hw21soln - 18 π 12 . OK here goes. I’ll put boxes around...

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Name: MAC 1114 Homework 21 § 7.7 Part II 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 cos(3 θ ) = 2 2 for θ , when - π 2 θ 3 π 2 . We first let α = 3 θ to simplify the problem, so we have cos α = 2 2 = 1 2 . Then draw the triangles below, since cos is positive in the first and fourth quadrants. 2 1 1 2 -1 Those are 45-45-90 triangles, so our angles are π 4 . In the first quadrant we have α = π 4 and in the fourth quadrant we have α = - π 4 . Now the period of cos is 2 π so we write 3 θ = α = π 4 ± 2 πk and 3 θ = α = - π 4 ± 2 πk. But we want to find θ , so we divide by 3 and find that θ = π 12 ± 2 π 3 k = π 12 ± 8 π 12 and θ = - π 12 ± 2 π 3 = - π 12 ± 8 π 12 . We found a common denominator to make the next step easier. Now we apply the restrictions, we write them with a denominator of 12: - 6 π 12 θ
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Unformatted text preview: 18 π 12 . OK here goes. I’ll put boxes around the answers that work. We add and subtract 8 π 12 until we are outside the restrictions. π 12 , π 12 + 8 π 12 = 9 π 12 , 9 π 12 + 8 π 12 = 17 π 12 , 17 π 12 + 8 π 12 = 25 π 12 TOO BIG. Ok, subtract, π 12-8 π 12 =-7 π 12 TOO SMALL. Now the second value.-π 12 ,-π 12 + 8 π 12 = 7 π 12 , 7 π 12 + 8 π 12 = 15 π 12 , 15 π 12 + 8 π 12 = 23 π 12 TOO BIG. Now subtract,-π 12-8 π 12 =-9 π 12 TOO SMALL. And so we have our six answers. 2. Solve 2 sin 2 ± θ 3 + π 4 ² = 4 for θ . Well, solving for sin gives sin ( θ 3 + π 4 ) = √ 2, but √ 2 > 1 and the range of sin is [-1 , 1] so this one is impossible. Full credit for every one, this will not be on the exam....
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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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