At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
The reference angle is the angle the terminal side of the angle makes with the x axis.
So let's begin. It helps to draw them
you mean 285º?
No. I meant 270. that's why there was 15 to go.
i see. thanks! what is the terminal side of the angle?
Ready for the next one?
OH. thank you so much! yes.
Same reference angle as the first one.
Now that we can draw them, we need to pay attention to the directions and the signs.
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.
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: \[\cot (285)=-\cot(75)\]
Are you with me?
i guess i'm a little confused with the quadrant thing...
Do you know that the cos is like x and the sin is like y?
what else should I know?
that tan = y/x cot=x/y sec=1/x csc=1/y
Knowing that will help you find the sign of the functions in the various quadrants.
ok, thank you... i don't think I had the hang of that either.
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.
So let's do the next one sec(-105)=
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
the secant will be negative right?
yes. So we would write: \[\sec(-105)=-\sec(75)\]
could you write it as sec(-75) or would that be wrong?
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.