1. Trigonometric Equations M 140 Precalculus V. J. Motto

2. Comments We begin by looking at solutions using the unit circle. This give you an opportunity to see that there are multiple solutions. Some of these examples will require a bit of algebra. We will also use the calculator to find solutions. The calculator will be your primary method for solving trigonometric equations.

3. A Review of the Unit Circle

4. We want to solve the equation: (Unit Circle Solution) Where on the unit circle is the sine value - 1/2? But if we want ALL solutions we could go another loop around the unit circle and come up with more answers. A loop around the circle is 2  so if we add 2  to our answers we'll get more answers. We can add another 2  and get more answers.

5. All solutions to the equation would be: What this means is as k goes from 0, 1, 2, etc. you would have the answer with another loop around the unit circle. First get the cos  by itself. What angles on the unit circle have this for a cos value? so ALL solutions would be these and however many loops around the circle you want.

6. We want to solve the equation: (Calculator Solution) This is an angle in the 4 th Quadrant. The corresponding positive coterminal is 2π – 0.52 or 5.76 radians This diagram tells us that the sine function is negative in the 3 rd Quadrant. So there is a solution there. It is π + 0.52 or 3.66 radians. You can find other solutions by adding integer multiples 2π to 5.76 or 3.66. Using our calculator we get:

7. First get the cos  by itself. What angles on the unit circle have this for a cos value? so ALL solutions would be these and however many loops around the circle you want. We want to solve the equation: (Unit Circle Solution)

8. We want to solve the equation: (Calculator Solution) First get the cos  by itself. This is an angle in the 2 nd Quadrant with a reference angle of π – 2.62 or 0.52 radians. This diagram tells us that the cosine function is negative in the 3 rd Quadrant. So there is a solution there. It is π + 0.52 or 3.66 radians. You can find other solutions by adding integer multiples 2π to 5.76 or 3.66.

9. This would mean only one loop around the circle. Get tan  alone  What angles on the unit circle have this value for tangent? Since it can be either plus or minus, there are 4 values. We don't add any to go around again because it says on the interval from 0 to 2 . You can find the solutions by using a calculator

10. We still ask where on the unit circle does the sine have this value. Notice that we have 2  NOT  Solve for  by dividing by 2 Since you divided the angle in half,  is smaller so you need to take another loop around the circle because you only want answers between 0 and 2  but by the time you divide by 2 you'll still be in that interval. The solution will be all 4 values because they are all still in [0, 2  ) (If you try another loop around you'll find yourself larger than 2  ). You can find the solutions by using a calculator!

11. If the values that these trig functions equal are NOT exact values on the unit circle you will need to use your calculator. You can use the inverse cosine button on your calculator (make sure mode is radians) but remember that the range is only the top half of the unit circle and we want the whole unit circle so you'll need to figure out the other value from the one given. from calculator : this value is somewhere in Quad I What other quadrant would have the same cosine value (same x value on the unit circle)? This angle is 2  minus angle from calculator. Quadrant IV

12. In the next few examples we'll explore various techniques to manipulate trigonometric equations so we can solve them. We'll find solutions on the interval from 0 to 2 . Often written as [0, 2  ).

13. Use the Pythagorean Identity to replace this with an equivalent expression using sine. Combine like terms, multiply by -1 and put in descending order Factor (think of sin  like x and this is quadratic) Set each factor = 0 and solve The first tip is to try using identities to get in terms of the same trig function.

14. When we don't have squared trig functions, we can't use the Pythagorean identities. If you have two terms with different trig functions you can try squaring both sides. re-order terms Get sine term alone 2 2 Square both sides. Must do whole side together NOT each term (so left side will need to be FOILed). Pythagorean Identity---this equals 1Double angle Identity Remember to do another loop when you have 2  Where is the sine -1?

15. Try to get equations in terms of one trig function by using identities. Be on the look-out for ways to substitute using identities. This simplifies many equations. Try to get trig functions of the same angle. If one term is cos2  and another is cos  for example, use the double angle formula to express first term in terms of just  instead of 2  Get one side equals zero and factor out any common trig functions See if equation is quadratic in form and will factor. (replace the trig function with x to see how it factors if that helps) If the angle you are solving for is a multiple of , don't forget to add 2  to your answer for each multiple of  since  will still be less than 2  when solved for. HELPFUL HINTS FOR SOLVING TRIGONOMETRIC EQUATIONS

16. Use a graphing utility to solve the equation. Express any solutions rounded to two decimal places. Graph this side as y 1 in your calculator Graph this side as y 2 in your calculator You want to know where they are equal. That would be where their graphs intersect. You can use the trace feature or the intersect feature to find this (or these) points (there could be more than one point of intersection). There are some equations that can't be solved by hand and we must use a some kind of technology.

17. After seeing the initial graph, lets change the window to get a better view of the intersection point and then we'll do a trace. Rounded to 2 decimal places, the point of intersection is x = 0.53 check: This is off a little due to the fact we approximated. If you carried it to more decimal places you'd have more accuracy.