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

2 Answers
Sep 30, 2015

#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).

enter image source here

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

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

Sep 30, 2015

#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).