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__NOTOC__ {| border="0" cellspacing="0" cellpadding="0" height="30" width="100%" | style="border-bottom:1px solid #797979" width="5px" | &nbsp; {{Vald flik|[[4.4 Trigonometriska ekvationer|Theory]]}} {{Ej vald flik|[[4.4 Övningar|Exercises]]}} | style="border-bottom:1px solid #797979" width="100%"| &nbsp; |} {{Info| '''Content:''' * The basic equations of trigonometry *Simple trigonometric equations }} {{Info| '''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 a basic equation. }} == 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 <math>\sin x = a</math>, <math>\cos x = a</math> and <math>\tan x = a</math>. 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). <div class="exempel"> ''' Example 1''' Solve the equation <math>\,\sin x = \frac{1}{2}</math>. Our task is to determine all the angles that have a sine with the value <math>\tfrac{1}{2}</math>. The unit circle helps us in this . Note that here the angle is designated as <math>x</math>. <center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/6 resp. 5π/6}}</center> In the figure, we have shown the two directions that give us points which have a ''y''- coordinate <math>\tfrac{1}{2}</math> on the unit circle, i.e. angles with a sine value <math>\tfrac{1}{2}</math>. The first is the standard angle <math>30^\circ = \pi / 6</math> and by symmetry the other angle makes <math>30^\circ</math> with the negative ''x''-axis, This means that the angle is <math>180^\circ – 30^\circ = 150^\circ</math> or in radians <math>\pi – \pi / 6 = 5\pi / 6</math>. These are the only solutions to the equation <math>\sin x = \tfrac{1}{2}</math> between <math>0</math> and <math>2\pi</math>. 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 {{Fristående formel||<math>\begin{cases} x &= \dfrac{\pi}{6} + 2n\pi\\ x &= \dfrac{5\pi}{6} + 2n\pi \end{cases}</math>}} where <math>n</math> is an arbitrary integer. This is called the general solution to the equation. The solutions also seen in the figure below where the graph of <math>y = \sin x</math> intersect the line <math>y=\tfrac{1}{2}</math>. <center>{{:4.4 - Figur - Kurvorna y = sin x och y = ½}}</center> </div> <div class="exempel"> ''' Example 2''' Solve the equation <math>\,\cos x = \frac{1}{2}</math>. We once again study the unit circle. <center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. -π/3}}</center> 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 {{Fristående formel||<math>x = \pm \pi/3 + n \cdot 2\pi\,\mbox{,}</math>}} where <math>n</math> is an arbitrary integer. </div> <div class="exempel"> ''' Example 3''' Solve the equation <math>\,\tan x = \sqrt{3}</math>. A solution to the equation is the standard angle <math>x=\pi/3</math>. 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 <math>x</math> with the positive ''x''-axis . <center>{{:4.4 - Figur - Två enhetscirklar med vinklar π/3 resp. π+π/3}}</center> 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>, {{Fristående formel||<math>x = \pi/3 + n \cdot \pi\,\mbox{,}</math>}} where <math>n</math> s an arbitrary integer. </div> == 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 equations, as in the example below where one uses <math>\cos 2x = 2 \cos^2\!x – 1</math>. <div class="exempel"> ''' Example 4''' Solve the equation <math>\,\cos 2x – 4\cos x + 3= 0</math>. Rewrite by using the formula <math>\cos 2x = 2 \cos^2\!x – 1</math> giving {{Fristående 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) {{Fristående 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 {{Fristående 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 {{Fristående formel||<math> x = 2n\pi \qquad (\,n \mbox{ arbitrary integer).}</math>}} </div> <div class="exempel"> ''' Example 5''' Solve the equation <math>\,\frac{1}{2}\sin x + 1 – \cos^2 x = 0</math>. 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 {{Fristående formel||<math>\tfrac{1}{2}\sin x + \sin^2\!x = 0\,\mbox{.}</math>}} Factorising out <math>\sin x</math> one gets {{Fristående formel||<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 {{Fristående formel||<math> \begin{cases} x &= n\pi\\ x &= -\pi/6+2n\pi\\ x &= 7\pi/6+2n\pi \end{cases} \qquad (\,n\ \text{godtyckligt heltal})\mbox{.}</math>}} </div> <div class="exempel"> ''' Example 6''' Solve the equation <math>\,\sin 2x =4 \cos x</math>. By rewriting the equation using the formula for double-angles one gets {{Fristående 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 {{Fristående 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 * <math>\cos x = 0</math>, * <math>\sin x = 2</math>. But <math>\sin x</math> can never be greater than 1, so the equation <math>\sin x = 2</math> has no solutions. That leaves just <math>\cos x = 0</math>, and using the unit circle gives the general solution <math>x = \pi / 2 + n \cdot \pi</math>. </div> <div class="exempel"> ''' Example 7''' Solve the equation <math>\,4\sin^2\!x – 4\cos x = 1</math>. Using the Pythagorean identity one can replace <math>\sin^2\!x</math> by <math>1 – \cos^2\!x</math>. Then we will have{{Fristående formel||<math> \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*}</math>}} This is a quadratic equation in <math>\cos x</math>, which has the solutions {{Fristående formel||<math> \cos x = -\tfrac{3}{2} \quad\text{och}\quad \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 {{Fristående formel||<math>\cos x = \tfrac{1}{2}\,\mbox{,}</math>}} that may be solved as in example 2. </div> [[4.4 Övningar|Exercises]] <div class="inforuta" style="width:580px;"> '''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 good idea to learns 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 <math>\sin x = a</math>, <math>\cos x = a</math> or <math>\tan x = a</math> (where <math>a</math> is a real number). It is also important to know that these equations typically have infinitely many solutions. '''Reviews''' For those of you who want to deepen your studies or need more detailed explanations consider the following references [http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/index.asp Learn more about trigonometric equations in Theducations gymnasielexikon ] [http://www.theducation.se/kurser/umaprep/4_trigonometri/44_trig_ekvationer/445_typ_asinx/index.asp Practise trigonometric calculations in Theducations gymnasielexikon ] '''Länktips''' [http://www.ies.co.jp/math/java/trig/ABCsinX/ABCsinX.html Experiment with the graph y = a sin b (x-c) ] [http://www.theducation.se/kurser/experiment/gyma/applets/ex45_derivatasinus/Ex45Applet.html Experiment with the derivative of sin x] </div>

• 4.4 - Bild - Die Kurven y = sin x und y = ½
• 4.4 - Bild - Zwei Einheitskreise mit Winkeln π/3 und -π/3,
• 4.4 - Bild - Zwei Einheitskreise mit Winkeln π/3 und π+π/3,
• 4.4 - Bild - Zwei Einheitskreise mit Winkeln π/6 und 5π/6,
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