Question 424858
sin(tan<sup>-1</sup>(13x))
This expression says "the sine of the angle whose tangent is 13x. To understand how we get an answer to your problem it will help greatly if we have a diagram to look at. So<ol><li>Draw a right triangle. (It doesn't matter what it looks like. Any right triangle will do.)</li><li>Pick one of the acute angles and label it A. (It doesn't matter which angle or how big or small the angle looks.)</li><li>Label the side opposite angle A as 13x.</li><li>Label the side adjacent to angle A as 1.</li><li>Label the hypotenuse as h.</li></ol>Knowing that tan is {{{opposite/adjacent}}} we should be able to see that tan(A) = {{{13x/1}}} or just 13x. So A is tan<sup>-1</sup>(13x). Since sin is {{{opposite/hypotenuse}}}, sin(A) = {{{(13x)/h}}}. Now we just need to find the hypotenuse. We can find the hypotenuse, in terms of x, using the Pythagorean Theorem:
{{{h^2 = (13x)^2 + (1)^2}}}
Solving...
{{{h^2 = 169x^2 + 1}}}
{{{h = sqrt(169x^2 + 1)}}}
So now
sin(tan<sup>-1</sup>(13x)) = sin(A) = {{{(13x)/sqrt(169x^2 + 1)}}}
This may be an acceptable answer. But often answers with square roots in the denominator are not acceptable. So we will go ahead and rationalize the denominator:
sin(tan<sup>-1</sup>(13x)) = sin(A) = {{{(13x)/sqrt(169x^2 + 1) = ((13x)/sqrt(169x^2 + 1))(sqrt(169x^2 + 1)/sqrt(169x^2 + 1)) = (13x*sqrt(169x^2 + 1))/(169x^2 + 1)}}}