# 14 Trigonometry (1) Case Study 14.1 Introduction to Trigonometry

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14 Trigonometry (1) Case Study 14.1 Introduction to Trigonometry
14.2 Trigonometry Ratios of Arbitrary Angles 14.3 Finding Trigonometric Ratios Without Using a Calculator 14.4 Trigonometric Identities 14.5 Trigonometric Equations 14.6 Graphs of Trigonometric Functions 14.7 Graphical Solutions of Trigonometric Equations Chapter Summary

Case Study How can we find the shape of the sound wave generated by a tuning fork? The sound wave generated can be displayed by using a CRO. The figure shows the sound wave generated by the tuning fork displayed on a cathode-ray oscilloscope (CRO). The pattern of the waveform of sound has the same shape as the graph of a trigonometric function. The graph repeats itself at regular intervals. Such an interval is called the period.

14.1 Introduction to Trigonometry
A. Angles of Rotation In the figure, the centre of the circle is O and its radius is r. Suppose OA is rotated about O and it reaches OP, the angle q formed is called an angle of rotation. OA: initial side OP: terminal side If OA is rotated in an anti-clockwise direction, the value of q is positive. If OA is rotated in a clockwise direction, then the value of q is negative.

14.1 Introduction to Trigonometry
A. Angles of Rotation Remarks: The figure shows the measures of two different angles: 130 and 230. However, they have the same initial side OA and terminal side OP. 2. The initial side and terminal side of 410 coincide with that of 50 as shown in the figure.

14.1 Introduction to Trigonometry
B. Quadrants In a rectangular coordinate plane, the x-axis and the y-axis divide the plane into four parts as shown in the figure. Each part is called a quadrant. Notes: The x-axis and the y-axis do not belong to any of the four quadrants. For an angle of rotation, the position where the terminal side lies determines the quadrant in which the angle lies. Thus, we can see that for an angle of rotation q, Quadrant I: 0  q  90 Quadrant II: 90  q  180 Quadrant III: 180  q  270 Quadrant IV: 270  q  360 Notes: 0, 90, 180 and 270 do not belong to any quadrant.

14.2 Trigonometric Ratios of Arbitrary Angles
A. Definition For an acute angle q, the trigonometric ratios between two sides of a right-angled triangle are We now introduce a rectangular coordinate plane onto DOPQ such that OP is the terminal side as shown in the figure. Suppose the coordinates of P are (x, y) and the length of OP is r. We can then define the trigonometric ratios of q in terms of x, y and r:

14.2 Trigonometric Ratios of Arbitrary Angles
A. Definition Now, we can extend the definition for angles greater than 90. For example: In the figure, P(–3 , 4) is a point on the terminal side of the angle of rotation q. We have x  3 and y  4. By definition:

14.2 Trigonometric Ratios of Arbitrary Angles
B. Signs of Trigonometric Ratios In the previous section, we defined the trigonometric ratios in terms of the coordinates of a point P(x, y) on the terminal side and the length r of OP. Since x and y may be either positive or negative, the trigonometric ratios may be either positive or negative depending upon the quadrant in which q lies. Sign of tan q Sign of cos q Sign of sin q Sign of y-coordinate x-coordinate Quadrant I II III IV

14.2 Trigonometric Ratios of Arbitrary Angles
B. Signs of Trigonometric Ratios The signs of the three trigonometric ratios in different quadrants can be summarized in the following diagram which is called an ASTC diagram. A : All positive S : Sine positive T : Tangent positive C : Cosine positive Notes: ‘ASTC’ can be memorized as ‘Add Sugar To Coffee’. IV III II I Sign of tan q Sign of cos q Sign of sin q Sign of y-coordinate x-coordinate Quadrant

14.2 Trigonometric Ratios of Arbitrary Angles
C. Using a Calculator to Find Trigonometric Ratios We can find the trigonometric ratios of given angles by using a calculator. For example, (a) sin 160  (cor. to 3 sig. fig.) (b) tan 245  (cor. to 3 sig. fig.) (c) cos(123)   (cor. to 3 sig. fig.) (d) sin(246)  (cor. to 3 sig. fig.)

14.3 Finding Trigonometric Ratios Without Using a Calculator
A. Angles Formed by Coordinates Axes If we rotate the terminal side OP with length r units (r  0) through 90 in an anti-clockwise direction, then the coordinates of P are (0, r). Thus, x  0 and y  r. , which is undefined.

14.3 Finding Trigonometric Ratios Without Using a Calculator
A. Angles Formed by Coordinates Axes Suppose we rotate the terminal side OP through 90, 180, 270 and 360 in an anti-clockwise direction. undefined 1 (0, r) 90 (r, 0) 0 tan q cos q sin q Coordinates of P q 180 (r, 0) 1 270 (0, r) 1 undefined 360 (r, 0) 1 Notes: The terminal sides OP of q  0 and 360 lie in the same position. Thus, their trigonometric ratios must be the same.

14.3 Finding Trigonometric Ratios Without Using a Calculator
B. By Considering the Reference Angles 1. Reference Angle For each angle of rotation q (except for q  90  n, where n is an integer), we consider the corresponding acute angle measured between the terminal side and the x-axis. It is called the reference angle b. Examples:  q  30  b  30  q  140  b  180  140  40  q  250  b  250  180  70  q  310  b  360  310  50

14.3 Finding Trigonometric Ratios Without Using a Calculator
B. By Considering the Reference Angles 2. Finding Trigonometric Ratios By using the reference angle, we can find the trigonometric ratios of an arbitrary angle. The following four steps can help us find the trigonometric ratio of any given angle q: Step 1: Determine the quadrant in which the angle q lies. Step 2: Determine the sign of the corresponding trigonometric ratio. According to the ASTC diagram Step 3: Find the trigonometric ratio of its reference angle b. Step 4: Find the trigonometric ratio of the angle q by assigning the sign determined in step 2 to the ratio determined in step 3.

14.3 Finding Trigonometric Ratios Without Using a Calculator
B. By Considering the Reference Angles For example, to find tan 240 and cos 240: Step 1: Determine the quadrant in which the angle 240 lies:  240 lies in quadrant III. Step 2: Determine the sign of the corresponding trigonometric ratio:  In quadrant III: tangent ratio: ve cosine ratio: ve \ tan q  tan b cos q  cos b Step 3: Find the trigonometric ratio of its reference angle b:  b  240  180  60 Step 4: Find the trigonometric ratio of the angle 240:  tan 240  tan 60 cos 240  cos 60

14.3 Finding Trigonometric Ratios Without Using a Calculator
C. Finding Trigonometric Ratios by Another Given Trigonometric Ratio In the last section, we learnt that the trigonometric ratios can be defined as where P(x, y) is a point on the terminal side of the angle of rotation q and is the length of OP. Now, we can use the above definitions to find other trigonometric ratios of an angle when one of the trigonometric ratios is given.

Example 14.1T 14.3 Finding Trigonometric Ratios
Without Using a Calculator C. Finding Trigonometric Ratios by Another Given Trigonometric Ratio Example 14.1T If , where 270  q  360, find the values of sin q and cos q. Solution: Since tan   0,  lies in quadrant II or IV. As it is given that 270    360,  must lie in quadrant IV where sin   0 and cos   0. P(12, 5) is a point on the terminal side of . By definition,

Example 14.2T 14.3 Finding Trigonometric Ratios
Without Using a Calculator C. Finding Trigonometric Ratios by Another Given Trigonometric Ratio Example 14.2T If , where 180  q  270, find the values of cos q and tan q. Solution: Since sin   0 and 180    270,  lies in quadrant III. Let P(x, 2) be a point on the terminal side of . We have y  2 and r  5. Since  lies in quadrant III, the x-coordinate of P must be negative.

14.4 Trigonometric Identities
With the help of reference angles in the last section, we can get the following important identities. For any acute angle q, since 180  q lies in quadrant II, we have sin (180  q)  sin q cos (180  q)  cos q tan (180  q)  tan q Since 180  q lies in quadrant III, we have sin (180  q)  sin q cos (180  q)  cos q tan (180  q)  tan q

14.4 Trigonometric Identities
Since 360  q lies in quadrant IV, we have sin (360  q)  sin q cos (360  q)  cos q tan (360  q)  tan q Notes: The above identities also hold if q is not an acute angle. They are useful in simplifying expressions involving trigonometric ratios. Remarks: The following identities also hold if q is not an acute angle: sin (90  q)  cos q cos (90  q)  sin q tan (90  q) 

Example 14.3T 14.4 Trigonometric Identities Solution:
Simplify the following expressions. (a) tan (180  q) sin (90  q) Solution: (a) tan (180  q) sin (90  q)

Example 14.4T 14.4 Trigonometric Identities Solution:
Simplify sin (90  q) cos (90  q)  2sin (180  q) cos q. Solution:

14.4 Trigonometric Identities
Example 14.5T Solution:

14.5 Trigonometric Equations
A. Finding Angles from Given Trigonometric Ratios In previous sections, we learnt how to find the trigonometric ratios of any angle. Now, we will study how to find the angle if a trigonometric ratio of the angle is given. For example: Given that , where 0  q  360. Step 1: Since sin q  0, q may lie in either quadrant III or quadrant IV. Step 2: Let b be the reference angle of q.  b  60 Step 3: Locate the angle q and its reference angle b in each possible quadrant. Step 4: Hence, if q lies in quadrant III, q  180  60  240. If q lies in quadrant IV, q  360  60  300.

14.5 Trigonometric Equations
A. Finding Angles from Given Trigonometric Ratios In general, for any given trigonometric ratio, it may correspond to more than one angle. q  120 q  120, 240, … Finding the trigonometric ratio Finding the corresponding angles

14.5 Trigonometric Equations
B. Simple Trigonometric Equations An equation involving trigonometric ratios of an unknown angle q is called a trigonometric equation. Usually, there are certain values of q which satisfy the given equation. The process of finding the solutions of the equation is called solving trigonometric equation. We will try to solve some simple trigonometric equations: a sin q  b, a cos q  b and a tan q  b, where a and b are real numbers.

Example 14.6T 14.5 Trigonometric Equations Solution:
B. Simple Trigonometric Equations Example 14.6T If (  1)sin q  2, where 0  q  360, find q. (Give the answers correct to 1 decimal place.) Solution: By using a calculator, the reference angle  . Hence,    or 180   (cor. to 1 d. p.)

14.5 Trigonometric Equations
C. Other Trigonometric Equations We now try to solve some harder trigonometric equations. Examples: Equation Technique 2sin q  3cos q  0 Using trigonometric identity 5sin2 q  4  0 Taking square root sin q  2sin q cos q  0 Taking out the common factor 2cos2 q  3sin q  0 Transforming into a quadratic equation

Example 14.7T 14.5 Trigonometric Equations Solution:
C. Other Trigonometric Equations Example 14.7T Solve the following equations for 0  q  360. (a) 7sin q  7cos q  0 Solution:

Example 14.8T 14.5 Trigonometric Equations Solution:
C. Other Trigonometric Equations Example 14.8T Solve the equation cos2 q tan q  cos q  0 for 0  q  360. Solution: Factorize the given expression and apply the fact that if ab  0, then a  0 or b  0.

Example 14.9T 14.5 Trigonometric Equations Solution:
C. Other Trigonometric Equations Example 14.9T Solve the equation 2cos2 q  sin q  1  0 for 0  q  360. Solution: Transform the equation into a quadratic equation with sin q as the unknown.

14.6 Graphs of Trigonometric Functions
A. The Graph of y  sin x Consider y  sin x. For every angle x, there is a corresponding trigonometric ratio y. Thus, y is a function of x. The following table shows some values of x and the corresponding values of y (correct to 2 decimal places if necessary) for 0 £ x £ 360. x 0 30 60 90 120 150 180 y 0.5 0.87 1 x 210 240 270 300 330 360 y 0.5 0.87 1 From the above table, we can plot the points on the coordinate plane.

14.6 Graphs of Trigonometric Functions
A. The Graph of y  sin x We can also plot the graph of y  sin x for 360 £ x £ 720, etc. The graph of y  sin x repeats itself in the intervals –360 £ x £ 0, 0 £ x £ 360, 360 £ x £ 720, etc. Remarks: A function repeats itself at regular intervals is called a periodic function. The regular interval is called a period. From the figure, we obtain the following results for the graph of y  sin x for 0 £ x £ 360: 1. The domain of y  sin x is the set of all real numbers. 2. The maximum value of y is 1, which corresponds to x  90. The minimum value of y is –1, which corresponds to x  270. 3. The function is a periodic function with a period of 360.

14.6 Graphs of Trigonometric Functions
B. The Graph of y  cos x The following table shows some values of x and the corresponding values of y (correct to 2 decimal places if necessary) for 0 £ x £ 360 for y  cos x. x 0 30 60 90 120 150 180 210 240 270 300 330 360 y 1 0.87 0.5 0.5 0.87 1 From the above table, we can plot the points on the coordinate plane.

14.6 Graphs of Trigonometric Functions
B. The Graph of y  cos x From the figure, we obtain the following results for the graph of y  cos x for 0 £ x £ 360: 1. The domain of y  cos x is the set of all real numbers. 2. The maximum value of y is 1, which corresponds to x  0 and 360. The minimum value of y is –1, which corresponds to x  180. Notes: If we plot the graph of y  cos x for –360 £ x £ 720, we can see that the graph repeats itself every 360. Thus, y  cos x is a periodic function with a period of 360.

14.6 Graphs of Trigonometric Functions
C. The Graph of y  tan x The following table shows some values of x and the corresponding values of y (correct to 2 decimal places if necessary) for 0 £ x £ 360 for y  tan x. x 0 30 45 60 75 90 105 120 135 150 y 0.58 1 1.73 3.73 Undefined 3.73 1.37 1 0.58 x 180 210 225 240 255 270 285 300 315 330 360 y 0.58 1 1.73 3.73 Undefined 3.73 1.37 1 0.58 The value of y is not defined when x  90 and 270. When an angle is getting closer and closer to 90 or 270, the corresponding value of tangent function approaches to either positive infinity or negative infinity.

14.6 Graphs of Trigonometric Functions
C. The Graph of y  tan x The graph of y  tan x is drawn as below. x 0 30 45 60 75 90 105 120 135 150 y 0.58 1 1.73 3.73 Undefined 3.73 1.37 1 0.58 x 180 210 225 240 255 270 285 300 315 330 360 y 0.58 1 1.73 3.73 Undefined 3.73 1.37 1 0.58

14.6 Graphs of Trigonometric Functions
C. The Graph of y  tan x From the figure, we obtain the following results for the graph of y  tan x: 1. For 0 £ x £ 180, y  tan x exhibits the following behaviours: From 0 to 90, tan x increases from 0 to positive infinity. From 90 to 180, tan x increases from negative infinity to 0. 2. y  tan x is a periodic function with a period of 180. 3. As tan x is undefined when x  90 and 270, the domain of y  tan x is the set of all real numbers except x  90, 270, ... .

14.6 Graphs of Trigonometric Functions
C. The Graph of y  tan x Given a trigonometric function, we can find its maximum and minimum values algebraically. For example, to find the maximum and minimum values of 3  4cos x: 1  cos x  1 4  4cos x  4 4  3  3  4cos x  4  3 1  3  4cos x  7 The maximum and minimum values are 7 and 1 respectively.

14.6 Graphs of Trigonometric Functions
D. Transformation on the Graphs of Trigonometric Functions In Book 4, we learnt the transformations such as translation and reflection of graphs of functions. Now, we will study the transformations on the graphs of trigonometric functions.

Example 14.10T 14.6 Graphs of Trigonometric Functions Solution:
D. Transformation on the Graphs of Trigonometric Functions Example 14.10T (a) Sketch the graph of y  cos x for 180 £ x £ 360. (b) From the graph in (a), sketch the graphs of the following functions. (i) y  cos x  2 (ii) y  cos (x  180) (iii) y  cos x y  cos x y  cos (x  180) Solution: (a) Refer to the figure. y  cos x  2 (b) The graph of the function (i) y  cos x  2 is obtained by translating the graph of y  cos x two units downwards. (ii) y  cos (x  180) is obtained by translating the graph of y  cos x to the left by 180. (ii) y  cos x is obtained by reflecting the graph of y  cos x about the x-axis.

14.7 Graphical Solutions of Trigonometric Equations
Similar to quadratic equations, trigonometric equations can be solved either by the algebraic method or the graphical method. We should note that the graphical solutions are approximate in nature.

Example 14.11T 14.7 Graphical Solutions of Trigonometric Equations
Consider the graph of y  cos x for 0 £ x £ 360. Using the graph, solve the following equations. (a) cos x  0.6 (b) cos x  0.7 y  0.6 Solution: (a) Draw the straight line y  0.6 on the graph. The straight line cuts the curve at x  54 and 306. y  0.7 So the solution of cos x  0.6 for 0 £ x £ 360 is 54 or 306. (b) Draw the straight line y  0.7 on the graph. The straight line cuts the curve at x  135 and 225. So the solution of cos x  0.7 for 0 £ x £ 360 is 135 or 225.

Example 14.12T 14.7 Graphical Solutions of Trigonometric Equations
Draw the graph of y  3cos x  sin x for 0 £ x £ 360. Using the graph, solve the following equations for 0 £ x £ 360. (a) 3cos x  sin x  0 (b) 3cos x  sin x  1.5 Solution: y  1.5 (a) From the graph, the curve cuts the x-axis at x  72 and 252. Therefore, the solution is 72 or 252. (b) Draw the straight line y  1.5 on the graph. The straight line cuts the curve at x  43 and 280. Therefore, the solution is 43 or 280.

Chapter Summary 14.1 Introduction to Trigonometry
In a rectangular coordinate plane, the x-axis and the y-axis divide the plane into four quadrants.

Chapter Summary 14.2 Trigonometric Ratios of Arbitrary Angles
The signs of different trigonometric ratios in different quadrants can be memorized by the ASTC diagram.

Chapter Summary 14.3 Finding Trigonometric Ratios Without
Using a Calculator If b is the reference angle of an angle q, then sin q  sin b, cos q  cos b, tan q  tan b, where the choice of the sign ( or ) depends on the quadrant in which q lies.

Chapter Summary 14.4 Trigonometric Identities
1. (a) sin (180 – q)  sin q (b) cos (180 – q)  –cos q (c) tan (180 – q)  –tan q 2. (a) sin (180  q)  –sin q (b) cos (180  q)  –cos q (c) tan (180  q)  tan q 3. (a) sin (360 – q)  –sin q (b) cos (360 – q)  cos q (c) tan (360 – q)  –tan q

Chapter Summary 14.5 Trigonometric Equations
Trigonometric equations can be solved by the algebraic method.

Chapter Summary 14.6 Graphs of Trigonometric Functions
1. Graph of y  sin x 2. Graph of y  cos x 3. Graph of y  tan x For any real value of x, 1  sin x  1 and 1  cos x  1. 5. The periods of sin x, cos x and tan x are 360, 360 and 180 respectively.

Chapter Summary 14.7 Graphical Solutions of Trigonometric Equations
Trigonometric equations can be solved by the graphical method.

Follow-up 14.1 14.3 Finding Trigonometric Ratios
Without Using a Calculator C. Finding Trigonometric Ratios by Another Given Trigonometric Ratio Follow-up 14.1 If , where 180  q  270, find the values of sin q and cos q. Solution: Since tan   0,  lies in quadrant I or III. As it is given that 180    270,  must lie in quadrant III where sin   0 and cos   0. P(4, 3) is a point on the terminal side of . By definition,

Follow-up 14.2 14.3 Finding Trigonometric Ratios
Without Using a Calculator C. Finding Trigonometric Ratios by Another Given Trigonometric Ratio Follow-up 14.2 If , where 90  q  180, find the values of sin q and tan q. Solution: Since cos   0 and 90    180,  lies in quadrant II. Let P(3, y) be a point on the terminal side of . We have x  3 and r  5. Since  lies in quadrant II, the y-coordinate of P must be positive.

Follow-up 14.3 14.4 Trigonometric Identities Solution:
Simplify the following expressions. (a) cos (180  q) tan (180  q) Solution: (a) cos (180  q) tan (180  q)

Follow-up 14.4 14.4 Trigonometric Identities Solution:
Simplify tan (90  q) sin (180  q)  4cos (180  q). Solution:

14.4 Trigonometric Identities
Follow-up 14.5 Solution:

Follow-up 14.6 14.5 Trigonometric Equations Solution:
B. Simple Trigonometric Equations Follow-up 14.6 Solve 5cos q  2, where 0  q  360. (Give the answers correct to 1 decimal place.) Solution: By using a calculator, the reference angle  . Hence,   180   or 180   (cor. to 1 d. p.)

Follow-up 14.7 14.5 Trigonometric Equations Solution:
C. Other Trigonometric Equations Follow-up 14.7 Solve the following equations for 0  q  360. (Give the answers correct to 1 decimal place.) (a) 6sin q  8cos q  0 (b) tan2 q  2  0 Solution: (cor. to 1 d. p.) (cor. to 1 d. p.)

Follow-up 14.8 14.5 Trigonometric Equations Solution:
C. Other Trigonometric Equations Follow-up 14.8 Solve the equation tan q  2sin q  0 for 0  q  360. Solution: As cos q  0, we can multiply the equation by cos q.

Follow-up 14.9 14.5 Trigonometric Equations Solution:
C. Other Trigonometric Equations Follow-up 14.9 Solve the equation 2sin2 q  3cos q  0 for 0  q  360. Solution: Note that 1  cos q  1. For details, refer to the graph of y  cos x in Section 14.6.

Follow-up 14.10 14.6 Graphs of Trigonometric Functions Solution:
D. Transformation on the Graphs of Trigonometric Functions Follow-up 14.10 The following figure shows the graph of the function y  tan x for 90 £ x £ 540. Sketch the graphs of the following functions on the figure. (i) y  tan x  1 (ii) y  tan x y  tan x Solution: y  tan x  1 (i) The graph of the function y  tan x  1 is obtained by translating the graph of y  tan x one unit downwards. (ii) The graph of the function y  tan x is obtained by reflecting the graph of y  tan x about the x-axis.

Follow-up 14.11 14.7 Graphical Solutions of Trigonometric Equations
Consider the graph of y  tan x for 0 £ x £ 360. Using the graph, solve the following equations. (a) tan x  3 (b) tan x  2 y  3 Solution: (a) Draw the straight line y  3 on the graph. The straight line cuts the curve at x  72 and 252. y  2 So the solution of tan x  3 for 0 £ x £ 360 is 72 or 252. (b) Draw the straight line y  2 on the graph. The straight line cuts the curve at x  117 and 297. So the solution of tan x  2 for 0 £ x £ 360 is 117 or 297.

Follow-up 14.12 14.7 Graphical Solutions of Trigonometric Equations
The figure shows the graph of y  acos x  bsin x for 0 £ x £ 360. (a) Find the values of a and b. (b) Using the graph, solve the equation 2cos x  3sin x  2. y  2 Solution: (a) Since the graph passes through (0, 2), we have (b) Draw the straight line y  2 on the graph. Since the graph passes through (90, 3), we have The straight line cuts the curve at x  66 and 180. Therefore, the solution is 66 or 180.