MAT 1275: Introduction to Mathematical Analysis. Graphs and Simplest Equations for Basic Trigonometric Functions. y=sin( x) Function

Transcription

1 MAT 275: Introduction to Mathematical Analsis Dr. A. Rozenblum Graphs and Simplest Equations for Basic Trigonometric Functions We consider here three basic functions: sine, cosine and tangent. For them, we will construct graphs, and solve three simplest equations: sin = a, cos = a, and tan = a. We call these equations simplest because solution of man more complicated equations can be reduced to simplest. We will use radian measure. Function =sin( ) Let s recall definition of sine for arbitrar angle: we draw an angle in standard position in sstem of coordinates with unit circle, and consider point of intersection of the terminal side of the angle with unit circle. Sine is the second (vertical) coordinate of this point. To draw graph of sine, we will move along unit circle starting with the right-most position and observe how vertical coordinate of points on the unit circle changes from quadrant to quadrant. Obvious that in the first quadrant vertical coordinate (i.e. sine) increases from zero to one: sin 0 For graph of sine, we will use another sstem of coordinates in which we mark angle on horizontal ais (we will use letter instead of ), and mark sin on vertical -ais. If ou pick up several values of angles in the first quadrant (i.e. from 0 to π /2), calculate sin, and plot points in the sstem of coordinates, ou will see that sine increases not along a strait line. Instead, it increases along the curve: = sin 0 π /2 In similar wa, in the second quadrant (from π /2 to π ), sine decreases from to 0:

2 2 Continue moving along unit circle, we see that in third quadrant (from π to 3 π /2) sine decreases from 0 to, and in fourth quadrant (from 3 π /2 to 2π ) sine increases from to 0. At this point, we get the graph of sine at one full ccle (we also sa, on one period interval): Graph of sine on one period interval If we continue to move arount unit cirlcle in either direction (positive or negative), we etend graph of sine to the entire number line, i.e. for all the values of from to + : Entire graph of function = sin You can see that domain of sine is interval (, + ) and range is [, ]. Sine is periodical function with the period 2π. It means that sine repeats itself on each interval of the length 2π. More formall, sin( + 2 π ) = sin( ) (Periodic propert of sine) Also, graph is smmetric with respect to origin. Algebraicall, it means that sin( ) = sin( ) (Odd propert of sine)

3 Solving Simplest Equation sin( ) = a on One Period Interval [ 0, 2π ) Notice that the right point 2π is not included in the interval. The reason is that this point corresponds to the angle of 0 o, which is alread taken for the left point 0. In this interval, equation sin( ) = a ma have zero, one or two solutions depending on the value of a. More precisel, the following statement is true. Proposition. Consider the equation sin( ) ) If a >, the equation does not have solutions. 2) If a =, the equation has one solution. 3) If a <, the equation has two solutions. = a in the interval [ 0, 2π ). Then It is eas to check all three statements using geometric interpretation of the equation sin( ) = a: its solutions are -coordinates of points of intersection of the horizontal line = a with the graph of sine. Drawing this line, we can see three different locations of it, i.e. three different values of number a. ) a >. This inequalit is equivalent to a > or a <. Horizontal line = a is located above or below the graph of sine, so no point of intersection, and no solutions. 2) a =. This equalit is equivalent to a = or a =. In both cases line = a touches the graph of sine onl in one point: For equation sin( ) =, the solution is = π /2. For equation sin( ) =, the solution is = 3 π / 2. 3) a <. This inequalit is equivalent to < a <. Line = a is located between lines = and = and intersects the graph of sine eactl into two points. Solutions of the equation sin( ) = a, depend on the sign of number a. Case : a is non-negative ( 0 a < ). One of the solutions we can find immediatel: = sin ( a). This is an acute angle. Another (obtuse) solution we can get using sin π = sin, so second solution is = π sin ( a). reduction formula ( ) Note: Another wa to get both solutions is to use definition of sine as vertical coordinate of points on unit circle. If ou mark a on vertical ais, ou will see two angles for which sine is a: sin ( a) in the first quadrant and π sin ( a) in the second: 3

4 4 π sin ( a) a sin ( a) Case 2: a is negative ( < a < 0). The value sin ( a) is negative and we cannot accept it as a root in the interval [ 0, 2π ). To find positive roots, we can use either reduction formulas or reference angle. We will use reference angle here. The angle sin ( a) is in fourth quadrant, and its reference angle denoted b r is = sin ( a). One root is 2π r and another is π + r. You can see this from the picture r π + r 2π r a (Reference angle) r Eample. Solve the equation 2sin( ) + 4 = 5 in the interval [ 0, 2π ). Solution. It is eas to reduce this equation to basic one b solving for sin( ) : sin( ) = / 2. This is case above: a = /2 >0. The equation has two roots. One of them we can find using calculator (or using special value /2): = sin (/ 2) = 30 = π / 6 Second root is supplement to the first: = π π /6= 5 π /6. Eample 2. Solve the equation 2sin( ) = 2 in the interval [ 0, 2π ). o. Solution. Solving for sin( ), we get basic equation sin( ) = 2 / 2. This is case 2 above: a = 2/2< 0. The equation has two roots. Using calculator (or using special o value 2/2), we have sin ( 2 / 2) = 45 = π / 4. We cannot accept this value as a root since it is negative. Reference angle for this angle is π /4. Two positive roots are = 2 π π /4= 7 π /4 and = π + π /4= 5 π /4.

5 5 Function =cos( ) We can proceed here similar to function sine. Let s do this in brief form hoping that the reader can restore details ourself. B definition, cosine is first (horizontal) coordinate of a point on unit circle that corresponds to given angle: 0 cos Moving around the unit circle from quadrant to quadrant, we can construct the graph of cosine b observing how horizontal coordinate is changing. For eample, in first quadrant when angle runs from 0 to π /2, cosine decreases from to 0: = cos 0 π /2 In second quadrant cosine continue to decrease from 0 to, in third quadrant it increases from to 0, and, finall, in fourth quadrant increases from 0 to. Here is the graph of cosine at one full ccle (on one period interval): Graph of cosine on one period interval π π 3π 2π 2 2 If we etend graph to the entire -ais, we get complete graph of cosine: Entire graph of function = cos

6 6 As for sine, the domain of cosine is (, + ), range is [, ], and cosine is periodical function with the same period 2π. Graph of cosine is smmetric with respect to -ais: cos( ) = cos( ) (Even propert of cosine). Solving Simplest Equation cos( ) = a on One Period Interval [ 0, 2π ) Number of solutions for this equation is eactl the same as for sine: Proposition 2. Consider the equation cos( ) = a on the interval [ 0, 2π ). Then ) If a >, the equation does not have solutions. 2) If a =, the equation has one solution. For equation cos( ) =, the solution is = 0. For equation cos( ) =, the solution is = π. 3) If a <, the equation has two solutions: = cos ( a) and Reasons are the same as for sine. = 2π cos ( a). Eample 3. Solve the equation 2cos( ) + 4 = 3 in the interval [ 0, 2π ). Solution. Solving this equation for cos( ), we have cos( ) = / 2. This equation has two solutions: = cos ( / 2) = 20 o = 2 π / 3, and = 2π 2 π /3= 4 π /3. Function =tan( ) On unit circle in the sstem of coordinates, we can interpret tangent like this. On the right side of unit circle, we draw vertical line and etend terminal side of the angle to meet with that line. Then tangent is the vertical coordinate of the point of intersection. Here are pictures of tangent when angle is located in each of the quadrants: tan tan Angle is in st quadrant Angle is in 2 nd quadrant

7 7 tan tan Angle is in 3 rd quadrant Angle is in 4 th quadrant We will draw graph of tangent in the wa similar to as we did for sine and cosine. Moving along unit circle in the first quadrant, notice that tangent increases from zero to infinit, and its graph in the st quadrant is this = tan 0 π /2 Line = π /2 becomes vertical asmptote. Continue moving in 2 nd quadrant, we get the picture = tan 0 π /2 π Moving in 3 rd and 4 th quadrants, we get graph of tangent on interval [ 0, 2π ):

8 8 Graph of tangent on [ 0, 2π ) interval Continue moving around unit circle in both directions, we can draw complete graph of tangent: Entire graph of function = tan We see that graph consists of infinite number of branches, also it has infinite number of vertical asmptotes. The graph is smmetric with respect to origin, so tangent is odd tan = tan. It repeats itself on π -length interval, so tangent has period function: ( ) ( ) π. Solving Simplest Equation tan( ) = a on Interval [ 0, 2π ) An horizontal line = a intersects the graph of tangent in [ 0, 2π ) interval alwas in two points, so the equation tan ( ) = a alwas has two solutions. Proposition 3. For an a, equation tan() = a has two solutions in the interval [ 0, 2π ). The solutions are ) If a 0, then = tan ( a) and = π + tan ( a) 2) If a < 0, then = 2π + tan ( a) and = π + tan ( a)

9 9 Notice that both solutions alwas differ b π (which is the period of tangent). If a > 0, angle tan (a) is acute and positive, and another solution π + tan (a) is obtuse angle. If a < 0, angle tan (a) is acute and negative, and we replace it with 2π + tan (a) (which is the same geometric angle). Another solution is π + tan (a) which is obtuse angle. Eample 4. Solve the equation 3tan() 2 3 = 3 in the interval [ 0, 2π ). Solution. Soving the equation for tan() we get basic equation tan() = solutions is tan ( 3) = 60! = π / 3. Another solution is π + π / 3 = 4π / 3. Eample 5. Solve the equation 3tan() = 3 3 in the interval [ 0, 2π ). 3. One of the Solution. Soving the equation for tan() we get basic equation tan() = 3 / 3. We have tan 3 / 3 ( ) = 30! = π / 6. This angle is negative and we replace it with 2π π / 6 =π / 6, which is one of the solutions. Second solution is π π / 6 = 5π / 6.