Trig (inverse functions)
posted by Dennis .
tan[arccsc(-5/3) + arctan(1/4)]
let arccsc(-5/3)=X and let arctan(1/4)=Y
where -pi/2<=X<=pi/2, X cannot be 0
and where -pi/2<Y<pi/2
so that cscX=-5/3 and tanY=1/4
The problem can now be written as
=tanX + tanY/1-tanXtanY which could also be written as
=sinXcosY + cosXsinY/cosXcosY - sinXsinY
I drew the respective triangles and came up with (for the X triangle) a=4, b=-3, c=5. (for the Y triangle) a=1, b=4,c=sqrt17
so sinX=-3/5 cosX=4/5
=[(-3/5)(1/sqrt17) + (4/5)(4/sqrt17)]/[(4/5)(1/sqrt17)-(-3/5)(4/sqrt17)]
This is not the correct answer. The book shows an answer of -8/19
Can someone please tell me where I went wrong. You don't have to give me the whole working, just where I started to go wrong.
Thank you so much. (hopefully all that typed stuff makes sense, it's hard without being able to use mathematical characters)
Trig (inverse functions) -
I found my mistake, please disregard.