Sunday
February 7, 2016

# Homework Help: Evaluation-repost

Posted by Jen on Monday, October 23, 2006 at 8:17am.

Reposting because it has gone down to the second page!

Evaluate

1. sin(tan inverse sqrt(x^2-2x))

2. tan (sec inverse 3y)

For 1. draw a right triangle and label the opposite leg sqrt(x^2-2x), the adjacent side 1 and the hypotenuse x-1. Do the calculations to see why this is so. Then the sin of that angle is
sqrt(x^2-2x)/(x-1)
For 2. label the hypotenuse 3y and the adjacent leg 1. The opposite leg is sqrt(9y^2 - 1) so the tan is sqrt(9y^2 - 1).

I didn't understand why we labelled so. :(
How is the adjacent side 1? I know we can find the third side if we have the other two sides. Also, what is the reason for labelling opposite leg sqrt(x^2-2x)?

When you have an inverse trig function you should first draw a right triangle. Every trig function is the ratio of two sides of the triangle, so you need to identify which side is the numerator and which is the denominator. Then you typically need to calculate the 3rd side from those expressions.
Here's an example:
If you have inverse sin x then you know that the sin is the ratio of the opposite to the hypotenuse. So you would draw a right triangle and label the opposite x and the hypotenuse 1. The 3rd leg, the adjacent one here, is sqrt(x^2 - 1). If you now wanted the tan of that angle it'd be x/sqrt(x^2 - 1). If you wanted the cos of that angle it'd be sqrt(x^2 - 1)/1 or just sqrt(x^2 - 1).
Does this help?