4.4 Trigonometrische Gleichungen
Aus Online Mathematik Brückenkurs 1
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The solutions can also be obtained in the figure below where the graph of <math>y = \sin x</math> intersects the line <math>y=\tfrac{1}{2}</math>. | The solutions can also be obtained in the figure below where the graph of <math>y = \sin x</math> intersects the line <math>y=\tfrac{1}{2}</math>. | ||
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Version vom 11:06, 21. 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.
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 Vorlage:Displayed math 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.
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
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 .
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, Vorlage:Displayed math
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
Vorlage:Displayed math
which can be simplified to the equation (after division by 2)
The left-hand side can factorised by using the squaring rule to give
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
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
Vorlage:Displayed math
Factorising out \displaystyle \sin x one gets Vorlage:Displayed math
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 Vorlage:Displayed math
Example 6
Solve the equation \displaystyle \,\sin 2x =4 \cos x.
By rewriting the equation using the formula for double-angles one gets
Vorlage:Displayed math
We divide both sides with 2 and factorise out \displaystyle \cos x, which gives Vorlage:Displayed 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
- \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 haveVorlage:Displayed math
This is a quadratic equation in \displaystyle \cos x, which has the solutions Vorlage:Displayed math
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 Vorlage:Displayed math
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
