You can take the arccos of both sides to get θ by itself.

Transcription

1 .7 SOLVING TRIG EQUATIONS Example on p. 8 How do you solve cos ½ for? You can tae the arccos of both sides to get by itself. cos - (cos ) cos - ( ½) / However, arccos only gives us an answer between 0 and. There are other angles that give us a cosine value of ½. Loo at the unit circle diagram. What other angle gives a cosine value of ½? cos / y x cos 5/ ½

2 We see from the unit circle that there are two possibilities for between 0 and for which cos ½: / and 5/ Since the values of cosine are repeated for integral multiples of, other values of that solve cos ½ are: / or 5/, where is any integer. NOW YOU DO # on p.5 Example : Solve the equation sin 0, 0 < Solution: First let x sin, and solve for x. Then solve for. The equation becomes x 0 x x Now - - replace x with and - sin Since sin < 0, we now can only be in Quadrant III or IV. y sin solve x for We choose / and 5/ because they meet the requirement of 0 < sin / - sin 5/ - NOW YOU DO #

3 So Example sin () ½, 0 < Don t convert this to the double-angle formula. Solve this the way it is. Find angle that has a sine value of ½, then divid that angle by. /6 or 5/6 Dividing both sides by gives us: OR Because dividing by gives us a repeating cycle of, there are multiple possible answers within the range of 0 and. We cannot use -, because that would give is a < 0, and we cannot use, because that would give us a >. So let s see what we get with 0 and : 0 : (0) : () the complete 0 solution 5 7,,, of sin( ), 0 < 5 (0) 5 () is : NOW YOU DO #7

4 Example 5 Solve the equation Let s see what ( - /) must be for the tangent function to be.we now that tan / and the tangent value repeats itself after a period of. 0, tan < 7, is so the complete solution for 0 7 () (0) 0 0, try So let's. would give us 0 and -would give us As before, so, the cycle repeats for For tangent, tan < > <

5 Example 6 When using a calculator to find the solution for, be aware that using the inverse trig function only gives you an answer within the range of the inverse function. Solve the equation sin 0., 0 < sin - (0.) will give you an answer with / / But what about Quadrants II and III? Sine is only positive in Quadrants I and II. From the supplementary angle theorem, sin( ) sin So So the complete solution is.0,.8 #0 sin - 0., 0 < sin - (-0.) will give you an answer with / < 0 since the sine is negative. sin - (-0.) What are the equivalent angles in the range 0 < that will give sin -.? We now sine is only negative in Quadrants III and IV. We also now sine repeats itself for cycles of. Since -.0, will still be <. y. rad x sin (. rad ) -.. rad. rad sin (-. rad ) rad 6.08 Solution:. rad, 6.08 rad

6 Snell s Law of Refraction (Applied Problem # on p.5) angle of incidence Incident ray s speed sin sin v v index of refraction angle of incidence Refracted ray s speed. The index of refraction of light passing from a vacuum into dense glass is.66. If the angle of incidence is 50, determine the angle of refraction. v 50, v sin sin.66 cross multiply to get.66sin sin 50 Divide both sides sin 50 sin Using the inverse sine function on the calculator we get sin sin( ) sin by.66 to get (.65) 7.5 Mae sure to change mode to Degrees in your calculator before computing sin 50

7 HOMEWORK p.5 #,5,7,9,,5,,

8 .8 Solving Trig Equations (II) Example Solve sin sin 0, 0 < How do we get sin by itself to solve for? Let s let x sin x x 0 We can factor to solve for x (x -)(x -) 0 The zero-product property that either factor must be 0 for the product to be 0. x 0 x x ½ sin ½ /6, 5/6 or x 0 x sin / Alternate Method: Use Quadratic formula If you have an equation in the form ax bx c 0 The solution is: b ± b ac x a x x 0 gives a,b -,c ( ) x OR ( ) x ( ) () ( ) () different real solutions if b -ac >0 repeated real solution if b ac 0 No real solutions if b -ac <0 ()() ()()

9 Example Solve the equation cos sin, 0 < This example has a combination of both sine and cosine functions, so maing xsin will not give me everything in terms of x. However, we can identities to get everything in terms of the same trig function, then convert to x. Let s use the fact that sin cos and get sin - cos We can not get everything in terms of cos cos (- cos ) cos - cos Putting everything on the left-hand-side and leaving righthand-side with 0 gives me: cos cos 0 Now let s mae x cos x x 0 Factoring we get (x )(x ) 0 x 0 x - x - ½ cos - ½ /, / OR x 0 x - cos -

10 Example 5 Solve the equation sin cos - ½, 0 < There is no identity to convert sin into cos, or vice versa, unless we used cos ± sin but that would mae the problem more complicated, so instead let s use sin() sin cos Notice that the term (sin cos ) is also in our equation to solve. Let s get it by itself in this identity by dividing both sides by : ½sin() sin cos Now let s substitute that for sin cos in our equation to solve: ½sin() - ½ Multiplying both sides by gives sin() - / Divide both sides by to get by itself / We cannot use -, because that would give is a < 0, and we cannot use, because that would give us a >. So let s see what we get with 0 and : 0: / (0) / : / () / / 7/

11 HOMEWORK p.58 #,,7,7,9,5,57