Question

What is sin(inverse tangent(12/5)?

Answer 1

`sin(arctan(12/5))=+-12/13`

Explanation:

`arctan(12/5)` implies one of the reference triangles pictured below
(with the hypotenuse caculated using the Pythagorean Theorem).

Since `s=("opposite")/("hypotenuse")`

`sin(arctan(12/5)) = color(red)(12/13) " or " color(blue)(-12/13)`

Answer 2

`sin(arctan(12/5)) = 12/13`

Explanation:

From the trig identity `sin^2(theta) + cos^2(theta) = 1`, we divide both sides by `sin^2(theta)`

`1 + cos^2(theta)/sin^2(theta) = 1/sin^2(theta)`

Since `sin(theta)/cos(theta) = tan(theta)` we can rewrite the second term

`1 + 1/tan^2(theta) = 1/sin^2(theta)`

Taking the least common multiple,

`(tan^2(theta) + 1)/(tan^2(theta)) = 1/sin^2(theta)`

Inverting both sides

`tan^2(theta)/(tan^2(theta) + 1) = sin^2(theta)`

Subsituting `theta = arctan(12/5)`

`(12/5)^2/((12/5)^2+1) = sin^2(arctan(12/5))`

`sin^2(arctan(12/5)) = 144/25 * 25/169 = 144/169`

Taking the root

`sin(arctan(12/5)) = +-12/13`

To pick the sign we look at the range of the arctangent, it only takes arguments on the first and fourth quadrants, during which the cosine is always positive. If, for `12/5` the cosine is positive and the tangent is positive, then the sine must be positive too.

`sin(arctan(12/5)) = 12/13`

Also, protip, you can use either the function with a "^-1" or put an arc before it to notate the inverse trig functions, but usually there's a lot less headache for everybody involved if you use the arc notation. (There's no grounds for mistaking it for other functions).