cosexample - = 1 2 is = 3 . Since the graph of cos is...

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Calculus 135 Trig Equation Example 1.1 The following trig equation from our 9/3 class example was taken from Strauss Bradley Smith text 1.1 page 9. Solving 2 cos θ sin θ = sin θ for θ led us to 2 cos θ sin θ - sin θ = 0 or sin θ (2 cos θ - 1) = 0 From which we get sin θ = 0 and cos θ = 1 2 . On [0 , 2 π ) , sin θ = 0 at θ = 0 , and θ = π while cos θ = 1 2 at θ = π 3 and θ = 5 π 3 . On [0 , 2 π ] , sin θ = 0 at θ = 0 , θ = π, and θ = 2 π while cos θ = 1 2 at θ = π 3 and θ = 5 π 3 . Notice, that the domain is important when deciding upon solutions to an equation. A common question was ”how did we ±nd the solution θ = 5 π 3 ?” Our knowledge of “30-60” right triangles, give us that the cosine of 60 degrees or π 3 radians is 1 2 . This give us the ±rst solution to cos
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Unformatted text preview: = 1 2 is = 3 . Since the graph of cos is periodic and completes a full cycle on [0 , 2 ] and is symmetric about the line = on [0 , 2 ], we know there must be another value of at which cos = 1 2 . 0 /3 /2 3/2 5/2 2 In fact, this value of would be the same distance from 2 as 3 is from 0 (by the symmetry of the graph) . Since 3 is 3 units to the right of 0, our second point would be 3 units to the left of 2 . 2 - 3 = 6 3- 3 = 5 3 . So our second point at which cos = 1 2 is 5 3 . 1...
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This note was uploaded on 09/25/2011 for the course CALCULUS 135 taught by Professor Augustarainsford during the Spring '11 term at Rutgers.

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