Expert: Chen Min - 4/4/2008

Question
Sin [csc^-1 (1/7x-8)]
sin [cot^-1 (1/x+8)]

How to do those? I know how to deal with sin[csc^-1 (1/2)] but with a variable I cant tell how to start this. If you can illustrate work please do so.

Answer
Hello! I saw your question in the pool and I think I may offer some help

For the first part,you must know sinA * cscA = 1,
Then assume that cscA = 1/7x - 8
Thus csc^-1( 1/7x - 8 ) = A
Sin [csc^-1 (1/7x-8)], therefore, is equal to sinA,
and sinA = 1/cscA = 7 / ( x - 56 )

For the second one, as sin and cot are not much related, you may need to construct a traiangle (The most direct method,anyway...)

Assume cotB = 1/x + 8
then sinB = 1/[( 1/x+8 )^2 + 1]^0.5

Is this the answer you desire?

Hope this helps