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

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

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).

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