Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions?

Presentation on theme: "Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions?"— Presentation transcript:

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2 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions? Do Now: A circle has a radius of 3 cm. Find the length of an arc cut off by a central angle of 270 0.

3 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Angles in Standard Position An angle on the coordinate plane is in standard position when its vertex is at the origin and its initial side coincides with the nonnegative ray of the x-axis. Quadrant IQ II Q IIIQ IV y x terminal side  t.s initial side 90 <  < 1800 <  < 90 180 <  < 270270 <  < 360 An angle formed by a counterclockwise rotation has a positive measure. t.s. Angles whose terminal side lies on one of the axes is a quadrantal angle. i.e. 90 0, 180 0, 270 0, 360 0, 450 0 etc.

4 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Co-terminal and Negative Angles Quadrant IQ II Q IIIQ IV y x initial side Angles in standard position having the same terminal side are co-terminal angles. An angle formed by a clockwise rotation has a negative measure 90 <  < 1800 <  < 90 180 <  < 270270 <  < 360  -- t.s. 60 0 300 0 =

5 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Angles whose terminal side rotates more than one revolution form angles with measures greater than 360 0. Angles Greater than 360 0 Quadrant IQ II Q IIIQ IV y x To find angles co-terminal with an another angle add or subtract 360 0. 125 0 and 485 0 are co-terminal 485 0 125 0 90 <  < 1800 <  < 90 180 <  < 270270 <  < 360

6 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems Find the measure of an angle between 0 0 and 360 0 co-terminal with a)385 0 b) 575 0 c) -405 0 In which quadrant or on which axis, does the terminal side of each angle lie? a) 150 0 b) 540 0 c) -60 0 25 o 215 o 315 o QII x- axis QIV

7 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Unit Circle y x 1 1 radius = 1 center at (0,0)  (x,y) hypotenuse side opp. side adj. cos , sin  cos 

8 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions? Do Now: Find the measure of an angle between 0 0 and 360 0 co-terminal with an angle whose measure is -125 0.

9 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Value of Sine & Cosine: Quadrant I y x 1 1 radius = 1 center at (0,0) 60 0 (x,y) cos 60 0, sin 60 0 What is the value of coordinates (x,y)? 30 0 -60 0 -90 0 triangle Hypotenuse = 2  shorter leg Longer leg =  shorter leg 3 Sine and Cosine values for angles in Quadrant I are positive.

10 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. (x,y) 60º is the reference angle (180º-120º) Value of Sine & Cosine: Quadrant II y x 1 1 60 0 cos 120 0, sin 120 0 What is the value of coordinates (x,y)? 120 0 1 What is the cosine/sine of a 120 0 angle? side adj. directed distance Sine values for angles in Quadrant II are positive. Cosine values for angles in Quadrant II are negative. Hypotenuse = 2  shorter leg Longer leg =  shorter leg 3 30 0 -60 0 -90 0 triangle A reference angle for any angle in standard position is an acute angle formed by the terminal side of the given angle and the x-axis.

11 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. (x,y) cos 240 0, sin 240 0 Value of Sine & Cosine: Quadrant III y x 1 1 What is the value of coordinates (x,y)? 1 What is the cosine/sine of a 240 0 angle? side adj. directed distance 240 0 60 0 Sine and Cosine values for angles in Quadrant III are both negative. side opp. directed dist. 60º is the reference angle (240º-180º)

12 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Value of Sine & Cosine: Quadrant IV y x 1 1 What is the value of coordinates (x,y)? 1 What is the cosine/sine of a 300 0 angle? 300 0 (x,y) 60 0 cos 300 0, sin 300 0 Sine values for angles in Quadrant IV are negative. Cosine values for angles in Quadrant IV are positive. side opp. directed dist. 60º is the reference angle (360º-300º)

13 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Unit Circle – 12 Equal Arcs

14 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Unit Circle – 8 Equal Arcs Periodic Negative Angles Identities

15 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Value of Sine & Cosine in Coordinate Plane y x Quadrant IQuadrant II Quadrant IIIQuadrant IV cos  is – sin  is + cos  is + sin  is + cos  is – sin  is – cos  is + sin  is – for any angle in standard position is an acute angle formed by the terminal side of the given angle and the x-axis. The reference angle:

16 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems Fill in the table Quad. Ref.  sin  cos  a)236 0 b)87 0 c)-160 0 d)-36 e)1332 0 f)-396 0

17 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Regents Prep On the unit circle shown in the diagram below, sketch an angle, in standard position, whose degree measure is 240 and find the exact value of sin 240 o. y x 1 1

18 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Aim: What good is the Unit Circle and how does it help us to understand the Trigonometric Functions? Do Now: Use the unit circle to find: a. sin 180 0 (  )b. cos 180 0

19 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems Use the unit circle to find: a. sin 180 0 (  )b. cos 180 0 (-1,0)  (x, y) = (-1, 0) sin 180 0 = y= 0 cos 180 0 = x= -1 180º - quadrantal angle

20 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Tan  y x 1 1 radius = 1 center at (0,0)  (x,y) cos , sin  cos  sin  1 (, )? ( 1, tan  ) tan  cos  = 1 sin 

21 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Trigonometric Values Quad. Sin  Cos  Tan  I ++ II +– reference angle Sin  = Sin(180–  ) Cos  = -Cos(180–  ) Tan  = -Tan(180–  ) III –– reference angle Sin  = -Sin(  -180) Cos  = -Cos(  -180) Tan  = Tan(  -180) IV –+ reference angle Sin  = -Sin(360-  ) Cos  = Cos(360-  ) Tan  = -Tan(360-  ) + – + –

22 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Trigonometric Values - A C T S x A All are + C Cosine is + T Tangent is + S Sine is + y Quadrant I 0 <  < 90 Q II 90 <  < 180 Q III 180 <  < 270 Q IV 270 <  < 360

23 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Need to Knows sin  = y cos  = x tan  = y/x When r = 1 y x 1 1 r x y  Reciprocal Functions csc  = 1/y sec  = 1/x cot  = x/y denominators  0 Negative Angles Identities

24 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. length of hypo. = Using the unit circle, find a.cos 45 0 (  /4) b.sin 45 0 c.tan 45 0 45 0 -45 0 -90 0 triangle y x 1 1 45 0 1 cos  = x sin  = y (x,y)(x,y) In a 45 0 -45 0 -90 0 triangle, the length of the hypotenuse is times the length of a leg. 2 A 45 0 -45 0 -90 0 triangle is an isosceles right triangle. therefore x = y Model Problems 2(x)(x) cos  = sin 

25 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. y x 1 1 45º 1 Using the unit circle, find a.cos 45º(  /4) b.sin 45º c.tan 45º cos  = x sin  = y (x,y)(x,y) Model Problems cos 45º = x = sin 45º = y = tan 45 = 1

26 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Trigonometric Values for Special Angles 90º  /2 0 tan  1 cos  0 sin  60º  /3 45º  /4 30º  /6 0º 0  1 1 0 UND. = slope Why is tan 90º undefined? What is the slope of a line perpendicular to the x-axis?

27 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems What is the tan 135º (3  /4)? 135º is in the 2nd quadrant 45º is reference angle (180 – 135 = 45) tan 45 º = 1 tangent is negative in 2nd quadrant tan 135º = -1 What is the cos 510º (17  /6)? 510º is in the 2nd quadrant (510 – 360 = 150) 30º is reference angle (180 – 150 = 30) cosine is negative in 2nd quadrant cos 30º = cos 510º = ≈ -.866…

28 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems

29 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems Given: sin 68 o = 0.9272 cos 68 o = 0.3746 Find cot 112 o A) -0.3746B) -2.4751 C) -0.404D) 1.0785 reference angle for 112 o is 68 o ; 112 o is in QII; tan and cot are negative in QII WHAT ELSE DO WE KNOW?

30 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problems Express sin 285º as the function of an angle whose measure is less than 45º. 285º in IV quadrant What do we know? the sine of a IV quadrant angle is negative reference angle for 285º is (360 – 285) = 75º > 45º -sin 75º complement of 75º is 15º sine and cosine are co-functions < 45º = -cos 15ºsin 285º = -sin 75º

31 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Trig Functions Using Radian Measures Algebraically: sin (π/3) π/360º Find: sin 60º = ≈.866… Using the calculator: Use the mode key: change setting from degrees to radians then hit: sin2ndπ÷ 3 ENTER Display:.8660254083 remember: π/3 radians

32 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Un-unit circle  is any angle in standard position with (x, y) any point on the terminal side of  and y x 1 1 1 unit circle r  1

33 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problem (-3, 4) is a point on the terminal side of . Find the sine, cosine, and tangent of . r = 5 3 4 Q II

34 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problem is a point on the terminal side of . Find , the sine, cosine, and tangent of . = 2 r Q III

35 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problem Tan  = -5/4 and cos  > 0, find sin  and sec  When tangent is negative and cosine is positive angle is found in Q IV.

36 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problem The terminal side of  is in quadrant I and lies on the line y = 6x. Find tan  ; find . y = mx + b - slope intercept form of equation m = slope of line y = 6x m = 6 = tan  Q I

37 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Model Problem The terminal side of  is in quadrant IV and lies on the line 2x + 5y = 0. Find cos . tan  = m = -2/5 y = mx + b slope intercept form of equation

38 Aim: Trig. Ratios for any Angle Course: Alg. 2 & Trig. Templates y x 1 1 45º 1