MA 1165 - Lecture 16 10/24/07
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Circles
x2 + y 2 = r 2 . (1)
A circle with radius r centered at the origin will have equation
For example, the circle with radius r = 2 is shown in Figure 1.
Figure 1: A circle with radius r = 2 centered at the origin. In
MA 1165 - Lecture 18 3/11/09
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Relationships between trig functions
Weve talked about all the trig functions in terms of a right triangle, and the sine, cosine, and tangent functions in terms of the unit circle. Figure 1 shows the unit circle and a righ
MA 1165 - Lecture 20 3/23/09
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The Law of Cosines
c2 = a 2 + b 2 ,
For right triangles, one of the basic formulas we've used is the Pythagorean theorem. (1)
where c is the length of the hypotenuse, the side opposite the right angle. If we look at a gene
MA 1165 - Lecture 19 3/13/09
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Graphs of the sine and cosine function
I think that most of what you need to know about the trig functions outside of your calculator is contained in the graphs for the trig functions. Well look at the sine function and co
MA 1165 - Lecture 23
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4/3/2009 These are the identities we've had so far. Expressing the other trig functions in terms of sine and cosine. tan() = sin() cos() sec() = 1 cos() (1) cos() cot() = sin() Relationships between squares of the trig functions. co
MA 1165 - Lecture 24 4/8/09
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Imaginary Numbers
One basic fact is that there is no real number that is the square root of -1. As a result, the quadratic equation x2 + 1 = 0 (1) has no real number solutions. This is a shortcoming of the real numbers, sin
MA 1165 - Practice Test II
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3/16/09 (Test II is Wednesday, 3/18/09) Use one of our log identities to change the form of the expresssion. 1. 2. 3. 4. 5. 6. 7. ln(7) ln(x). log2
x 7
.
ln(2) ln(x). x log3 (2). ln(x) + ln(2). log
2 x
.
ln(x) ln(3).
Solve the
MA 1165 - Practice Test III (Answers corrected) 4/13/09 (Test III is Wednesday, 4/15/09) For problems 1-5, nd all solutions between 0 and 360. 1. 2. 3. 4. 5. sin() = .75. sin() = .75. cos() = .75. cos() = .75. cos() =
3 2 .
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For problems 6-9, use the la