solve without calculator

jennielynn_1980

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Joined
Sep 5, 2008
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7
Hi everyone! I have three of these questions and I can't quite figure it out.

It says solve without your calculator:
a) sin (-5pi/4)
the answer in this case is the square root of 2 over 2. I have no idea how to get to this answer. The only thing I can think of is that at 45[sup:275pmsji]o[/sup:275pmsji] sin is = to pi over 4 and that has something to do with the answer. A hint in the right direction would be appreciated.
These are the other two questions.
b) cos(5pi/6)
c) tan(pi/3)

Thanks again :)
Jenn
 
jennielynn_1980 said:
I have no idea how to get to this answer.
So your class hasn't covered the periodicity of trig functions or the unit circle...?

Eliz.
 
Umm, I don't think so. Maybe? This is kind of the refresher part of the course but I haven't done math for 10 years or so. I forget alot of basic math skills.
 
jennielynn_1980 said:
Hi everyone! I have three of these questions and I can't quite figure it out.

It says solve without your calculator:
a) sin (-5pi/4)
the answer in this case is the square root of 2 over 2. I have no idea how to get to this answer. The only thing I can think of is that at 45[sup:3rcc3wc4]o[/sup:3rcc3wc4] sin is = to pi over 4 and that has something to do with the answer. A hint in the right direction would be appreciated.
These are the other two questions.
b) cos(5pi/6)
c) tan(pi/3)

Thanks again :)
Jenn
You must have met these concepts:

1) Quadrant
2) Reference Angle

Yes?
 
WE briefly covered the unit circle r[sup:39x94n9h]2[/sup:39x94n9h] = (x -x[sub:39x94n9h]o[/sub:39x94n9h])[sup:39x94n9h]2[/sup:39x94n9h] + (y-y[sub:39x94n9h]0[/sub:39x94n9h])[sup:39x94n9h]2[/sup:39x94n9h]. I am guessing the quadrant and reference angles are how you define trig functions in relation to themselves as in tan = sin/cos? We did use the unit circle to identify trig identities and values but I am still unsure of how to apply this information. I have a chart of radians in relation to degrees and trig functions. But again, I am lost.
I will review the concepts and see if I can get any further.
 
jennielynn, we have a problem. You do not seem to have sufficient background. Something definitely is missing from this picture.

1) r^2 = (x -xo)^2 + (y-y0)^2 -- That's not a unit circle. It's an arbitrary circle. THE Unit Circle is this one: 1 = x^2 + y^2, but that hardly is significant for this problem.

2) Quadrant is a fundamental idea of the Cartesian Coordinate System. You must be able to identify Quadrants I, II, III, and IV, just so you can talk about things.

3) Radian to degree conversions may be beneficial, but not if you have no idea what you are seeing.

Here is a brief overview.

http://www.themathpage.com/aTrig/unit-circle.htm

Give it a read and see if it makes any sense.
 
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