When we solve equations we need to be aware of the domains and ranges of the functions involved. This is particularly so in the case of the inverse trigonometric functions because they each have restrictions on one or both of their domain and range.
Sine and arcsine
The sine function has the set of all real numbers as its domain. Its range is the interval $[-1,1]$[−1,1]. If we restrict the domain to the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$[−π2,π2], the function becomes a one-to-one function and it is then possible to define its inverse.
The inverse sine function, $\sin^{-1}$sin−1 or $\arcsin$arcsin, has domain $[-1,1]$[−1,1] and range $[-\frac{\pi}{2},\frac{\pi}{2}]$[−π2,π2].
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Cosine and arccosine
The cosine function has the same domain and range as the sine function. However, in this case, we make it a one-to-one function by restricting the domain to the interval $[0,\pi]$[0,π]. With this restricted domain, the function is invertible.
The inverse cosine function, $\cos^{-1}$cos−1 or $\arccos$arccos, has domain $[-1,1]$[−1,1] and range $[0,\pi]$[0,π].
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Tangent and arctangent
The tangent function is one-to-one if its domain is restricted to the open interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2). Its range is the set of real numbers.
We define the inverse tangent function, $\tan^{-1}$tan−1 or $\arctan$arctan over the real numbers and its range is the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2).
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Solving equations
It is possible to write down equations involving the inverse trigonometric functions that have no solutions. For example, the equation $\arcsin2x=\pi$arcsin2x=π has no solution because $\pi$π is not in the range of the $\arcsin$arcsin function.
On the other hand, the equation $3\arcsin2x=\pi$3arcsin2x=π does have a solution because it implies $\arcsin2x=\frac{\pi}{3}$arcsin2x=π3 and since $-\frac{\pi}{2}<\frac{\pi}{3}<\frac{\pi}{2}$−π2<π3<π2, the right hand side is within the range.
The equation is equivalent to the equation $2x=\sin\frac{\pi}{3}$2x=sinπ3. (We took the sine of both sides.) Therefore, $2x=\frac{\sqrt{3}}{2}$2x=√32 and so, the solution is $x=\frac{\sqrt{3}}{4}$x=√34.
Worked examples
Example 1
Solve, if possible, the equation $\arctan x=\frac{7\pi}{4}$arctanx=7π4. Give all solutions.
As we have defined it, the $\arctan$arctan function is one-to-one. Hence, there can be at most one solution. However, since $\frac{7\pi}{4}$7π4 is greater than $\frac{\pi}{2}$π2, the right-hand side is outside the range of the function and therefore there is no solution.
We might argue that the tangent function is periodic with period $\pi$π and so, we could look instead for a solution to $\arctan x=\frac{7\pi}{4}-\pi=-\frac{\pi}{4}$arctanx=7π4−π=−π4. This equation does have a solution, $x=-1$x=−1.
We can say that $\frac{7\pi}{4}$7π4 is an angle whose tangent is $-1$−1 but we cannot say $\frac{7\pi}{4}$7π4 is the angle whose tangent is $-1$−1. Our definition of the $\arctan$arctan function with its domain limited to the interval $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$(−π2,π2) allows us to claim that $-\frac{\pi}{4}$−π4 is the angle whose tangent is $-1$−1 because we are confining our attention to a limited domain.
Example 2
Consider the two equations
$\arcsin(\cos x)=\frac{\pi}{6}$arcsin(cosx)=π6
$\cos(\arcsin y)=\frac{1}{2}$cos(arcsiny)=12
In the first equation, the right-hand side is within the range of the $\arcsin$arcsin function and so a solution could exist. We also check that the range of the $\cos$cos function is within the domain of the $\arcsin$arcsin function. This condition is satisfied, so it should be possible to solve the equation.
| $\arcsin(\cos x)$arcsin(cosx) | $=$= | $\frac{\pi}{6}$π6 |
| $\therefore\ \ \sin\left[\arcsin(\cos x)\right]$∴ sin[arcsin(cosx)] | $=$= | $\sin\frac{\pi}{6}$sinπ6 |
| $\therefore\ \ \cos x$∴ cosx | $=$= | $\frac{1}{2}$12 |
| $\therefore\ \ x$∴ x | $=$= | $\arccos\frac{1}{2}$arccos12 |
| $=$= | $\frac{\pi}{3}$π3 |
In the second of the two equations, the right-hand side, $\frac{1}{2}$12, is within the range of the $\cos$cos function and the range of the $\arcsin$arcsin function is in the domain of the $\cos$cos function. We can proceed to solve the equation.
| $\cos(\arcsin y)$cos(arcsiny) | $=$= | $\frac{1}{2}$12 |
| $\therefore\ \ \arccos\left[\cos(\arcsin y)\right]$∴ arccos[cos(arcsiny)] | $=$= | $\arccos\frac{1}{2}$arccos12 |
| $\therefore\ \ \arcsin y$∴ arcsiny | $=$= | $\frac{\pi}{3}$π3 |
| $\therefore\ \ \sin(\arcsin y)$∴ sin(arcsiny) | $=$= | $\sin\frac{\pi}{3}$sinπ3 |
| $\therefore\ \ y$∴ y | $=$= | $\frac{\sqrt{3}}{2}$√32 |
Practice questions
Question 1
Consider the equation $-6\sin^{-1}x=\pi$−6sin−1x=π.
Solve the equation for $\sin^{-1}x$sin−1x.
Consider your answer from part a. Does the equation have a solution?
Not enough information
AYes
BNo
CSolve $\sin^{-1}x=-\frac{\pi}{6}$sin−1x=−π6 for $x$x.
Question 2
Consider the equation $\sin^{-1}x-\tan^{-1}1=-\frac{\pi}{4}$sin−1x−tan−11=−π4.
Solve for $\sin^{-1}x$sin−1x.
Consider your answer from part (a). Does the equation have a solution?
Yes
ANot enough information
BNo
CSolve $\sin^{-1}x=0$sin−1x=0 for $x$x.
Question 3
Solve exactly for $x$x:
$\cos^{-1}x=\sin^{-1}\frac{5}{13}$cos−1x=sin−1513
Question 4
Solve $y=5\tan3x$y=5tan3x for $x$x in terms of $y$y, where $x$x is in $($($-\frac{\pi}{6}$−π6, $\frac{\pi}{6}$π6$)$).
Question 5
Solve $y=\tan\left(3x-1\right)$y=tan(3x−1) for $x$x, where $x$x is in $($($\frac{1}{3}-\frac{\pi}{6}$13−π6, $\frac{1}{3}+\frac{\pi}{6}$13+π6$)$).
Solving Trigonometric equations
Given a particular value of a trigonometric function, we may wish to find the values of the domain variable that are mapped to that function value. Or, given two distinct functions defined on the same domain, we may ask what values of the domain variable make the two functions equal.
These situations are what is meant by the idea of solving trigonometric (and other) equations. We are finding the values of the variable that make the equations true.
Worked examples
Example 3
Given the function defined by $2\sin x$2sinx for $x\in[0,2\pi]$x∈[0,2π], find the values of $x$x for which the function attains the value $1$1.
The problem is to solve the equation $2\sin x=1$2sinx=1, giving the solutions that are between $0$0 and $2\pi$2π. The equation is rearranged to read $\sin x=\frac{1}{2}$sinx=12 and we proceed by asking what values of $x$x produce this function value.
From previous work we recall that $\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6=12 and consequently that $\sin\left(\pi-\frac{\pi}{6}\right)=\sin\frac{5\pi}{6}=\frac{1}{2}$sin(π−π6)=sin5π6=12. The sine function is negative between $\pi$π and $2\pi$2π. So, these two are the only values that satisfy the equation.
We can give the solution as $x=\frac{\pi}{6}$x=π6, $\frac{5\pi}{6}$5π6.
If we had not known that $\sin\frac{\pi}{6}=\frac{1}{2}$sinπ6=12, it would have been necessary to use the inverse sine function. The steps are as follows:
| $\sin x$sinx | $=$= | $\frac{1}{2}$12 |
| $x$x | $=$= | $\arcsin\frac{1}{2}$arcsin12 |
| $\approx$≈ | $0.5236$0.5236 |
The number $0.5236$0.5236 is obtained from a hand-held calculator or a desktop computer or from a printed table of sines. It is an approximation to the true value $\frac{\pi}{6}$π6 which cannot be written down with a finite decimal string.
A calculator gives only one solution since the inverse sine function is defined only on the interval $[-\frac{\pi}{2},\frac{\pi}{2}]$[−π2,π2]. We subtract the calculator value from $\pi$π to obtain the second solution, $\pi-0.5236\approx2.6180$π−0.5236≈2.6180.
Another way to consider the solution is to think of $2$2 functions $y=2\sin x$y=2sinx and $y=1$y=1. To solve $2\sin x=1$2sinx=1, is the same as graphically finding the intersection of these two separate lines.
You can see both solutions mentioned above here.
Example 4
Solve $4\cos x+1=3$4cosx+1=3 for $x\in[0,2\pi]$x∈[0,2π].
The equation is rearranged to $\cos x=\frac{1}{2}$cosx=12. In the domain $[0,2\pi]$[0,2π], we can expect to find a solution somewhere in the interval $[0,\frac{\pi}{2}]$[0,π2] and another in the interval $[\frac{3\pi}{2},2\pi]$[3π2,2π] as these are the intervals in which the cosine function is positive.
Again, from previous work, we know that $\cos x=\frac{1}{2}$cosx=12 when $x=\frac{\pi}{3}$x=π3. The next solution must occur when $x=2\pi-\frac{\pi}{3}$x=2π−π3. That is, when $x=\frac{5\pi}{3}$x=5π3.
Thus, the solutions are $x=\frac{\pi}{3},\frac{5\pi}{3}$x=π3,5π3.
The graphs for this can also be used to find the solution. Graph both y=4\cos{x} + 1 and y=3 on the same set of axes. The intersections of these curves are the solutions.
Trigonometric equations do not always have easy solutions. It can be necessary to use a numerical method to obtain an approximate solution.
Practice questions
Question 6
Find all values of $\theta$θ in the interval $[$[$0,2\pi$0,2π$)$) that satisfy $\sin\theta=\frac{1}{2}$sinθ=12.
Write all values on the same line separated by a comma.
Question 7
Solve $6\cos x-3\sqrt{2}=0$6cosx−3√2=0 over the interval $\left[0,2\pi\right)$[0,2π).
Question 8
Solve for $x$x over the interval $[$[$0$0, $2\pi$2π$)$).
$\sin\left(\frac{x}{2}\right)=\frac{\sqrt{3}}{2}$sin(x2)=√32