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Solving Trig Equations (Day 1)

Solving Trig Equations (Day 1). MHF4UI Wednesday November 7 th , 2012. Application of Special Acute Angles Example.

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Solving Trig Equations (Day 1)

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  1. Solving Trig Equations (Day 1) MHF4UI Wednesday November 7th, 2012

  2. Application of Special Acute Angles Example You are flying a kite at the end of a 80 metre string which currently makes an angle of with the ground. The wind picks up and the kite soars higher to make an angle of with the ground. What is the horizontal distance that the kite moves between the two positions?

  3. Quick Review of Related Angles Your Related angle, , is always drawn from the Terminal arm to the x-axis 0 2 If the Terminal Arm is in the first Quadrant then the angle is its own related angle

  4. Equations with Infinite Solutions So far we have found the trig ratios of angles with the restriction that . Which means that we have had at most two possible solutions for each trig ratio. Let’s examine the following angles: When we draw these angles in standard position, the Terminal Arm will be in the same position for each angle. Each of these angles have the exact same Trig Ratios. Therefore, if we have no restriction on , there will be an infinite number of solutions.

  5. Applying the CAST Rule The CAST Rule is very important when solving Trig equations because it tells us two important pieces of information: - How many solutions exist for any Trig Equation -Which Quadrants the Terminal Arms for solutions are found in So far we have found the trig ratios of angles with the restriction that With this restriction, we will have two solutions for each Trig Ratio. This is because each Trig Ratio is Positive in 2 Quadrants and each Trig Ratio is Negative in 2 Quadrants.

  6. Solving for : Example 1 Solve the equation is positive, so the Terminal Arms are found in Quadrants I and II

  7. Solving for : Example 2 Solve the equation -1 is negative, so the Terminal Arms are found in Quadrants II and IV

  8. These Examples Were Simple Cases In the past two examples we were able to easily determine our solutions , because when is drawn in standard position, the R.A.A.() that is created is one of the ‘special’ angles. Let’s now take a look at some examples where we are not working with any of the Special Acute Angles. For these cases, we must find , before find our solutions

  9. Solving for : Example 3 Solve the equation 0.8472 is positive, so the Terminal Arms are found in Quadrants I and II

  10. Solving for : Example 4 Solve the equation -0.7362 is negative, so the Terminal Arms are found in Quadrants II and III

  11. Solving for : Example 5 Solve the equation -0.4215 is negative, so the Terminal Arms are found in Quadrants III and IV

  12. How do we Solve Trig Equations for our Reciprocal Ratios? When you are asked to solve an equation for a Reciprocal Trig Ratio you must first convert it to a Primary Trig Ratio. Ex/ Solve the equation

  13. Homework Questions: • Textbook Chapter 4.2 (Page 216) • Part B: 9, 10 Solve the following equations for when sec

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