How do you convert this into an algebraic equation?

Question

sin(arcsin u - arctan v)

Answer 1

A few keys to working through this problem:

1. The problem contains 'arcsin u' and 'arctan v'. Remember that the arcsin u is the angle which has u as its sine. Similarly, the arctan v is the angle which has v as its tangent.
2. Looking at the problem, you're going to take the difference of these two angles and then take the sine of that difference. Fortunately, there exist formulas (trigonometric identities) for the sine of the sum and difference of two angles. I googled "sin(x-y)" to find: sin(x-y) = (sin x)(cos y) - (sin y)(cos x)
3. To apply the identity, let x = arcsin u. Let y = arctan v. When you apply the identity, you will have terms like: "cos(arctan v)". So to deal with this and similar terms, drawing a small right triangle will be helpful to visualize what's going on.
4. Let's consider cos(arctan v). Pick one of the two smaller angles of the triangle. It doesn't matter which one. That angle will be arctan v. For that angle, the opposite over adjacent sides must equal to v. An easy way to do that is to let the opposite side be of length v and the adjacent side be 1, so that the ratio is v/1 = v.
5. We can now calculate the hypotenuse from the Pythagorean relationship. So h² = 1² + v². So the hypotenuse h is sqrt(1 + v²)
6. So now that we have all three sides of the triangle, we can take the cosine of our chosen angle, so that the cos(arctan v) = adjacent / hypotenuse = 1/sqrt(1 + v²).
7. The other terms will follow a similar pattern. You'll be left with an algebraic expression which does not contain any trigonometric functions. You'll likely have to simplify the algebraic expression, but that's not too tough.
8. If you see a trigonometric function of its inverse, you're in luck! For example, sin(arcsin(x)) = x. In words, that means the sine of an angle whose sine is x is just x itself. That's the idea of the sine and arcsin functions being inverses to each other.

Answer 2

Let α be the angle whose sine = u, and let the hypotenuse = 1. The adjacent side of this triangle A is √(1 - u²).

Let β be the angle whose tangent = v, and let the adjacent side = 1. The hypotenuse of this triangle B is √(1 + v²).

Use the difference identity sin(α-β) = sin α cos β - cos α sin β, to get sin(α-β) = (u/1)[1/√(1+v²)] - [√(1-u²)/1][v/√(1+v²)]. Then sin(arcsin u - arctan v) = [u - v√(1-u²)] / √(1+v²).

Answer 3

arctanx is an angle whose tangent function =x/1

Considering the sides of the right triangle. We have opposite side =x, adjacent side =1 and hypotenuse =√x2+1

Therefore the sine of this angle =opposite side hypotenuse=x/√x^2+1