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**Trigonometric Equations I**

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**We want to solve the equation:**

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.

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**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. What angles on the unit circle have this for a cos value? First get the cos by itself. so ALL solutions would be these and however many loops around the circle you want.

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** 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.

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**We still ask where on the unit circle does the sine have this value.**

Notice that we have 2 NOT 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. Solve for by dividing by 2 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).

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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. this value is somewhere in Quad I from calculator: What other quadrant would have the same cosine value (same x value on the unit circle)? Quadrant IV This angle is 2 minus angle from calculator.

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Acknowledgement I wish to thank Shawna Haider from Salt Lake Community College, Utah USA for her hard work in creating this PowerPoint. Shawna has kindly given permission for this resource to be downloaded from and for it to be modified to suit the Western Australian Mathematics Curriculum. Stephen Corcoran Head of Mathematics St Stephen’s School – Carramar

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When trying to figure out the graphs of polar equations we can convert them to rectangular equations particularly if we recognize the graph in rectangular.

When trying to figure out the graphs of polar equations we can convert them to rectangular equations particularly if we recognize the graph in rectangular.

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