Download Algebra/Trig Review (2006)(en)(98s) by Dawkins P. PDF

By Dawkins P.

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Sample text

Recall that −1 ≤ cos ( x ) ≤ 1 . So, 1 divided by something less than 1 will be greater than 1. Also, 1 secant. R sec (ω x ) ≥ R ±1 = ±1 and so we get the following ranges out of and R sec (ω x ) ≤ − R 8. y = csc ( x ) Solution 1   For this graph we will have to avoid x’s where sine is zero  csc x =  . So, the sin x   graph of cosecant will not exist for x =  , −2π , −π , 0, π , 2π , Here is the graph of cosecant. Cosecant will have the same range as secant. R csc (ω x ) ≥ R and R csc (ω x ) ≤ − R 9.

Aspx Algebra/Trig Review 4 x + 5 < −3 4 x < −8 x < −2 or 4x + 5 > 3 4 x > −2 1 x>− 2 1 So the solution to this inequality will be x’s that are less than -2 or greater than − . 2 Now, as I mentioned earlier you CAN NOT write the solution as the following double inequality. 1 −2 > x > − 2 When you write a double inequality (as we have here) you are saying that x will be a number that will simultaneously satisfy both parts of the inequality. In other words, in writing this I’m saying that x is some number that is less than -2 and AT THE 1 SAME TIME is greater than − .

From quick inspection we can see that t = π is a solution. However, as I have shown 6 on the unit circle there is another angle which will also be a solution. aspx Algebra/Trig Review determine what this angle is. When we look for these angles we typically want positive angles that lie between 0 and 2π . This angle will not be the only possibility of course, but by convention we typically look for angles that meet these conditions. To find this angle for this problem all we need to do is use a little geometry.

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