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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 11 1 √ 21 Now we follow the same steps as above. Since our triangle is a 454590 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.
 Spring '11
 Gentimis
 Trigonometry

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