This preview shows page 1. Sign up to view the full content.
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 128 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 128 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....
View
Full
Document
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

Click to edit the document details