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Calculus/Inverse Trig function?

Question
Hi! Thank you so much for doing this.

I just cannot figure out how to solve this problem
csc(tan^-1(-2))

the -2 is messing me up! thanks!

Hello Jillian,

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.

OK?

Abe

Questioner's Rating
 Rating(1-10) Knowledgeability = 10 Clarity of Response = 10 Politeness = 10 Comment that was great! THANK YOU SO MUCH. this guy is great

Calculus

Volunteer

Abe Mantell

Expertise

Hello, I am a college professor of mathematics and regularly teach all levels from elementary mathematics through differential equations, and would be happy to assist anyone with such questions!

Experience

Over 15 years teaching at the college level.

Organizations
NCTM, NYSMATYC, AMATYC, MAA, NYSUT, AFT.

Education/Credentials
B.S. in Mathematics from Rensselaer Polytechnic Institute
M.S. (and A.B.D.) in Applied Mathematics from SUNY @ Stony Brook