How do you write sin(arctan(a)) as an expression of "a" not involving any trigonometric functions or their inverses?

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Answer 1 · 2015-06-23T21:31:44

Let #alpha = arctan(a)#.

Then #-pi/2 < alpha < pi/2#, and #cos alpha > 0#.

So #cos alpha = sqrt(1-sin^2(alpha))#

Hence #sin(alpha) = a/sqrt(1+a^2)#

Having let #alpha = arctan(a)#, we have:

#a = tan(alpha) = sin(alpha)/cos(alpha)#

#=sin(alpha)/sqrt(1-sin^2(alpha))#

Multiply both ends by #sqrt(1-sin^2(alpha))# to get:

#a sqrt(1-sin^2(alpha)) = sin(alpha)#

Square both sides to get:

#sin^2(alpha) = a^2(1-sin^2(alpha))#

#=a^2-a^2 sin^2(alpha)#

Note that this will introduce a spurious solution to be eliminated later.

Add #a^2 sin^2(alpha)# to both ends to get:

#(1+a^2)sin^2(alpha) = a^2#

Divide both sides by #(1+a^2)# to get:

#sin^2(alpha) = a^2/(1+a^2)#

So #sin(alpha) = +-a/sqrt(1+a^2)#

Now remember that #-pi/2 < alpha < pi/2#, so #cos alpha > 0#.

Therefore #sin(alpha)# must have the same sign as #tan(alpha) = a#

So the correct solution is #sin(alpha) = a/sqrt(1+a^2)#