tan(sin^-1(2/5))

sin^-1(cot(11pi/4))

what are you trying to do? solve for the value?

on the first one, work inside out, start with a triangle. remember sin^1 is asking "the angle whose sine is 2/5" and remember sine is opposite/hypotenuse. label the two sides accordingly (2 and 5). Use pythagorean theorem to find the adjacent side to the angle, and use soh cah toa to find tangent of said angle. computing the angle is unnecessary

as for the second, remember 11pi/4 is the same as 3pi/4 (because 8pi/4 would be one lap around the circle). That's a pi/4 reference angle in the 3rd quadrant. The cot value of 3pi/4 then is -1. I trust you can evaluate that on your own. Inverse sine of -1 is asking "Where is sine = to -1?" Well that's 3pi/2.

I'm thinking quick so double check for clumsiness

This post was edited by Derkaderk on Jul 24 2013 08:25pm

tan (arcsine (2/5))
Ѳ = arcsine (2/5) {just a substitution}
sin Ѳ = 2/5 {given in problem}

(sin Ѳ)^2 + (cos Ѳ)^2 =1 {trig identity/common knowledge}
(cos Ѳ)^2 = 1 - (sin Ѳ)^2 {just rewriting}
(cos Ѳ)^2 = 1 - 4/25 {substitution}
(cos Ѳ)^2 = 21/25 {simplification}
cos Ѳ = [sqrt 21]/5 {square root of both sides}
sec Ѳ = 5/ [sqrt 21] {identity}

(tan Ѳ)^2 + 1 = (sec Ѳ)^2 {trig identity}
(tan Ѳ)^2 = (sec Ѳ)^2 - 1 {rewrite}
(tan Ѳ)^2 = 25/21 - 1 {substitution}
(tan Ѳ)^2 = 4/21 {simplify}
tan Ѳ = 2/[sqrt 21] {square root both sides}


I didn't consider the quadrants and I didn't rationalize the denominator, but this is how it's done; more or less.

This post was edited by Azrad on Jul 24 2013 08:46pm

what is this mojo with identities? seems unnecessary

heh well your method uses them too, you just aren't labeling them explicitly.

oh that's crafty. guess i never invested a lot of thought into the meanings and proofs of the identities

Hey, your method gets the job done, and that is all that matters.

oh this one is much easier, just realize that:
cot(11pi/4) = cot(11pi/4 - 2pi) = cot(3pi/4) = 1/[tan (3pi/4)] = -1/[tan (pi/4)] = -1/1 = -1
should be easy from there!

This post was edited by Azrad on Jul 25 2013 11:39am