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A Review of the Trigonometry of Algebra II 7-1 Measurement of Angles 1 A) Convert –80˚ to radians in terms of π 5π B) Convert 12 radians to degrees. C) Convert –18˚ to radians 2 Find two angles, one positive and one negative, which are co-terminal with the given angle. A) –100˚ 5π B) 12 rds 3 pp.261, 262 1, 5, 9, 13, 17, 19, 25 7-2 Sectors of Circles 4 A sector of a circle is the region bounded by a central angle and the intercepted arc. For example, sector ABD, as pictured below, is the region bounded by segments AB and AD and arc BD. B s A D r A) A sector of a circle has a central angle of 2 radians and a radius of 5 inches. Find the length of the sector's arc. B) The angle of a sector is 100˚ and the sector's arc is 5 inches. Find the radius of the sector to the nearest thousandth of an inch. 1 6 A sector of a circle has a central angle of π/2 radians and a radius of 2 inches. Find the length of the intercepted arc to the nearest thousandth. 7-3 The Sine and Cosine Functions 9 Name the quadrant described: cosW > 0 and tanW < 0 10 Evaluate with no calculator: 5π A) cos(0) B) sin( 2 ) π 7π C) tan (– 2 ) D) sin(– 2 ) π π 11 Find x, in terms of π, if one knows that x lies in the interval (2 , π): cosx = –cos 15 12 pp. 272-274: 1-17 (odd) 21, 27, 33-41 (odd) 7-4 Evaluating and Graphing Sine and Cosine 13 14 A) Graph y = cosx for one period if the left end-point of the domain is π/2. 1 B) How many solutions does the equation x = cosx have for the domain of part A? 2 π A) Graph y = sinx on the domain [– 2 , 2π] B) i) T/F: sin1.1 > sin1.3 ii) T/F: sin1.6 > sin1.7 iii) T/F: sin3.5 > sin4 15 We say that both the sine and cosine are periodic (or cyclic) functions because their graphs keep repeating. What is the length of one period (or cycle) of y = sinx? 16 What is the amplitude of y = sinx? 17 pp. 279-281: 1, 3, 11-17(odd) 7-5 The Other Trigonometric Functions 2 5 18 If W is an acute angle and tanW = 3 then find the exact answer of secW. 2 19 If cscW = 3 then express tanW in radical form knowing that W is in quadrant II 7π B) sec( 2 ) 20 Evaluate without the calculator: A) sec(π) 11π 21 Express in terms of a reference angle: A) cot 14 4π B) csc 5 22 Find the exact values without the calculator: π A) tan 3 π B) sec3 9π C) cot 4 23 pp. 285-286: 1-7 (odd), 13-17 (odd), 23-27 (odd) 8-1 Simple Trigonometric Equations 24 Find the values of x, in terms of π, in the interval [0,2π] for which 6sinx – 3 3 = 0 25 Solve for x on the interval [0,2π] in terms of π. No calculator. A) 2cosx – 3 = 0 B) 5cscx – 3 = 7 C) tanx = – 1 D) cscx = –1 2 3 E) cscx = – 3 26 pg. 299: 7-21 (odd) 8-2 Sine and Cosine Curves 27 Graph each of the following for one period beginning at x = 0. A) y = cos2x B) y = cos 2πx 3 5π C) tan 6 28 Graph the following for one period beginning at x = –π/2 1 y = sin2 x 29 pp. 305-306: 1-19 (odd) 8-4 Relationships Among the Functions 30 Each of the following can be simplified to one of the six trig. functions. Simplify completely, A) 2 – 2cos2x 2sinx cosx B) cosx (secx + 1) (secx – 1) sinx 31 Simplify completely: A) cotx + tanx B) (sinx + cosx)2 – 2sinxcosx cosx sinx 1+cosx C) 1+cosx + sinx 32 Simplify completely: secy – tany A) 1–siny tanx 1+secx B) 1+secx + tanx 9-1 Solving Right Triangles 33 pp. 334-337: 1-9 (odd), 15, 16, 23, 33, 35 4 9-2 The Area of a Triangle 34 Find the area, to the nearest thousandth, of ∆WRT if w= 4, t = 8 and R = 98˚ 35 pp. 342-344: 1-15 (odd), 19 Challenges. Entire solutions are in the back, not just the answers. 1 p265 9 You should wind up with an equation that you can solve on the calculator. 2 Find the quadrant that z is in if cscz < 0 and cos(–z) < 0 3 Evaluate without the calculator: sec( – 35π 6 ) 4 p286 5 Solve for x in terms of π on the interval [0,2π]: 3secx +5=3 3 6 p305 7 8 18 but just find cot 16 cscx cscx Simplify: cscx –1 + cscx + 1 (Don't forget that there are three Pythagorean Identities) p353 13 5 Answers 1A) -4π/9 1B) 75˚ 1C) -0.314 rds. 4A) 10 inches 4B) 2.865 inches 2A) 260˚,–460 2B) 29π/12, 19π/12 – 3) back of text 6) 3.142 in. 9) IV 10A) 1 10B) 1 14π 11) 15 12) text 13A) 13B) zero 10C) undef. 10D) 1 3 2 1 .5š 1š 1 .5š 2š 2 .5š -1 -2 -3 14A) 14B) i. F ii. T iii. T iv. F 3 2 1 -.5 š .5š 1š 1 .5š -1 -2 -3 15) 2π 29 18) 3 21B) csc 16) 1 19) – 5 17) text 8 8 21C) tan 6 20A) –1 20B) undef. 22A) 22B) 2 3 6 3π 21A) – cot14 22C) 1 23) text π 11π 24) π/3, 2π/3 25A) 6 , 6 26) text 25B) 5 , 6 6 3π 7π 3π 25C) 4 , 4 25D) 2 27A) 2 1 .1š .2š .3š .4š .5š .6š .7š .8š .9š .2 .3 .4 .5 .6 .7 .8 .9 -1 -2 27B) 2 1 .1 -1 -2 28) P=4 π; Amp=1 29) text 30A) tanx 31A) 30B) tanx 1 sin x cos x 32A) secy 33) text 31B) secx 31C) 2cscx 32B) 2cscx 34) 15.844 sq. units 35) text 7 4π 5π 25E) 3 , 3 Solutions to Challenge Problems 1) 1.5cm or 2 cm 2) III Step #1 cscz < 0 means that z is in III or IV y x x Step #2 Perimeter is 2x+y Thus, 2x+y =7 Circumference of entire circle is 2πx y Sector is of the circle's area πx2 2πx y Thus, we have 2πx . πx2 = 3 y Thus, 2 x = 3 but y = 7–2x (7–2x) Thus, 2 x = 3 on calculator gives 1.5 or 2 z –z Angles z and –z will have points on the circle which have the same x-coordinates. Therefore, cos(–z) = cosz Step #3 cos(–z) < 0 becomes cosz <0 thus z is in II or III Step #4 Putting Steps 1 and 3 together, z must be in quad III 12π π 3) 2/ 3 ; Keep adding 2π until you get rid of negative. I keep adding 6 till I get to 6 and π 2 sec6 = 3 4) 2 6 π 3π 2 < x < 2 means that the choices for x are II and III quadrants but csc is negative Thus, x must be in III; cot must be positive Use a right triangle whose hypotenuse is 5 and where a leg is 1. 5π 7π 5) 6 , 6 2 3 3 Multiply both sides by 3 and solve for sec to get secx = – 3 Thus, cosx = – 2 3 3 Rationalize denominator to get: cosx = – 2 8 6) amp is 5. Period is 2 y = –5cos(πx) Period is 2 can be seen by noting that one cycle goes from one max to the next max and that the length of this cycle is 2. I know answer uses cos because y-intercept is not 0 y = A cos(kx) Amp is 5. Thus A is 5 or –5. I know A is negative because y-intercept is negative. Thus A = –5 2π k =2 Solve to get k =π 7) 2sec2x Combine: cscx(cscx+1)+cscx(cscx–1) = (csx–1)(cscx+1) csc2x + cscx + csc2x – cscx = csc2x–1 2csc2x cot2x 2 2 = sin 2 x cos x sin2 x = 2 = 2sec2x cos2x 8) 9.74 cm and 13.31 cm We are given the following parallelogram: B C 6 A D 70 ° 10 To find diagonal BD work in ∆ABD, law of cosines deals with SAS. Let a be the measure of BD 2 a 2 36 100 – 2(6)(10)cos70 o Thus a 94.9575828 Thus, a 9.74 Now work in ∆ACD, let d = measure of AC DC= 6 D= 180–70 =110 2 2 o d 36 100 – 2(6)(10)cos110 Thus, d 177.0424172 Thus, d 13.31 9 Trig Review: Chapter 12 1) Draw an angle of 550˚ in standard position 11) Solve for x on the interval 0˚ ≤ x ≤ 360˚ 2 3 A) tanx = – 3 B) cscx = – 3 2) Give the measure of an angle in standard position that has the following two C) tanx = 1 D) sinx = 1 properties i) it is co-terminal to the 550˚ angle 12) Solve for x, to the nearest degree, on the ii) the measure is positive but smaller than interval 0˚ ≤ x ≤ 360˚ 550˚ A) secx = – 2.5 B) cotx = 4 3) Give the measure of an angle in standard position that has the following two properties i) it is co-terminal to the 550˚ angle ii) the measure is negative 13) Find the measure of angle E to the nearest degree where F is a right angle. G 5.3 4) Draw the angle of #3. 5) Assume all numbers are in degrees. We have discussed the co-function: cosx = sin(90–x) cscx = sec(90–x) cotx = tan(90–x) E 14) Find AD to the nearest unit if we know that C is a right angle and A, D and C are collinear. B A) Write the remaining three co-function formulas. A B) Solve for x where x is the measure of an acute angle: cos48 = sinx C) If tanx = K then express the cot(90–x) in terms of K 6) Sample: Write sin179˚ as a function of an acute angle: The answer is sin1˚ Write each of the following as a function of an acute angle. A) sec(127˚) B) cot(190˚) C) sin(302˚) F 4 23o 38o D 6 C 15) In ∆ABD, a=5, b = 7 and d = 10.5. find the measure of angle D to the nearest degree 16)In ∆ABD, a = 5, b = 7 while D = 26˚. Find d to the nearest tenth. 17) In∆ABD, A = 34˚, D = 50˚ while b = 7. Find d to the nearest tenth. Answers 2)190˚ 3)–170˚ 5B)42 C) K 6A) –sec53˚ B) cot10˚ C) –sin58˚ 2 7)A)=sec150˚ = – B)= tan300˚ = – 3 7) Evaluate. Your answer must be exact. 3 A) sec(870˚) B) tan(–420˚) 5 22 10) A) – B) 3 11A) 120˚,300˚ 22 8) Explain why csc(450˚) =1 11B) 240˚,300˚ 11C) 45˚,225˚ 11D) 90˚ 9) Explain why cot(–180˚) is undefined 12A) 114˚, 246˚ B) 14˚, 194˚ 5 10) If 90˚ < x < 270˚ and if cscx = – then 13 41˚ 14) 5 15) 121˚ 16) 3.3 17) 5.4 3 find A) secx B) cotx 10 Review of Chapter 13 Please do in your blue book or follow your teacher’s instructions. Show all work and give good explanations. Remember that you will find this useful when it is returned next year. 18) Change 100˚ to radians, in terms of π 19) Change 5 radians to degrees. 20) Evaluate. Your answer should be left in radical form. A) sec 5π 6 B) csc 3π 2 C) tan(– π ) 4 21) Solve for x on [0,2π). Your answer should be in radians, in terms of π. 3 2 A) tanx = – 3 B) sinx = – 2 C) cosx = –1 22) Graph for one period and state the amplitude: 1 A) y = 3sin(2x) B) y = –2cos(2 x) 23) Simplify completely: A) sinx + cosxcotx Answers 5π 18) 9 900 19) π ˚ 2 3 20) A) – 3 B) –1 C) –1 5π 11π 5π 7π 21) A) 6 , 6 B) 4 , 4 B) secx– cosx tan2 C) π 22) check on the calculator 23) A) cscx B) cosx 11