Solving Trigonometry Functions

Question

Rewrite the expression as an algebraic function of x and y.

sin(cos⁻¹ x − tan⁻¹ y)

Answer 1

sin(u-v) = sinucosv - sinvconu

u= cos^-1(x)

v= tan^-1(y)

sin(u-v) = sin(cos^-1(x))cos(tan^-1(y)) - sin(tan^-1(y))cos(cos^-1(cos(x))

sin(u-v) = (sqr(1-x^2))(1/(sqr(1+y^2) - [y/(sqr(1+y^2)](x)

sin(u-v) = sqr((1-x^2)/(1+y^2)) -xy/sqr(1+y^2)

sin(u-v) = [(sqr(1-x^2)-xy]/sqr(1+y^2)

Answer 2

Rewriting the cos^-1 term as cos^-1(x/1), we are looking for the angle (α) where the adjacent side is "x" and the hypotenuse is 1:

Triangle #1:

and using the Pythagorean theorem to solve for the missing side we get it as √(1 - x²)

Rewriting the tan⁻¹ term as tan^-1(y/1), we are looking for the angle (β) where the opposite side is "y" and the adjacent side is 1:

Triangle #2:

Using the Pythagorean theorem to solve for the hypotenuse we get that it is √(1 + y²)

To get the sine of the difference of two angles, we use the sine angle subtraction identity:

sin(α - β) = sin(α)cos(β) - cos(α)sin(β)

sin(α) using Triangle #1 is √(1 - x²)/1 = √(1 - x²)

cos(β) using Triangle #2 is 1/√(1 + y²)

cos(α) using Triangle #1 is x/1 = x

sin(β) using Triangle #2 is y/√(1 + y²)

Putting these together sin(α - β) =

√(1 - x²)/√(1 + y²) - xy/√(1 + y²)

The simplification of this is somewhat arbitrary. There is already a common denominator so it can easily by put into a single expression. Rationalizing the denominator is another possible simplification step.