Calculus/Inverse Trig function?
Hi! Thank you so much for doing this.
I just cannot figure out how to solve this problem
the -2 is messing me up! thanks!
I will use "arctan" instead of "tan^-1" since it is easier to read and type.
Thus, the problem is: csc(arctan(-2))
arctan(-2) is an angle...also, since arctan(x) is between -pi/2 and +pi/2,
i.e. in quadrant I or IV, we know arctan(-2) must be an angle in quadrant IV (where tan<0).
Now we can just focus on arctan(2)...let A=arctan(2), thus tan(A)=2...we can construct
a right triangle from this, since tan(A)=2/1=opposite/adjacent...so, we now have a right
triangle with sides 1, 2, and sqrt(5) - from the Pythagorean Thrm.
Now all we need is csc(A), which is 1/sin(A), sin(A)=opposite/hypotenuse=2/sqrt(5)
But, angle A is in quadrant IV, so csc(A) is negative.
Answer: csc(arctan(-2))=-sqrt(5)/2 or about -1.118