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sinx + circle

sinx + circle. 0. -360. -180. -270. -90. 270. 360. 180. 90. -450 o. -270 o. -90 o. 0 o. 90 o. -360 o. -180 o. 450 o. 180 o. 360 o. 270 o. 0. -270. -90. -180. 270. 90. 180. -360. 360. 90 o. The Trigonometric Ratios for any angle. 180 o. 0 o. 270 o. 1. 270 o. 0 o.

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sinx + circle

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

  2. SOHCATOA 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 900. Sin CosTan H O  A ?  1 O Trigonometry 5: Sine, cosine and tangent for any angle In order to be able to do this we have to define sine, cosine and tangent in a different way. For angles greater than 90o we can establish a connection between the trigonometric ratios and the moving point on the circumference of a circle of unit radius.

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

  4. y P (cos ,sin ) 1  sin  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. x O A cos  y y y y P P 310o 120o 245o 60o x x x x P P Trigonometry 5: Sine, cosine and tangent for any angle We are now in a position to define the sine and cosine of any angle including angles greater than 90o (-0.5,0.87) (0.5,0.87) (-0.42,-0.91) (0.64,-0.77) cos 60o = 0.5 sin 60 = 0.87 cos 120o = -0.5 sin 120 = 0.87 cos 245o = - 0.42 sin 245 = - 0.91 cos 310o = 0.64 sin 310 = - 0.77

  5. y y y y y y y y y y y y 120o 180o 150o 30o 90o 60o x x x x x x x x x x x x 210o 240o 360o 270o 300o 330o y = sin x 1 7 4 3 6 9 8 2 5 Sin  The sine curve is symmetrical above the x axis about 90o and below the x axis about 270o Use your calculator to plot the graph of y = sin x on the grid below. y 1.0 -0.5 0.0 -0.87 -1.0 -0.87 0.5 0.5 -0.5 0.00 0.87 0.87 0.0 1 0.8 0.6 0.4 0.2 x 0 270 90 180 360 - 0.2 sin 600 = sin 120o sin 300 = sin 150o - 0.4 - 0.6 In general sin x =sin(180 - x) - 0.8 - 1

  6. y y y y y y y y y y y y 120o 180o 150o 30o 90o 60o x x x x x x x x x x x x 210o 240o 360o 270o 300o 330o y = cos x 7 4 1 9 3 6 8 5 2 Cos  Use your calculator to plot the graph of y = cos x on the grid below. y 1 0.8 0.6 0.4 1.00 0.87 -0.87 -0.5 0.5 0.0 0.0 -0.5 0.2 1.0 0.87 -0.87 -1.0 0.5 x 0 270 90 180 360 - 0.2 - 0.4 - 0.6 - 0.8 - 1

  7. y y y y y y y y y y y y 120o 180o 150o 30o 90o 60o x x x x x x x x x x x x 210o 240o 360o 270o 300o 330o y = cos x Use your calculator to plot the graph of y = cos x on the grid below. y 1 0.8 0.6 cos 300 = - cos 150o 0.4 0.2 x 0 270 90 180 360 - 0.2 - 0.4 In general cos x = - cos(180o - x) - 0.6 cos 600 = - cos 120o - 0.8 - 1

  8. y = Sin x 1 1 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. x x -360 -360 90 90 -90 -90 -180 -180 0 0 270 270 180 180 -270 -270 360 360 -1 -1 y = Cos x O -60o

  9. 1 1 135o -315o -225o x x -360 -360 90 90 -90 -90 -180 -180 0 0 270 270 180 180 -270 -270 360 360 -1 -1 O -60o 292o -292o -68o y = Sin x If sin 45o = 0.71, use the graph above to find three more angles whose sine also has this value. If cos 68o = 0.37, use the graph below to estimate three more angles whose cosine also has this value. y = Cos x

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

  11. 3 y = f(x) 2 1 x -360 90 -90 -180 0 270 180 -270 360 -1 -2 -3 3Cosx 2Cosx ½Cosx Cosx

  12. 0 -360 -180 -270 -90 270 360 180 90 150o 150o 30o 30o 50o 50o -50o -50o 0 -270 -90 -180 270 90 180 -360 360 Some relationships between the sine and cosine ratios in different quadrants y y 30o x x Sin (-50o)= -Sin 50o Sin(180o - 30o) = Sin 150o = Sin 30o Sin (-) = -Sin  Sin (180 - ) = Sin  y y x 30o x Cos (-50o)= Cos 50o Cos (180o - 30o) = Cos 150o = -Cos30o Cos (-) = Cos  Cos (180 - ) = -Cos 

  13. P  Q R Clue 1 Clue 3 Clue 2 The Tangent Ratio for Angles greater than 90 Degrees 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 ? This is true for all angles of .

  14. y y y y y y 120o 180o 150o 30o 90o 60o x x x x x x y y y y y y 210o x x x x x x 240o 360o 270o 300o 330o y S+ S+ Quadrant 1 Quadrant 2 C- C+ What about the tangent ratio? Quadrant 4 Quadrant 3 C+ S- S- C- 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. T- T+ T+ T-

  15. y y y y y y 120o 180o 150o 30o 90o 60o x x x x x x y y y y y y 210o x x x x x x 240o 360o 270o 300o 330o y S+ S+ C- C+ C+ S- S- 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+ All Sin+ Science A S Quadrant 1 Quadrant 2 T- T+ Quadrant 4 C Quadrant 3 T Teachers T+ T- Care Cos+ Tan+

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

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

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

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

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

  21. 30o Solving Equations Involving the Trigonometric Functions Example 1: Find all solutions to 2 sinx = 1 in the range -360o to 360o. Solve the equation for x using your calculator if necessary. Step 1 2sinx = 1 So the solutions are 30o , 150o , - 210o and - 330o.  sinx = 0.5  x = 30o Step 2 (Graph not given) 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. -330o 150o -210o

  22. 2 y = 2Sin x 1 90 180 270 0 360 -90 -180 -270 -360 -1 -2 Solving Equations Involving the Trigonometric Functions Example 1: Find all solutions to 2 sinx = 1 in the range -360o to 360o. Solve the equation for x using your calculator if necessary. Step 1 2sinx = 1 So the solutions are 30o , 150o , - 210o and - 330o.  sinx = 0.5  x = 30o Step 2 (Graphgiven) Draw line y = 1 and locate the first solution (x = 30o). Then by considering the symmetry of the situation, read off all other solutions at the intersections with the graph.

  23. 41.4o Solving Equations Involving the Trigonometric Functions Question 1: Find all solutions to 4cosx = 3 in the range -360o to 360o. Solve the equation for x using your calculator if necessary. Step 1 4cosx = 3 So the solutions are 41.4o , 318.6o , - 41.4o and - 318.6o.  cosx = 0.75  x = 41.4o Step 2 (Graph not given) 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. -270 -48.6 270 + 48.6 318.6o -318.6o -41.4o

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

  25. 1 0 120 240 60 180 300 y = Sin 3x -1 Solving Equations Involving the Trigonometric Functions Example 1: Use the graph below to solve sin3x = sin 60o for all values of x in the range 0o to 270o. Solve the equation for x using your calculator if necessary. Step 1 sin3x = sin 60o  3x = 60o  x = 20o(sin 60o = 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: 20o, 40o, 140o, 160o and 260o.

  26. 1 0 120 240 60 180 -1 Solving Equations Involving the Trigonometric Functions Example 1: Use the graph below to solve sin3x = sin 60o for all values of x in the range 0o to 240o. Solve the equation for x using your calculator if necessary. Step 1 sin3x = sin 45o  3x = 45o  x = 15o(sin 45o = 0.71) Step 2 Draw the line y = 0.71 on the graph and read off the solutions at the points of intersection. y = Sin 3x From the graph solutions are: 15o, 45o, 135o and 165o.

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

  28. Historical Note P Radius Half Chord PM (Sinus) Angular Bisector M Chord O Q OM (Cosinus) 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 (90o – A) (Sin 60o = Cos 30o, Sin 70o = Cos 20o etc)

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

  30. 1 1 0.8 0.8 0.6 0.6 0.4 0.4 0.2 0.2 x x 0 0 270 270 90 90 180 180 360 360 - 0.2 - 0.2 - 0.4 - 0.4 - 0.6 - 0.6 - 0.8 - 0.8 - 1 - 1

  31. 1 1 x x -360 -360 90 90 -90 -90 -180 -180 0 0 270 270 180 180 -270 -270 360 360 -1 -1

  32. 3 y = f(x) 2 1 x -360 90 -90 -180 0 270 180 -270 360 -1 -2 -3

  33. 4 2 3 y = 2Sin x y = 4Cos x 2 1 1 90 90 180 180 270 270 0 0 360 360 -90 -90 -180 -180 -270 -270 -360 -360 -1 -1 -2 -3 -2 -4

  34. 1 1 0 0 120 120 240 240 60 60 180 180 300 300 y = Sin 3x y = Sin 3x -1 -1

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