anonymous
  • anonymous
How do you express trigonometric functions of degrees in terms of a reference angle? Please use the following as examples: 4a. cot 285º b. sec -105º c. csc 600º d. tan 3º 6a tan 160º b. csc 115º c. sec 235º d. cot 5º
Mathematics
katieb
  • katieb
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Mertsj
  • Mertsj
The reference angle is the angle the terminal side of the angle makes with the x axis.
Mertsj
  • Mertsj
So let's begin. It helps to draw them
Mertsj
  • Mertsj
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anonymous
  • anonymous
you mean 285º?
Mertsj
  • Mertsj
|dw:1328759118643:dw|
Mertsj
  • Mertsj
No. I meant 270. that's why there was 15 to go.
anonymous
  • anonymous
i see. thanks! what is the terminal side of the angle?
Mertsj
  • Mertsj
|dw:1328759258167:dw|
Mertsj
  • Mertsj
Ready for the next one?
anonymous
  • anonymous
OH. thank you so much! yes.
Mertsj
  • Mertsj
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Mertsj
  • Mertsj
Same reference angle as the first one.
anonymous
  • anonymous
ok.
Mertsj
  • Mertsj
Now that we can draw them, we need to pay attention to the directions and the signs.
Mertsj
  • Mertsj
285 degrees. That angle is in quadrant 4. The reference angle is 75 degrees. So the cotangent of 285 is the same as the cotangent of 75 except for the sign.
Mertsj
  • Mertsj
All functions are positive in the first quadrant and all reference angles will be first quadrant angles. However, the cotangent is negative in the fourth quadrant so we have to write: \[\cot (285)=-\cot(75)\]
Mertsj
  • Mertsj
Are you with me?
anonymous
  • anonymous
yes.
anonymous
  • anonymous
i guess i'm a little confused with the quadrant thing...
Mertsj
  • Mertsj
Do you know that the cos is like x and the sin is like y?
anonymous
  • anonymous
I didn't...
anonymous
  • anonymous
what else should I know?
Mertsj
  • Mertsj
that tan = y/x cot=x/y sec=1/x csc=1/y
Mertsj
  • Mertsj
Knowing that will help you find the sign of the functions in the various quadrants.
anonymous
  • anonymous
ok, thank you... i don't think I had the hang of that either.
Mertsj
  • Mertsj
For example: since the sec is the reciprocal of the cos, it will be positive in quadrants 1 and 4 because that is where x is positive.
anonymous
  • anonymous
I see.
Mertsj
  • Mertsj
So let's do the next one sec(-105)=
Mertsj
  • Mertsj
We have seen that -105 is a third quadrant angle. Will the secant be positive or negative in the third quadrant? Remember the sec agrees with the cos that the cos is like x
anonymous
  • anonymous
the secant will be negative right?
Mertsj
  • Mertsj
yes. So we would write: \[\sec(-105)=-\sec(75)\]
anonymous
  • anonymous
could you write it as sec(-75) or would that be wrong?
anonymous
  • anonymous
also, i just rechecked the book and 4d is actually in radians (tan=3) and 6d is also in radians (cot=5) how do we do those? I understand the degrees now... but i don't understand the radians.

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