How do you evaluate #cot(arcsin(-5/8)#?

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Answer 1 · 2015-04-22T01:20:42

Let #t=arcsin(-5/8)#

Then #t# is in #[- pi/2, pi/2]# and #sint = -5/8#

Find #cost# (I'm sure you have done this kind of problem by now.)

Because #-pi/2 < t < pi/2#, we know that #cost > 0#

#cost = sqrt39 / 8#

And #cot t = cost/sint =- sqrt39 / 5#

Method 2

Use #cot^2t =csc^2t -1# together with #-pi/2 < t < pi/2#, so we know that #cos t > 0# and so #cot t < 0#

#csc t = -8/5#, so #cot^2t =(-8/5)^2 -1 = 39/25# and so on