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JMerrill, 2010 It will be imperative that you know the identities from Section 5.1. Concentrate on the reciprocal, quotient, and Pythagorean identities. The Pythagorean identities are crucial! Solve sin x = ½ Where on the circle does the sin x = ½ ? Solve for [0,2π] x 5 , 6 6 Particular Solutions Find all solutions General solutions 5 x 2n , 2n 6 6 Solve 2sin 1 0 for [0 ,360 ) 2 2sin 1 0 2 2sin 1 2 1 sin 2 2 2 sin 2 o o There are 4 solutions because sin is positive in 2 quadrants and negative in 2 quadrants. 45o ,135o ,225o ,315o Find all solutions to: sin x + 2 = -sin x sinx sinx 2 0 2sinx 2 sin x x 2 2 5 2n 4 and x 7 2n 4 Solve 3tan2 x 1 0 for [0,2] 3 tan x 3 5 7 11 x , , , 6 6 6 6 Solve sin x tan x 3sin x for [0, 2 ) sin x tan x 3sin x 0 Round to nearest hundredth sin x(tan x 3) 0 sin x 0 or tan x 3 x 0,3.14 x 1.25,4.39 Solve cot x cos2 x 2cot x in [0,2) These 2 solutions are true because of the interval specified. If we did not specify and interval, you answer would be based on the period of tan x which is π and your only answer would be the first answer. cot x cos2 x 2cot x 0 cot x(cos x 2) 0 2 cot x 0 x 3 , 2 2 cos2 x 2 0 cos2 x 2 cos x 2 DNE (Does Not Exist) No solution Verify graphically Quick review of Identities Reciprocal Identities sin 1 csc 1 cos sec 1 tan cot Also true: 1 csc sin 1 sec cos 1 cot tan Quotient Identities sin tan cos cos cot sin Pythagorean Identities sin cos 1 2 2 1 cot csc 2 2 tan 1 sec 2 2 These are crucial! You MUST know them. sin2 cos2 1 tan2 sec2 cot2 csc2 (Add the top of the triangle to = the bottom) Strategies Change all functions to sine and cosine (or at least into the same function) Substitute using Pythagorean Identities Combine terms into a single fraction with a common denominator Split up one term into 2 fractions Multiply by a trig expression equal to 1 Factor out a common factor x 3x 4 0 2 ( x 1)( x 4) 0 ( x 1) 0 or ( x 4) 0 x 1 x4 sin x sin x cos x 2 2 Hint: Make the words match so use a Pythagorean identity sin x sin x 1 sin x 2 2 sin x sin x 1 sin x 0 2 2 2sin x sin x 1 0 2 Quadratic: Set = 0 Combine like terms Factor—(same as 2x2-x-1) (2sin x 1)(sin x 1) 0 1 sin x or sin x 1 2 x 7 11 , 2 6 , 6 Solve 2sin cos for [0 ,360 ) o 2sin cos cos 2 sin o 2 cot 1 tan 2 26.6 , 206.6 o o You cannot divide both sides by a common factor, if the factor cancels out. You will lose a root… sin x tan x 3sin x sin x tan x 3sin x sin x sin x tan x 3 Common factor— lost a root 2sin cos 2 cos sin 2 cot 1 tan 2 No common factor = OK Sometimes, you must square both sides of an equation to obtain a quadratic. However, you must check your solutions. This method will sometimes result in extraneous solutions. Solve cos x + 1 = sin x in [0, 2π) There is nothing you can do. So, square both sides (cos x + 1)2 = sin2x cos2x + 2cosx + 1 = 1 – cos2x 2cos2x + 2cosx = 0 Now what? Remember—you want the words to match so use a Pythagorean substitution! 2cos2x + 2cosx = 0 2cosx(cosx + 1) = 0 2cosx = 0 cosx + 1 = 0 cosx = 0 cosx = -1 x 3 , 2 2 x cos x 3 , 2 2 1 sin 2 0 1 1 x 2 3 3 cos 1 sin 2 2 0 1 1 cos 1 sin 1 1 0 Solve 2cos3x – 1 = 0 for [0,2π) 2cos3x = 1 cos3x = ½ Hint: pretend the 3 is not there and solve cosx = ½ . Answer: 1 1 x cos 2 But…. 5 x , 3 3 In our problem 2cos3x – 1 = 0 What is the 2? amplitude What is the 3? frequency This graph is happening 3 times as often as the original graph. Therefore, how many answers should you have? 6 1 x cos 2 5 x , 3 3 1 Add a whole circle to each of these 7 11 , 3 3 And add the circle once again. 13 17 , 3 3 Final step: Remember we pretended the 3 wasn’t there, but since it is there, x is really 3x: 3x 5 7 11 13 17 3 , 3 , 3 , , , 3 3 3 5 7 11 13 17 So, x , , , , , 9 9 9 9 9 9 Work the problems by yourself. Then compare answers with someone sitting next to you. Round answers: o o 191.5 ,348.5 1. csc x = -5 (degrees) 2. 2 tanx + 3 = 0 (radians) 3. 2sec2x + tanx = 5 (radians) 2.16, 5.30 2.16, 5.30, .79, 3.93 4. 3sinx – 2 = 5sinx – 1 7 11 , 6 6 5. cos x tan x = cos x 3 5 , , , 2 2 4 4 6. cos2 - 3 sin = 3 3 2