Solving Trigonometric Equations

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1 Solving Trigonometric Equations CHAT Pre-Calculus Section 5. The preliminary goal in solving a trig equation is to isolate the trig function first. Eample: Solve 1 cos. Isolate the cos term like you would isolate any variable term in an algebraic equation. 1 cos cos cos 1 1 We know that if our angle is from to π, we have 5 But there are many angles that are coterminal with these angles. Every time we add a multiple of π we get a coterminal angle. Thus, the solution should be 5 n n 1

2 Section 5. Eample: Solve sin 1 sin. sin 1 sin sin 1 1 sin Since sine has a period of π, find the solutions on the interval from to π first (i.e. on the interval [, π)) Every time we add a multiple of π to these solutions, we get another solution. Thus, we write the solution as 7 11 n n 6 6

3 Section 5. Eample: Solve tan. tan tan tan We had to isolate the trig function and then take the square root of both sides. Because the period of tangent is π, we will first find all the solutions on the interval from to π. We write this interval as [, π). Our solution is then Every time we add a multiple of π to either of these solutions we get an angle that has the same tangent. Thus, our solution is n n

4 Section 5. Checking a Solution Graphically F the equation tan we got the solution n n When solving any equation graphically, we graph the left side and right side separately, and then see where the graphs intert. If the equation is written to =, we can graph it on the calculat and see where it crosses the -ais to find the solutions. This can be done by using the [CALC] [zero] feature the [TRACE] feature. Graph y tan. Use the [CALC] [zero] feature to find the approimate values of the -intercepts. These are the solutions of the equation. Or, use [TRACE] to find the approimate values of the -intercepts. 4

5 Section 5. Eample: Solve csc csc. Do not divide both sides by csc. This would eliminate possible solutions. Instead, when the equation involves me than one trig function, collect all terms on one side and try to separate the functions by facting using identities. csc csc csc csc csc ( 1) csc 1 csc 1 There is no angle whose coant is, since csc = means sin = 1, which is undefined (and never happens f the sine function). So we look at the other part. If = 1, then cos = 1 1 = 1. This happens at, π, 4π, etc. We write the solution as n 5

6 Section 5. Equations of Quadratic Type Eample: Solve cos cos 1 on the interval [, π). (Think of this as u u 1.) cos (cos cos cos 1 cos 1)(cos 1) 1 1 cos 5,, cos 1 1 6

7 Section 5. Eample: Solve cos sin on the interval [, π). Befe attempting this problem, we must write it as a single trig function. To do this, remember that we have the Pythagean identity that says sin cos 1 which can also be written as cos 1 sin. Substitute in the above equation f cos. cos (1 sin sin sin sin sin 1 (sin 1)(sin 1) sin 1 ) sin sin 1 sin sin 1 5,, 6 6 sin 1 7

8 Section 5. Eample: Solve 1 tan. Secant and tangent are related, but only if you have tan. If we square both sides of the equation, we will have a tan on the right. 1 tan 1 tan 1 tan Now substitute the Pythagean identity tan 1. 1 tan We know that if = -1, then cos = 1 n =-1. Thus, *Remember to check your solution, because squaring both sides sometimes introduces etraneous solutions!! 8

9 Section 5. Funtions Involving Multiple Angles If the argument of the trig function is a multiple angle of the fm sin ku and cos ku, then first solve the equation f ku and then solve f u. Eample: Solve sin t 1 on the interval [, π). sin t sin t sin t 1 Ask the question, The sine of what angle is? The answeris 6 6. So t 6 6. Since t π, then t 4π. This means we are actually looking f answers from to 4π. So we have t ,,, Divide thru by to get t ,,,

10 Section 5. Eample : Solve the equationcot 1 on the interval [, ). cot cot 1 1 The cotangent = -1 at 4 so must = 4. Take 4 and solve f. 4 () () 4 1

11 Section 5. Using Inverse Functions Eample: Solve 1 f all values of. 1 ( 5)( ) 5 5 We can solve these equations by taking the -1 of both sides, but we need to remember that the angle returned f arcant is between and π (the range of arcant). F = 5, the angle must be in Quadrant I and IV. Taking the -1 of both sides will give us the Quadrant I angle. The angle in Quadrant IV will be the opposite of that angle. 5 1 ( ) 1 (5) 1 (5) The Quadrant I angle is = The Quadrant IV angle is =

12 Section 5. F = -, the angle must be in Quadrant II and III. Taking the -1 of both sides will give us the Quadrant II angle. The angle in Quadrant III will be the opposite of the Quadrant II angle. 1 ( ) 1 ( ) 1 ( ) The Quadrant II angle is = -1 (-) = cos -1 (-1/)=. The Quadrant III angle is = -1 (-)=. Since we were asked to find all angles that are solutions, we must add nπ to each angle. Thus, our solution is 1 (5) n 1 (5) n n n 1

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