hw20soln - 5 π 3 which is also out of the range. Same goes...

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Name: MAC 1114 Homework 20 § 7.7 Part I 5 Show your work, if you’re not sure you’ve shown enough, try writing a few english sentences. Make sure you indicate what rule you are using in each step!! 1. Solve csc θ = 2 3 for θ , when - π 2 θ 3 π 2 . 2 2 3 3 -1 1 The triangles are 30-60-90 triangles and our angle is opposite the 3 so it is 60 o or π 3 . The angle in the first quadrant is θ = π 3 , and the angle in the second quadrant is θ = π - π 3 = 2 π 3 . Now since the period of cos is 2 π , we write θ = π 3 ± 2 πk and θ = 2 π 3 ± 2 πk. OK, now we use the restrictions. π 3 is in the range, but π 3 + 2 π = 7 π 3 which is outside of the range and π 3 - 2 π = -
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Unformatted text preview: 5 π 3 which is also out of the range. Same goes for 2 π 3 , so those are our answers. 2. Solve 7 cot β + 35 = 42 for β . First we solve for tan θ and we find that tan θ = 1. √ 2 1-1 1 √ 2-1 Now we follow the same steps as above. Since our triangle is a 45-45-90 triangle, we know the angles are 45 o or π 4 . The angle in the the first quadrant is then θ = π 4 and the angle in the fourth quadrant is θ = π + π 4 = 5 π 4 . Now since the period of tan is π we write θ = π 4 ± πk and θ = 5 π 4 ± πk....
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This note was uploaded on 12/27/2011 for the course MAC 1114 taught by Professor Gentimis during the Spring '11 term at University of Florida.

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