Question

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

Answer 1

`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`