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(6 – 1) Angle and their Measure Learning target: To convert between decimals and degrees, minutes, seconds forms To find the arc length of a circle To convert from degrees to radians and from radians to degrees To find the area of a sector of a circle To find the linear speed of an object traveling in circular motion Vocabulary: Initial side & terminal side: Initial side Initial side Initial side Drawing an angle Positive angles: Counterclockwise Negative angles: Clockwise 360 200 90 -90 150 -135 Another unit for an angle: Radian Definition: An angle that has its vertex at the center of a circle and that intercepts an arc on the circle equal in length to the radius of the circle has a measure of one radian. r r One radian From Geometry: C 2 r The entire circle has 360 360 2 r 180 r or r 180 r 180 Therefore: using the unit circle r = 1 = 180 So, one revolution 360 = 2 Converting from degrees to radians & from radians to degrees Degrees Radians Degree multiply by Radians Degrees Radian multiply by 180 180 180 60 3 3 5 150 150 radians 180 6 I do: Convert from degrees to radians or from radians to degrees. (a) -45 (b) 3 2 You do: Convert from degrees to radians or from radians to degrees. (a) 90 (c) radians 4 (b) 270 (d) 3 radians Special angles in degrees & in radians degrees 0 30 45 60 90 180 360 0 6 4 3 2 2 Radians Finding the arc length & the sector area of a circle Arc length (s): S s r is the central angle. r Area of a sector (A): Important: is in radians. 1 2 A r 2 (ex) A circle has radius 18.2 cm. Find the arc length and the area if the central angle is 3/8. Arc length: s r 3 s (18.2) 8 5406 21.4cm 8 Area of the sector: 1 2 A r 2 1 2 3 A (18.2) 2 8 3 (165.62) 8 195.1cm 2 You do: (ex) A circle has radius 18.2 cm. Find the arc length and the area if the central angle is 144. Arc length: 1. Convert the degrees to radians 2. s r Area of the sector: 1 2 A r 2 Trigonometric functions & Unit circle Learning target: To find the values of the trigonometric functions using a point on the unit circle (6 – 2) To find the exact values of the trig functions in different quadrants To find the exact values of special angles To use a circle to find the trig functions Vocabulary: Unit circle is a circle with center at the origin and the radius of one unit. Unit circle Recall: trig ratio from Geometry Opposite a sin A Hypotenuse c cos A adjacent b Hypotenuse c opposite tan A adjacent SOH CAH TOA Also, Two special triangles 3 30, 60, 90 triangle 2 2 1 60 1 90 30 6 3 3 45, 45, 90 triangle 2 2 45 1 1 45 90 1 4 4 1 Using the unit circle y sin r r y x cos r x y tan x Finding the values of trig functions Now we have six trig ratios. Opposite y sin h ypotenuse r 1 hypotenuse r csc sin opposite y adjacent x cos h ypotenuse r 1 hypotenuse r sec cos adjacent x opposite y tan adjacent x 1 adjacent x cot tan opposite y Sin is positive when is in QI. y 0 sin 0 0 r 1 sin 30 sin sin 45 sin 6 4 sin 60 sin sin 90 sin = = Find the exact value of the trig ratios. 3 2 = Sin is positive when is in QII y sin r 2 0 2 sin120 sin 3 3 sin135 sin 4 3 2 5 sin150 sin 6 sin180 sin y sin r 7 sin 210 sin 6 Sin is negativewhen is in QIII 2 0 3 2 5 sin 225 sin 4 4 sin 240 sin 3 3 sin 270 sin 1 2 Sin is negative when is in QIV y sin r 5 sin 300 sin 3 7 sin 315 sin 4 11 sin 330 sin 6 sin 360 sin 2 1 x cos r cos is positive when is in QI cos 30 cos 6 cos is negative when is in QII 3 cos135 cos 4 cos is negative when is in QIII 4 cos 240 cos 3 cos is positive when is in QIV 11 cos 330 cos 6 y tan x tan is positive when is in QI (+, +) tan 60 tan 3 cos is negative when is in QII(-, +) 3 tan135 tan 4 cos is negative when is in QIII(-, -) 4 tan 240 tan 3 cos is positive when is in QIV(+, -) 11 tan 330 tan 6 Find the exact values of the trig ratios. csc 60 csc 3 sec30 sec cot 45 cot 6 4 11 csc 330 csc 6 5 sec 225 sec 4 cot 90 cot 2 csc180 csc sec90 sec 2 4 cot 240 cot 3 (6 – 3) Properties of trigonometric functions Learning target: To learn domain & range of the trig functions To learn period of the trig functions To learn even-odd-properties Signs of trig functions in each quadrant in Q. I II III IV sin + + - cos + + tan + + - csc + + - sec + + cot + + - (sin)(csc) = 1 (cos)(sec) = 1 cos cot (tan)(cot) = 1 sin sin tan cos cos cot sin The formula of a circle with the center at the origin and the radius 1 is: x2 y 2 1 y y sin y r 1 x x cos x r 1 sin 2 cos2 y 2 x2 1 Therefore, sin 2 cos 2 1 Fundamental Identities: (1) Reciprocal identities: 1 csc sin 1 cot tan 1 sec cos (2) Tangent & cotangent identities: sin cos tan cot cos sin (3) Pythagorean identities: sin cos 1 tan 1 sec 2 2 2 2 cot 2 1 csc 2 Even-Odd Properties sin( ) sin csc( ) csc cos( ) cos sec( ) sec tan( ) tan cot( ) cot Co-functions: Find the period, domain, and range y = sinx • Period: 2 • Domain: All real numbers •Range: -1 y 1 y = cosx • Period: 2 • Domain: All real numbers •Range: -1 y 1 y = tanx • Period: • Domain: All real number but (2n 1) x 2 •Range: - < y < y = cotx • Period: • Domain: All real number but x n •Range: - < y < y = cscx y = cscx y = sinx • Period: • Domain: All real number but x n •Range: - < y -1 or 1 y < y = secx • Period: • Domain: All real number but (2n 1) x 2 Range: - < y -1 or 1 y < Summary for: period, domain, and range of trigonometric functions Functions y = sinx Period Domain 2 All real #’s Range -1 y 1 y = cosx 2 All real #’s -1 y 1 y = tanx All real #’s but (2n 1) - < y < x 2 y = cotx All real #’s but - < y < y = cscx All real #’s but - < y -1 or 1 y < y = secx All real #’s but (2n 1) x 2 - < y -1 or 1 y < x n x n (6 – 4) Graph of sine and cosine functions Learning target: To graph y = a sin (bx) & y = a cos (bx) functions using transformations To find amplitude and period of sinusoidal function To graph sinusoidal functions using key points To find an equation of sinusoidal graph Sine function: Notes: a function is defined as: y = a sin(bx – c) + d Period : P Amplitude: a 2 b Period and amplitude of y = sinx graph 2 2 3 2 2 5 2 Graphing a sin(bx – c) +d a: amplitude = |a| is the maximum depth of the graph above half and below half. bx – c : shifting along x-axis Set 0 bx – c 2 and solve for x to find the starting and ending point of the graph for 1 perid. d: shifting along y-axis 2 Period: one cycle of the graph P b I do (ex) : Find the period, amplitude, and sketch the graph y = 3 sin2x for 2 periods. Step 1:a = |3|, b = 2, no vertical or horizontal shift Step 2: Amplitude: |3| Period: 2 P b Step 3: divide the period into 4 parts equally. Step 4: mark one 4 points, and sketch the graph y = 3 sin2x a = |3| P: 3 -3 y = cos x 2 2 3 2 2 5 2 Graphing a cos (bx – c) +d a: amplitude = |a| is the maximum depth of the graph above half and below half. bx – c : shifting along x-axis Set 0 bx – c 2 and solve for x to find the starting and ending point of the graph for 1 perid. d: shifting along y-axis 2 Period: one cycle of the graph P b We do: Find the period, amplitude, and sketch the graph y = 2 cos(1/2)x for 1 periods. Step 1:a = |2|, b = 1/2, no vertical or horizontal shift Step 2: Amplitude: |2| Period: P 2 b Step 3: divide the period into 4 parts equally. Step 4: mark the 4 points, and sketch the graph 2 -2 You do: Find the period, amplitude, and sketch the graph y = 3 sin(1/2)x for 1 periods. I do: Find the period, amplitude, translations, symmetric, and sketch the graph y = 2 cos(2x - ) - 3 for 1 period. Step 1:a = |2|, b = 2 Step 2: Amplitude: |2| Period: P 2 b Step 3: shift the x-axis 3 units down. Step 4: put 0 2x – 2 , and solve for x to find the beginning point and the ending point. Step 5: divide one period into 4 parts equally. Step 6: mark the 4 points, and sketch the graph. y = 2 cos(2x - ) – 3 a: |2| Horizontal shift: /2 x 3/2, P: Vertical shift: 3 units downward We do: Find the period, amplitude, translations, symmetric, and sketch the graph y = -3 sin(2x - /2) for 1 period. Step 1: graph y = 3 sin(2x - /2) first Step 2:a = |3|, b = 2, no vertical shift Step 3: Amplitude: |3| 2 Period: P b Step 4: put 0 2x – /2 2 , and solve for x to find the beginning point and ending point. Step 5: divide one period into 4 parts equally. Step 6: mark the 4 points, and sketch the graph with a dotted line. Step 7: Start at -3 on the starting x-coordinates. y = -3 sin(2x - /2) a=3 P= 3 /4 x 5/4 No vertical shift 0 -3 You do: Find the period, amplitude, translations, symmetric, and sketch the graph y = 3 cos(/4)x + 2 for 1 period. Step 1: graph y = 3 cos(/4)x first Step 2:a = |3|, b = /4 Step 3: Shift 2 units upward Step 4: Amplitude: |3| 2 Period: P b Step 5: Step 5: divide one period into 4 parts equally. Step 6: mark the 4 points, and sketch the graph with a dotted line. (6 – 5) Graphing tangent, cotangent, cosecant, and secant functions Learning target: To graph functions of the form y = a tan(bx) + c and y = a cot(bx) + c To graph functions of the form y = a csc(bx) + c and y = a sec(bx) + c The graph of a tangent function • Period: • Domain: All real number but (2n 1) x 2 •Range: - < y < interval: 2 x 2 Tendency of y = a tan(x) graph y = 2 tan(x) y = tan(x) y = ½ tan(x) To graph y = a tan(bx + c): (1)The period is b and c (2) The phase shift is b (3) To find vertical asymptotes for the graph: solve for x that shows the one period 2 bx c 2 I do: Find the period and translation, and sketch the graph y = ½ tan (x + /4) a=½, b = 1, c = /4 P= b 2 x Interval: 4 -3/4 2 3 x 4 4 One half of the interval is the zero point. /4 We do: Find the period and translation, and sketch the graph 1 y tan x 3 2 1 Graph y tan x 3 2 a=1 b=½ c = /3 P= b Interval: - /2< (1/2)x + /3 < /2 first 1 y tan x 3 2 1 y tan x 3 2 a=1 P = 2 Interval: -5/3 < x < /3 You do: Find the period and translation, and sketch the graph y tan( x ) 4 a=1 P= Interval: The graph of a cotangent function • y = cot(x) • Period: • interval: x n 0<x< • Domain: All real number but x n •Range: - < y < The tendency of y = a cot(x) 1 y cot( x) 2 As a gets smaller, the graph gets closer to the asymptote. y cot( x) y 2 cot( x) Graphing cosecant functions • Period: • Interval: 0 < x < • Domain: all real numbers, but x n • Range: |y| 1 or y -1 or y 1 (-, -1] [1, ) Step 1: y = cos(x), graph y = sin(x) Step 2: draw asymptotes x-intercepts Step 3: draw a parabola between each asymptote with the vertex at y = 1 Graphing secant functions • Period: • Interval: /2 < x < 3/2 • Domain: all real numbers, but (2n 1) x 2 • Range: |y| 1 or y -1 or y 1 (-, -1] [1, ) Graphing secant functions Step 1: graph y = cos(x) Step 2: draw asymptotes x-intercepts Step 3: draw a parabola between each asymptote with the vertex at y = 1 I do (ex) Find the period, interval, and asymptotes and sketch the graph. y csc(2 x ) Graph y = sin(2x - ) •Period: P = 2/|b| • Interval: 0 <2x - < 2 1 • draw the asymptotes -1 •Draw a parabola between the asymptotes You do: Find the period, interval, and asymptotes and sketch the graph. y sec x 2 Graph y = cos(x - /2) •Period: P = 2/|b| • Interval: 0 <x - /2 < 2 • draw the asymptotes •Draw a parabola between the asymptotes