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cosexample

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

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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 find 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 first solution to cos
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Unformatted text preview: θ = 1 2 is θ = π 3 . Since the graph of cos θ is periodic and completes a full cycle on [0 , 2 π ] and is symmetric about the line θ = π on [0 , 2 π ], we know there must be another value of θ at which cos θ = 1 2 . 0 π/3 π/2 π 3π/2 5π/2 2π In fact, this value of θ would be the same distance from 2 π as π 3 is from 0 (by the symmetry of the graph) . Since π 3 is π 3 units to the right of 0, our second point would be π 3 units to the left of 2 π . 2 π-π 3 = 6 π 3-π 3 = 5 π 3 . So our second point at which cos θ = 1 2 is 5 π 3 . 1...
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