2. Whiteboardmaths.com © 2004 All rights reserved 5 7 2 1

3. sinx + circle 90 o 180 o  0o0o 270 o 1 The Trigonometric Ratios for any angle 0 90 180 360270-90 -180 -270 -360 0 90180 270 -90 -180 -270 -360360 450 o 0o0o 90 o 180 o 270 o 360 o -90 o -180 o -270 o -360 o -450 o 360 o 0o0o 90 o 180 o 270 o

4. Trigonometry 5: Sine, cosine and tangent for any angle The trigonometric ratios have previously been used to solve problems in right-angled triangles only. We need to develop an approach that will solve problems involving triangles with angles greater than 90 0. SOHCATOA O A H  Sin Cos Tan  ? In order to be able to do this we have to define sine, cosine and tangent in a different way. For angles greater than 90 o we can establish a connection between the trigonometric ratios and the moving point on the circumference of a circle of unit radius. O 1

5. Trigonometry 5: Sine, cosine and tangent for any angle x y O 1 P  As the Point P moves in an anti-clockwise direction around the circumference of the circle, the angle  changes from 0 o to 360 o. A In this triangle the distance OA = x. Consider the right-angled triangle formed by the vertical line PA. x y The distance OP = y. So point P has co-ordinates (x,y). (x,y) So the co-ordinates of P are (cos , sin  ). cos  sin  (cos ,sin  ) Therefore x = cos  and y = sin . x O 1 P  A y

6. Trigonometry 5: Sine, cosine and tangent for any angle cos  sin  (cos ,sin  ) x O 1 P  A y We are now in a position to define the sine and cosine of any angle including angles greater than 90 o (0.5,0.87) (-0.5,0.87) (-0.42,-0.91) (0.64,-0.77) x y x y x y x y We can now state the values of the sine and cosine for any angle. Consider the co-ordinates of the point P as it moves around the circle through the angles shown below. P 60 o P 120 o P 245 o P 310 o cos 60 o = 0.5 sin 60 = 0.87 cos 120 o = -0.5 sin 120 = 0.87 cos 245 o = - 0.42 sin 245 = - 0.91 cos 310 o = 0.64 sin 310 = - 0.77

7. Use your calculator to plot the graph of y = sin x on the grid below. x y x y x y x y x y x y x y x y x y x y x y x y y 1 - 1 360 x 90 180 270 0 0.2 0.4 0.6 0.8 - 0.2 - 0.4 - 0.6 - 0.8 60 o 30 o 90 o 120 o 150 o 180 o 210 o 240 o 270 o 300 o 330 o 360 o y = sin x 0.000.50.87 1.0 0.870.5 0.0-0.5 -0.87-0.87-0.5 7 8 9 4 5 6 1 2 3 Sin  0.0 The sine curve is symmetrical above the x axis about 90 o and below the x axis about 270 o sin 30 0 = sin 150 o sin 60 0 = sin 120 o In general sin x =sin(180 - x)

8. Use your calculator to plot the graph of y = cos x on the grid below. x y x y x y x y x y x y x y x y x y x y x y x y y 1 - 1 360 x 90 180 270 0 0.2 0.4 0.6 0.8 - 0.2 - 0.4 - 0.6 - 0.8 60 o 30 o 90 o 120 o 150 o 180 o 210 o 240 o 270 o 300 o 330 o 360 o 7 8 9 4 5 6 1 2 3 Cos  y = cos x 1.00 0.87 0.5 0.0 -0.5 -0.87 -0.87 -0.5 0.0 0.5 0.871.0

9. Use your calculator to plot the graph of y = cos x on the grid below. x y x y x y x y x y x y x y x y x y x y x y x y y 1 - 1 360 x 90 180 270 0 0.2 0.4 0.6 0.8 - 0.2 - 0.4 - 0.6 - 0.8 60 o 30 o 90 o 120 o 150 o 180 o 210 o 240 o 270 o 300 o 330 o 360 o y = cos x cos 30 0 = - cos 150 o cos 60 0 = - cos 120 o In general cos x = - cos(180 o - x)

10. 90 180 x 0 270 360 -90 -180 -270 -360 1 y = Sin x O - 60 o 90 180 x 0 270 360 -90 -180 -270 -360 1 y = Cos x The graphs of the sine and cosine functions are also defined for angles that are negative. This corresponds to the point P going around the circle in a clockwise direction.

11. 90 180 x 0 270 360 -90 -180 -270 -360 1 y = Sin x O - 60 o 90 180 x 0 270 360 -90 -180 -270 -360 1 y = Cos x -315 o -225 o 135 o If cos 68 o = 0.37, use the graph below to estimate three more angles whose cosine also has this value. If sin 45 o = 0.71, use the graph above to find three more angles whose sine also has this value. -292 o -68 o 292 o

12. 90 180 x 0 270 360 -90 -180 -270 -360 1 2 -2 Sinx 2Sinx 3 -3 3Sinx Amplitude  1 Period 360 o Amplitude  2 Period 360 o Amplitude  3 Period 360 o

13. 90 180 x y = f(x) 0 270 360 -90 -180 -270 -360 1 2 -2 3 -3 Cosx ½Cosx 2Cosx 3Cosx

14. Some relationships between the sine and cosine ratios in different quadrants 30 o 150 o Sin(180 o - 30 o ) = Sin 150 o = Sin 30 o Sin (180 -  ) = Sin  50 o -50 o y x y x Sin (-50 o )= -Sin 50 o Sin (-  ) = -Sin  30 o 150 o x Cos (180 o - 30 o ) = Cos 150 o = -Cos30 o Cos (180 -  ) = -Cos  y 50 o -50 o y x Cos (-50 o )= Cos 50 o Cos (-  ) = Cos  0 90 180 360270-90 -180 -270 -360 0 90180 270 -90 -180 -270 -360360

15. The Tangent Ratio for Angles greater than 90 Degrees P Q R  Tan  can be expressed in terms of the sine and cosine functions. This relationship can be determined by considering the three ratios in a right- angled triangle. Can you use the ratios above to write tan  in terms of sin  and cos  ? Clue 1Clue 2Clue 3 This is true for all angles of .

16. Quadrant 1 Quadrant 2 Quadrant 3 Quadrant 4 The sine and cosine functions can have values that are either positive or negative depending on the size of . It is useful to identify the quadrants in which their values are positive or negative. S+S+ C+C+ S+S+ C-C- S-S- C-C- S-S- C+C+ What about the tangent ratio? T+T+ T-T- T+T+ T-T-

17. Quadrant 1 Quadrant 2 Quadrant 3 Quadrant 4 S+S+ C+C+ S+S+ C-C- S-S- C-C- S-S- C+C+ T+T+ T-T- T+T+ T-T- ALL + Sin + Tan + Cos + A S T C It is very useful to be able to recollect all this information. One way is just to remember the quadrants in which the ratios are positive. There is a well known mnemonic to help you remember this. Whether you agree with it or not is another matter! All Science Teachers Care

18. Origin of the Tangent Function 1  cos  sin  Some Tangents!

19. tan  Origin of the Tangent Function Some Tangents! 1  cos  sin  This is the reason that this ratio is called the tangent.

20. A S T C sin + cos + tan + sin + cos - tan - sin - cos - tan + sin - cos + tan - tan  1 cos  sin   tan    and is not defined when cos  = 0 Cos 90 o = 0 Cos 270 o = 0

21. 35 o 1.In the 2 nd quadrant sin is positive. sin 35 o 2. In the 3 rd quadrant cos is negative. -cos 70 o 3. In the 4 th quadrant tan is negative. 60 o -tan 60 o 4.In the 3 rd quadrant sin is negative. -sin 40 o 5. In the 4 th quadrant cos is positive. 30 o cos 30 o Obtuse and reflex angles can be written in terms of an acute angle. A S T C 1. Sin 145 o 2. cos 250 o 3. tan 300 o 4. sin 220 o 5. cos 330 o 6. tan 210 o We will write each of the following in terms of an acute angle with the aid of diagrams and by considering symmetry. 6. In the 3 rd quadrant tan is positive. tan 30 o 70 o 40 o 30 o 145 o 250 o 70 o 35 o 300 o 220 o 40 o 330 o 30 o 210 o

22. y = tan  0o0o 90 o 180 o 270 o 360 o -90 o -180 o -270 o -360 o The Graph of the Tangent Function 450 o -450 o Tan  is not defined when cos  = 0. That is, for angles of 90 o, 270 o, 450 o …. and -90 o, -270 o, -450 o ….

23. Solving Equations Involving the Trigonometric Functions Example 1: Find all solutions to 2 sinx = 1 in the range -360 o to 360 o. Step 1 Solve the equation for x using your calculator if necessary. 2sinx = 1  sinx = 0.5  x = 30 o Step 2 Picture the quadrants in your mind’s eye or make a rough sketch and go from there, considering the symmetry of the situation and the positive and negative movement of the point. 30 o (Graph not given) 150 o -210 o -330 o So the solutions are 30 o, 150 o, - 210 o and - 330 o.

24. Solving Equations Involving the Trigonometric Functions Example 1: Find all solutions to 2 sinx = 1 in the range -360 o to 360 o. Step 1 Solve the equation for x using your calculator if necessary. 2sinx = 1  sinx = 0.5  x = 30 o Step 2 (Graph given) Draw line y = 1 and locate the first solution (x = 30 o ). Then by considering the symmetry of the situation, read off all other solutions at the intersections with the graph. 90 180 0 270 360 -90 -180 -270 -360 y = 2Sin x 1 2 -2 So the solutions are 30 o, 150 o, - 210 o and - 330 o.

25. Solving Equations Involving the Trigonometric Functions Question 1: Find all solutions to 4cosx = 3 in the range -360 o to 360 o. Step 1 Solve the equation for x using your calculator if necessary. 4cosx = 3  cosx = 0.75  x = 41.4 o Step 2 Picture the quadrants in your mind’s eye or make a rough sketch and go from there, considering the symmetry of the situation and the positive and negative movement of the point. (Graph not given) 318.6 o -41.4 o -318.6 o So the solutions are 41.4 o, 318.6 o, - 41.4 o and - 318.6 o. 41.4 o 270 + 48.6 -270 -48.6

26. Draw line y = 3 and locate the first solution (x = 41.4 o ). Then by considering the symmetry of the situation, read off all other solutions at the intersections with the graph. 90 180 0 270 360 -90 -180 -270 -360 y = 4Cos x 1 2 3 4 -2 -3 -4 Solving Equations Involving the Trigonometric Functions Question 1: Find all solutions to 4cosx = 3 in the range -360 o to 360 o. Step 1 Solve the equation for x using your calculator if necessary. 4cosx = 3  cosx = 0.75  x = 41.4 o Step 2 (Graph given) So the solutions are 41.4 o, 318.6 o, - 41.4 o and - 318.6 o.

27. Solving Equations Involving the Trigonometric Functions Example 1: Use the graph below to solve sin3x = sin 60 o for all values of x in the range 0 o to 270 o. Step 1 Solve the equation for x using your calculator if necessary. sin3x = sin 60 o  3x = 60 o  x = 20 o (sin 60 o = 0.87) Step 2 Draw the line y = 0.87 on the graph and read off the solutions at the points of intersection. From the graph solutions are: 20 o, 40 o, 140 o, 160 o and 260 o. 1 0 60 120 180 240 300 y = Sin 3x

28. Solving Equations Involving the Trigonometric Functions Example 1: Use the graph below to solve sin3x = sin 60 o for all values of x in the range 0 o to 240 o. Step 1 Solve the equation for x using your calculator if necessary. sin3x = sin 45 o  3x = 45 o  x = 15 o (sin 45 o = 0.71) Step 2 Draw the line y = 0.71 on the graph and read off the solutions at the points of intersection. From the graph solutions are: 15 o, 45 o, 135 o and 165 o. 1 0 60 120 180 240 y = Sin 3x

29. x y x y x y x y x y x y x y x y x y x y x y x y 60 o 30 o 90 o 120 o 150 o 180 o 210 o 240 o 270 o 300 o 330 o 360 o By considering the movement of the point on the above diagrams you should be able to deduce the following identities: sin (360 + x) = sin x cos (360 + x) = cos xtan(360 + x) = tan x sin (180 - x) = sin x cos (180 - x) = -cos x tan(180 - x) = -tan x sin (- x) = -sin xcos (- x) = cos x tan (- x) = -tan x A S T C

30. The origins of trigonometry are closely tied up with problems involving circles. One particular problem is that of finding the lengths of chords subtended by different angles at the centre of a circle. The Arabs called the half chord “ardha-jya”. This became mis-interpreted and mis-translated over the centuries and eventually ended up as “sinus” in Latin, meaning cove or bay. Other derivations include: bulge, bosom, sinus, cavity, nose and skull. The cosinus simply means the compliment of the sinus, since SinA = Cos (90 o – A) (Sin 60 o = Cos 30 o, Sin 70 o = Cos 20 o etc) Chord MAngular Bisector Radius P O Q Half Chord PM (Sinus) OM (Cosinus) Historical Note

31. oo P O 1 M Chord Radius P O Q Half Chord PM Sinus Sin  = O/H = PM/1 = PM Cos  = A/H = OM/1 = OM Tan  = PT/1 = PT Cosinus P M O 1  P O T  1 The following diagrams show the relationships between the three trigonometric ratios for a circle of radius 1 unit. Tangent means “to touch”. MAngular Bisector oo Tangent T M OP 1 

32. Worksheets Various 1 - 1 360 x 90 180 270 0 0.2 0.4 0.6 0.8 - 0.2 - 0.4 - 0.6 - 0.8 1 - 1 360 x 90 180 270 0 0.2 0.4 0.6 0.8 - 0.2 - 0.4 - 0.6 - 0.8

33. 90 180 x 0 270 360 -90 -180 -270 -360 1 90 180 x 0 270 360 -90 -180 -270 -360 1

34. 90 180 x y = f(x) 0 270 360 -90 -180 -270 -360 1 2 -2 3 -3

35. 90 180 0 270 360 -90 -180 -270 -360 y = 2Sin x 1 2 -2 90 180 0 270 360 -90 -180 -270 -360 y = 4Cos x 1 2 3 4 -2 -3 -4

36. 1 0 60 120 180 240 300 y = Sin 3x 1 0 60 120 180 240 300 y = Sin 3x