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Analytic Trigonometry

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Presentation on theme: "Analytic Trigonometry"β€” Presentation transcript:

1 Analytic Trigonometry
Chapter 7 Analytic Trigonometry

2 Section 7 Trigonometric Equations

3 Almost always we will specifically be looking for solutions of trig equations for 0 ≀ πœƒ ≀ 2πœ‹ The goal in solving trig equations is to get the trig function and what you are taking that trig function of by itself.

4 Examples 1-4: 1 οƒ  2sinπœƒ + 3 = 2 2 οƒ  cos(2πœƒ) = Β½ 3οƒ  tan(πœƒ/2 + πœ‹/3) = 1 4 οƒ  4cos2πœƒ = 1 What is the function and what are you taking the function of? Is it by itself? 1 οƒ  2 οƒ  3 οƒ  4 οƒ  For the examples above, you need to get the ”function of” part alone FIRST, then solve for πœƒ.

5 Important Notes: Make sure trig function of ( ) is by itself, everything else is on the other side of the equation Then pretend nothing is happening to πœƒ and think where does cos πœƒ = ?, sin πœƒ = ?, tan πœƒ = ? happen on the unit circle Take what is in ( ) = unit circle value + period x k Solve for πœƒ οƒ  use -1, 0, 1, etc. for k and evaluate until you no longer have answers between 0 - 2πœ‹

6 Example 1: 2sinπœƒ + 3 = 2, 0 ≀ πœƒ ≀ 2πœ‹

7 k represents the number of complete revolutions 2πœ‹ = period = length of when sine and cosine repeat For the previous example, because you are not doing anything directly to πœƒ, you want to be able to change k and stay between 0 < πœƒ < 2πœ‹ so your only answers are: If the question was to solve in general, then you would leave you answer as:

8 Example 2: cos(2πœƒ) = Β½, 0 ≀ πœƒ ≀ 2πœ‹

9

10 Example 3: tan(πœƒ/2 + πœ‹/3) = 1, 0 ≀ πœƒ ≀ 2πœ‹

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12 EXIT SLIP


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