How do you evaluate # tan (arcsin (3/4))#?

1 Answer
Apr 12, 2018

#tan(arcsin(3/4)) = (3sqrt(7))/7#

Explanation:

Note that since #3/4 > 0#, the angle #theta = arcsin(3/4)# will be in Q1. So we can take it to be an acute angle of a right angled triangle.

Remember:

#sin theta = "opposite"/"hypotenuse"#

#tan theta = "opposite"/"adjacent"#

So consider a right angled triangle with hypotenuse #4#, one leg #3# and the other #sqrt(4^2-3^2) = sqrt(16-9) = sqrt(7)#. This last leg is the one adjacent to #theta#.

So:

#tan(arcsin(3/4)) = "opposite"/"adjacent" = 3/sqrt(7) = (3sqrt(7))/7#

Alternatively, using:

#cos^2 theta + sin^2 theta = 1#

we find:

#cos(theta) = +-sqrt(1-sin^2 theta) = +-sqrt(1-(3/4)^2) +-sqrt(1-9/16) = +-sqrt(7)/4#

Then since #arcsin(theta) in (-pi/2, pi/2]#, where #cos > 0#, we require the #+# sign and find:

#cos(theta) = sqrt(7)/4#

Then:

#tan(arcsin(3/4)) = tan(theta) = sin(theta)/cos(theta) = (3/4)/(sqrt(7)/4) = 3/sqrt(7) = (3sqrt(7))/7#