+0

Need Help Finding Angles

+2
838
3
+1006

Going by a unit circle, how do $$\underset{\,\,\,\,^{{360^\circ}}}{{cos}}^{\!\!\mathtt{-1}}{\left({\mathtt{\,-\,}}{\frac{{\sqrt{{\mathtt{2}}}}}{{\mathtt{2}}}}\right)}$$ and $$\underset{\,\,\,\,^{{360^\circ}}}{{tan}}^{\!\!\mathtt{-1}}{\left({\frac{{\sqrt{{\mathtt{3}}}}}{{\mathtt{3}}}}\right)}$$ get their opposite angles? I know that the primary angles are 135 deg. and 30 deg. respectively, but there is a second angle, and I am not sure how to find them. I know how to find asin opposite angles already. Also, just need clarification, but is cosecant just an inverse of sine?

And finally, how do you deal with a negative angle? I got one while plugging in some numbers.

May 13, 2014

#3
+114818
+5

I knew your answer would  be already in Chris - Now how did I know that!

May 13, 2014

#1
+121063
+5

OK...let's look at the first one....notice that it's a 45 degree reference angle in the 2nd quadrant. Now, let's ask ourselves....where else is the cosine negative??...yup...in the 3rd quadrant

So this angle is a 45 degree reference angle in the 3rd quadrant, i.e., 225 degrees.

As for the second one, it's a little trickier......the tangent is positive in the 1st quadrant and in the 3rd quadrant, because, in the 3rd quadrant, both x and y are negative. So this angle is a 30 degree reference angle in quadrant 3....... i.e., 210 degrees.

Cosecant is the reciprocal of the sine....not the inverse!!  If the sine is defined as y/r, then the csc is just r/y.

Remember that the basic functions return angle values. The inverse functions return the angles themselves.

As for dealing with negative angles.......let me know what you're specifically having trouble with and I'll see what I can do.

I hope I've answered what you've asked....

May 13, 2014
#2
+114818
+5

I am sure other mathematicians will look at what i am telling you but

For inverse trig there is no second angle - at least not if the inverse is a part of the question, as opposed to something your have introduced as a part of an answer.

I'll show you some graphs if i get up to it but

$$0\le acos(ratio)\le 180^0\\ -90^0\le atan(ratio) \le 90^0\\ -90^0\le asin(ratio) \le 90^0$$

Here is the graphs - remember,

90degrees is pi/2 radians.

180 degrees is pi radians.

now an anle of -30 degress is in the 4th quadrant.  If you needed the equivalent angle between 0 and 360 it would  be 360-30=330degrees.

I am not really sure how the unit circle ties into your question.

Have i answered you to your satisfaction?  You might need to think about it first.

May 13, 2014
#3
+114818
+5