
3. There are infinite number of such restricted domains possible.
• We say that:
♦ Each restricted domain gives a corresponding branch of the cos-1 function.
♦ The restricted domain $\left[0, \pi \right]$ gives the principal branch of the cos-1 function.
4. In class 11, we plotted the cosine 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.
• $\left(\frac{\pi}{3} , \frac{1}{2} \right)$ 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 $\left(\frac{\pi}{3} , \frac{1}{2} \right)$ in
the set f will become $\left(\frac{1}{2} , \frac{\pi}{3} \right)$ 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.18.9 below:
 |
| Fig.18.9 |
(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 red curve is related to the $\left[0, \pi \right]$ branch.
• The cyan curve is related to the $\left[\pi, 2 \pi \right]$ branch.
• The magenta curve is related to the $\left[- \pi, 0 \right]$ branch.
5. The cos-1 function is also known as arc cosine function.
In the next section, we will see cosec-1 function.
Previous
Contents
No comments:
Post a Comment