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)?
Please clarify? Thanks. :)
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?
Yes, that helps!
In the given expression, sin(tan inverse sqrt(x^2-2x)), we have the inverse of the tangent function (tan inverse) with an argument of √(x^2 - 2x).
To understand why we label the sides of the right triangle as mentioned, let's break it down step by step:
1. We start by drawing a right triangle. The angle whose tangent we are taking the inverse of will be one of the acute angles in the triangle.
2. We label the opposite leg, adjacent side, and hypotenuse of the triangle. Here's how:
- The opposite leg represents the side opposite to the angle of interest. In this case, it is labeled as √(x^2 - 2x).
- The adjacent side represents the side adjacent to the angle of interest. Since we don't have any specific information about it, we label it as 1 (for simplicity).
- The hypotenuse is the side opposite the right angle, which is typically not labeled in these problems. We label it as x - 1 to represent that side.
3. With the given labels, we can now calculate the actual lengths of the sides of the triangle using the Pythagorean theorem.
- The Pythagorean theorem states that for a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
- In our case, we have (x - 1)^2 = (√(x^2 - 2x))^2 + 1^2. Solving this equation will give us the value of the hypotenuse.
Once we have the lengths of the sides of the triangle, we can substitute them into the trigonometric expression sin(tan inverse sqrt(x^2-2x)) to get:
sin(tan inverse sqrt(x^2-2x)) = √(x^2 - 2x)/(x - 1)
Similarly, for the second expression tan(sec inverse 3y), we follow the same process of labeling the sides of the right triangle based on the inverse functions given.
I hope this explanation clarifies the reasoning behind labeling the sides of the right triangles. Let me know if you have any further questions!