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sec and cot of an angle resulting in different signs for the answer

Posted on 2011-03-13
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I've got a question which gives the sec and cot of an angle (theta).

On working out the angle for each expression, the answer is the same for both, but the sign is different (i.e. one positive and one negative).

Both answers fall within the range given for theta (greater than negative pi, less than pi).

What, then, is the "correct" answer?
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Question by:purplesoup
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7 Comments
 
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Expert Comment

by:aburr
ID: 35125060
I have not the faintest idea. What is the question?
-
"On working out the angle for each expression"
What is the expression on which you are working?
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LVL 37

Expert Comment

by:TommySzalapski
ID: 35125425
As you well know, there are an infinite number of angles that have the same sine/cosine/secant/tangent/etc. Obviously -2pi, 0, 2pi, 4pi, etc are all the same for all of them. But there are others that match as well.
If you look at the secant graph (and the others) which you can find here:
http://home.windstream.net/okrebs/page74.html
You will see that +x and -x will always have the same secant, no matter what x is. Even if you are restricted to the range of -pi to pi there are still duplicates.
But if you know the sign of the cotangent as well, then you know which it is. See here:
http://www.mathwizz.com/algebra/help/help29.htm
Since sec = 1/cos they always have the same sign. Likewise tan = 1/cot so they have the same sign.
You can use the figure in the link to see which quadrant the angle came from and thus find the original angle.
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LVL 37

Expert Comment

by:TommySzalapski
ID: 35125431
So the correct answer is the one that falls in the correct quadrant given the signs of the cos and tan functions.
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Author Comment

by:purplesoup
ID: 35129744
ok I think I've got it - please confirm.

The two values are

sec (theta) = 2^(1/2)
cot (theta) = -1

Where pi < theta < pi

For sec this gives theta as pi/4
For cot this gives theta as 7pi/8

Since given the range of theta, theta must be pi/4.

Correct?

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LVL 18

Expert Comment

by:deighton
ID: 35130484
sec (theta) = 2^(1/2)
cos (theta) = 1/ 2^(1/2)

this is a standard provable by pythagoras
an by considering the unit circle

theta = pi/4
theta = -pi/4

the latter is equivalent to 2pi - pi/4 = 7pi/4

cot (theta) = -1
cos(theta) / sin (theta) = -1

cos(theta) = -sin(theta)

by considering the unit circle, this occurs at -pi/4 and and 3pi/4  (in the second and 4th quadrants where 1 is negative and the other positive)
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Expert Comment

by:ozo
ID: 35130893
sec (theta) = 2^(1/2)
cot (theta) = -1

Where pi < theta < pi

assuming you mean -pi < theta < pi
sec(-pi/4) = 2^(1/2)
cot(-pi/4) = -1
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Accepted Solution

by:
TommySzalapski earned 500 total points
ID: 35135056
You have a positive sec and a negative cot.
This means you have a positive cos and negative tan.
This means theta is in quadrant IV which means it is in the range of -pi/2 to 0. Your options are pi/4 and 7pi/4 but these don't fit in your range. However, you know you can always add or subtract 2pi and all the trig functions will be the same. pi/4 - 2pi = -7/4pi which is still outside the known range.
7pi/4 - 2pi = -pi/4 which is in the range and must be the answer.
Remember, any time they say sec(theta) = x, then there are an infinite number of values for theta that will work. You just have to find the one in the correct range.
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