4.4 Trigonometrische Gleichungen
Aus Online Mathematik Brückenkurs 1
K (hat „4.4 Trigonometric equations“ nach „4.4 Trigonometrische Gleichungen“ verschoben: Page title translated) |
K (Robot: Automated text replacement (-{{Displayed math +{{Abgesetzte Formel)) |
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However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles with a value of the sine <math>\tfrac{1}{2}</math> are | However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles with a value of the sine <math>\tfrac{1}{2}</math> are | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\begin{cases} |
x &= \dfrac{\pi}{6} + 2n\pi\\ | x &= \dfrac{\pi}{6} + 2n\pi\\ | ||
x &= \dfrac{5\pi}{6} + 2n\pi | x &= \dfrac{5\pi}{6} + 2n\pi | ||
| Zeile 64: | Zeile 64: | ||
We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other direction in the unit circle, which produces the same value for the cosine is the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution | We know that cosine is <math>\tfrac{1}{2}</math> for the angle <math>\pi/3</math>. The only other direction in the unit circle, which produces the same value for the cosine is the angle <math>-\pi/3</math>. Adding an integral number of revolutions to these angles we get the general solution | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}} |
where <math>n</math> is an arbitrary integer. | where <math>n</math> is an arbitrary integer. | ||
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Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution <math>\pi/3</math>, <math>\pi/3 +\pi</math>, <math>\pi/3+ \pi +\pi</math> and so on. The general solution can be obtained by using the solution <math>\pi/3</math> and adding or subtracting multiples of <math>\pi</math>, | Therefore, we see that the solutions to <math>\tan x = \sqrt{3}</math> repeat themselves every half revolution <math>\pi/3</math>, <math>\pi/3 +\pi</math>, <math>\pi/3+ \pi +\pi</math> and so on. The general solution can be obtained by using the solution <math>\pi/3</math> and adding or subtracting multiples of <math>\pi</math>, | ||
| - | {{ | + | {{Abgesetzte Formel||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}} |
where <math>n</math> s an arbitrary integer. | where <math>n</math> s an arbitrary integer. | ||
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Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math> giving | Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math> giving | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,}</math>}} |
which can be simplified to the equation (after division by 2) | which can be simplified to the equation (after division by 2) | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\cos^2\!x - 2 \cos x +1 =0\,\mbox{.}</math>}} |
The left-hand side can factorised by using the squaring rule to give | The left-hand side can factorised by using the squaring rule to give | ||
| - | {{ | + | {{Abgesetzte Formel||<math>(\cos x-1)^2 = 0\,\mbox{.}</math>}} |
This equation can only be satisfied if <math>\cos x = 1</math>. The basic equation <math>\cos x=1</math> can be solved in the normal way and the complete solution is | This equation can only be satisfied if <math>\cos x = 1</math>. The basic equation <math>\cos x=1</math> can be solved in the normal way and the complete solution is | ||
| - | {{ | + | {{Abgesetzte Formel||<math> |
x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}} | x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}} | ||
</div> | </div> | ||
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According to the Pythagorean identity <math>\sin^2\!x + \cos^2\!x = 1</math>, i.e. <math>1 – \cos^2\!x = \sin^2\!x</math>, the equation can be written as | According to the Pythagorean identity <math>\sin^2\!x + \cos^2\!x = 1</math>, i.e. <math>1 – \cos^2\!x = \sin^2\!x</math>, the equation can be written as | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}} |
Factorising out <math>\sin x</math> one gets | Factorising out <math>\sin x</math> one gets | ||
| - | {{ | + | {{Abgesetzte Formel||<math> |
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}} | \sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}</math>}} | ||
From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in example 1. The solutions turn out to be | From this factorised form of the equation, we see that the solutions either have to satisfy <math>\sin x = 0</math> or <math>\sin x = -\tfrac{1}{2}</math>, which are two basic equations of the type <math>\sin x = a</math> and can be solved as in example 1. The solutions turn out to be | ||
| - | {{ | + | {{Abgesetzte Formel||<math> |
\begin{cases} | \begin{cases} | ||
x &= n\pi\\ | x &= n\pi\\ | ||
| Zeile 147: | Zeile 147: | ||
By rewriting the equation using the formula for double-angles one gets | By rewriting the equation using the formula for double-angles one gets | ||
| - | {{ | + | {{Abgesetzte Formel||<math>2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.}</math>}} |
We divide both sides with 2 and factorise out <math>\cos x</math>, which gives | We divide both sides with 2 and factorise out <math>\cos x</math>, which gives | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.}</math>}} |
As the product of factors on the left-hand side can only be zero if one of the factors is zero, we have reduced the original equation into two basic equations | As the product of factors on the left-hand side can only be zero if one of the factors is zero, we have reduced the original equation into two basic equations | ||
| Zeile 166: | Zeile 166: | ||
| - | Using the Pythagorean identity one can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then we will have{{ | + | Using the Pythagorean identity one can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then we will have{{Abgesetzte Formel||<math> |
\begin{align*} | \begin{align*} | ||
4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\ | 4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\ | ||
| Zeile 175: | Zeile 175: | ||
This is a quadratic equation in <math>\cos x</math>, which has the solutions | This is a quadratic equation in <math>\cos x</math>, which has the solutions | ||
| - | {{ | + | {{Abgesetzte Formel||<math> |
\cos x = -\tfrac{3}{2} \quad\text{and}\quad | \cos x = -\tfrac{3}{2} \quad\text{and}\quad | ||
\cos x = \tfrac{1}{2}\,\mbox{.}</math>}} | \cos x = \tfrac{1}{2}\,\mbox{.}</math>}} | ||
Since the value of <math>\cos x</math> is between <math>–1</math> and <math>1</math> the equation <math>\cos x=-\tfrac{3}{2}</math> has no solutions. That leaves only the basic equation | Since the value of <math>\cos x</math> is between <math>–1</math> and <math>1</math> the equation <math>\cos x=-\tfrac{3}{2}</math> has no solutions. That leaves only the basic equation | ||
| - | {{ | + | {{Abgesetzte Formel||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}} |
that may be solved as in example 2. | that may be solved as in example 2. | ||
Version vom 08:13, 22. Okt. 2008
Contents:
- The basic equations of trigonometry
- Simple trigonometric equations
Learning outcomes:
After this section, you will have learned how to:
- Solve the basic equations of trigonometry
- Solve trigonometric equations that can be reduced to basic equations.
Basic equations
Trigonometric equations can be very complicated, but there are also many types of trigonometric equations which can be solved using relatively simple methods. Here, we shall start by looking at the most basic trigonometric equations, of the type \displaystyle \sin x = a, \displaystyle \cos x = a and \displaystyle \tan x = a.
These equations usually have an infinite number of solutions, unless the circumstances limit the number of possible solutions (for example, if one is looking for an acute angle).
Example 1
Solve the equation \displaystyle \,\sin x = \frac{1}{2}.
Our task is to determine all the angles that have a sine with the value \displaystyle \tfrac{1}{2}. The unit circle helps us in this. Note that here the angle is designated as \displaystyle x.
![[Image]](/wikis/2009/bridgecourse1-TU-Berlin/img_auth.php/metapost/1/1/2/112f5660b91bccb8295414c925bf9198.png)
In the figure, we have shown the two directions that give us points which have a y-coordinate \displaystyle \tfrac{1}{2} on the unit circle, i.e. angles with a sine value \displaystyle \tfrac{1}{2}. The first is the standard angle \displaystyle 30^\circ = \pi / 6 and by symmetry the other angle makes \displaystyle 30^\circ with the negative x-axis. This means that the angle is \displaystyle 180^\circ – 30^\circ = 150^\circ or in radians \displaystyle \pi – \pi / 6 = 5\pi / 6. These are the only solutions to the equation \displaystyle \sin x = \tfrac{1}{2} between \displaystyle 0 and \displaystyle 2\pi.
However, we can add an arbitrary number of revolutions to these two angles and still get the same value for the sine . Thus all angles with a value of the sine \displaystyle \tfrac{1}{2} are
\displaystyle \begin{cases}
x &= \dfrac{\pi}{6} + 2n\pi\\
x &= \dfrac{5\pi}{6} + 2n\pi
\end{cases}
|
where \displaystyle n is an arbitrary integer. This is called the general solution to the equation.
The solutions can also be obtained in the figure below where the graph of \displaystyle y = \sin x intersects the line \displaystyle y=\tfrac{1}{2}.
![[Image]](/wikis/2009/bridgecourse1-TU-Berlin/img_auth.php/metapost/c/a/5/ca5f64ed4d32214ab240f7cc3a95f4a5.png)
Example 2
Solve the equation \displaystyle \,\cos x = \frac{1}{2}.
We once again study the unit circle.
![[Image]](/wikis/2009/bridgecourse1-TU-Berlin/img_auth.php/metapost/5/7/7/577c29f5440e418620c30ead2b1e871b.png)
We know that cosine is \displaystyle \tfrac{1}{2} for the angle \displaystyle \pi/3. The only other direction in the unit circle, which produces the same value for the cosine is the angle \displaystyle -\pi/3. Adding an integral number of revolutions to these angles we get the general solution
| \displaystyle x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,} |
where \displaystyle n is an arbitrary integer.
Example 3
Solve the equation \displaystyle \,\tan x = \sqrt{3}.
A solution to the equation is the standard angle \displaystyle x=\pi/3.
If we study the unit circle then we see that tangent of an angle is equal to the slope of the straight line through the origin making an angle \displaystyle x with the positive x-axis .
![[Image]](/wikis/2009/bridgecourse1-TU-Berlin/img_auth.php/metapost/3/0/e/30e856a6ebc45b43349b481848cc1276.png)
Therefore, we see that the solutions to \displaystyle \tan x = \sqrt{3} repeat themselves every half revolution \displaystyle \pi/3, \displaystyle \pi/3 +\pi, \displaystyle \pi/3+ \pi +\pi and so on. The general solution can be obtained by using the solution \displaystyle \pi/3 and adding or subtracting multiples of \displaystyle \pi,
| \displaystyle x = \pi/3 + n \cdot \pi\,\mbox{,} |
where \displaystyle n s an arbitrary integer.
Somewhat more complicated equations
Trigonometric equations can vary in many ways, and it is impossible to give a full catalogue of all possible equations. But let us study some examples where we can use our knowledge of solving basic equations.
Some trigonometric equations can be simplified by being rewritten with the help of trigonometric relationships. This, for example, could lead to a quadratic equation, as in the example below where one uses \displaystyle \cos 2x = 2 \cos^2\!x – 1.
Example 4
Solve the equation \displaystyle \,\cos 2x – 4\cos x + 3= 0.
Rewrite by using the formula \displaystyle \cos 2x = 2 \cos^2\!x – 1 giving
| \displaystyle (2 \cos^2\!x – 1) – 4\cos x + 3 = 0\,\mbox{,} |
which can be simplified to the equation (after division by 2)
| \displaystyle \cos^2\!x - 2 \cos x +1 =0\,\mbox{.} |
The left-hand side can factorised by using the squaring rule to give
| \displaystyle (\cos x-1)^2 = 0\,\mbox{.} |
This equation can only be satisfied if \displaystyle \cos x = 1. The basic equation \displaystyle \cos x=1 can be solved in the normal way and the complete solution is
\displaystyle
x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}
|
Example 5
Solve the equation \displaystyle \,\frac{1}{2}\sin x + 1 – \cos^2 x = 0.
According to the Pythagorean identity \displaystyle \sin^2\!x + \cos^2\!x = 1, i.e. \displaystyle 1 – \cos^2\!x = \sin^2\!x, the equation can be written as
| \displaystyle \tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.} |
Factorising out \displaystyle \sin x one gets
\displaystyle
\sin x\,\cdot\,\bigl(\tfrac{1}{2} + \sin x\bigr) = 0 \, \mbox{.}
|
From this factorised form of the equation, we see that the solutions either have to satisfy \displaystyle \sin x = 0 or \displaystyle \sin x = -\tfrac{1}{2}, which are two basic equations of the type \displaystyle \sin x = a and can be solved as in example 1. The solutions turn out to be
\displaystyle
\begin{cases}
x &= n\pi\\
x &= -\pi/6+2n\pi\\
x &= 7\pi/6+2n\pi
\end{cases}
\qquad (\,n\ \text{ arbitrary integer})\mbox{.}
|
Example 6
Solve the equation \displaystyle \,\sin 2x =4 \cos x.
By rewriting the equation using the formula for double-angles one gets
| \displaystyle 2\sin x\,\cos x – 4 \cos x = 0\,\mbox{.} |
We divide both sides with 2 and factorise out \displaystyle \cos x, which gives
| \displaystyle \cos x\,\cdot\,( \sin x – 2) = 0\,\mbox{.} |
As the product of factors on the left-hand side can only be zero if one of the factors is zero, we have reduced the original equation into two basic equations
- \displaystyle \cos x = 0,
- \displaystyle \sin x = 2.
But \displaystyle \sin x can never be greater than 1, so the equation \displaystyle \sin x = 2 has no solutions. That leaves just \displaystyle \cos x = 0, and using the unit circle gives the general solution \displaystyle x = \pi / 2 + n \cdot \pi.
Example 7
Solve the equation \displaystyle \,4\sin^2\!x – 4\cos x = 1.
Using the Pythagorean identity one can replace \displaystyle \sin^2\!x by \displaystyle 1 – \cos^2\!x. Then we will have
\displaystyle
\begin{align*}
4 (1 – \cos^2\!x) – 4 \cos x &= 1\,\mbox{,}\\
4 – 4 \cos^2\!x – 4 \cos x &= 1\,\mbox{,}\\
–4\cos^2\!x – 4 \cos x + 4 – 1 &= 0\,\mbox{,}\\
\cos^2\!x + \cos x – \tfrac{3}{4} &= 0\,\mbox{.}\\
\end{align*}
|
This is a quadratic equation in \displaystyle \cos x, which has the solutions
\displaystyle
\cos x = -\tfrac{3}{2} \quad\text{and}\quad
\cos x = \tfrac{1}{2}\,\mbox{.}
|
Since the value of \displaystyle \cos x is between \displaystyle –1 and \displaystyle 1 the equation \displaystyle \cos x=-\tfrac{3}{2} has no solutions. That leaves only the basic equation
| \displaystyle \cos x = \tfrac{1}{2}\,\mbox{,} |
that may be solved as in example 2.
Study advice
Basic and final tests
After you have read the text and worked through the exercises, you should do the basic and final tests to pass this section. You can find the link to the tests in your student lounge.
Remember:
It is a good idea to learn the most common trigonometric formulas (identities) and practice simplifying and manipulating trigonometric expressions.
It is important to be familiar with the basic equations, such as \displaystyle \sin x = a, \displaystyle \cos x = a or \displaystyle \tan x = a (where \displaystyle a is a real number). It is also important to know that these equations typically have infinitely many solutions.
Useful web sites
