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anonymous
 4 years ago
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
anonymous
 4 years ago
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

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Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1The reference angle is the angle the terminal side of the angle makes with the x axis.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1So let's begin. It helps to draw them

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1No. I meant 270. that's why there was 15 to go.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i see. thanks! what is the terminal side of the angle?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0OH. thank you so much! yes.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Same reference angle as the first one.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Now that we can draw them, we need to pay attention to the directions and the signs.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1285 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
 4 years ago
Best ResponseYou've already chosen the best response.1All 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: \[\cot (285)=\cot(75)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i guess i'm a little confused with the quadrant thing...

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Do you know that the cos is like x and the sin is like y?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0what else should I know?

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1that tan = y/x cot=x/y sec=1/x csc=1/y

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1Knowing that will help you find the sign of the functions in the various quadrants.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ok, thank you... i don't think I had the hang of that either.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1For 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.

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1So let's do the next one sec(105)=

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1We 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
 4 years ago
Best ResponseYou've already chosen the best response.0the secant will be negative right?

Mertsj
 4 years ago
Best ResponseYou've already chosen the best response.1yes. So we would write: \[\sec(105)=\sec(75)\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0could you write it as sec(75) or would that be wrong?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0also, 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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