Question

# Write the trigonometric expression cos(sin^(-1) x - cos^(-1) y) as an algebraic expression (that is, without any trigonometric functions). Assume that x and y are positive and in the domain of the given inverse trigonometric function.

Trigonometric Functions
Write the trigonometric expression $$\displaystyle{\cos{{\left({{\sin}^{{-{1}}}{x}}-{{\cos}^{{-{1}}}{y}}\right)}}}$$ as an algebraic expression (that is, without any trigonometric functions). Assume that x and y are positive and in the domain of the given inverse trigonometric function.

2020-11-27
The difference between the two angles based on the cosine terms is given by,
$$\displaystyle{\cos{{\left(\alpha-\beta\right)}}}={\cos{\alpha}}{\cos{\beta}}+{\sin{\alpha}}{\sin{\beta}}$$
Substituting the values of alpha and beta in the formula,
$$\displaystyle{\cos{{\left(\alpha-\beta\right)}}}={\cos{\alpha}}{\cos{\beta}}+{\sin{\alpha}}{\sin{\beta}}$$
$$\displaystyle{\cos{{\left({{\sin}^{{-{1}}}{x}}-{{\cos}^{{-{1}}}{y}}\right)}}}={\cos{{\left({{\sin}^{{-{{1}}}}{x}}\right)}}}{\cos{{\left({{\cos}^{{-{1}}}{y}}\right)}}}+{\sin{{\left({{\sin}^{{-{1}}}{x}}\right)}}}{\sin{{\left({{\cos}^{{-{1}}}{y}}\right)}}}$$
Hence,
$$\displaystyle{\sin{{\left({{\sin}^{{-{1}}}{x}}\right)}}}={x}{\quad\text{and}\quad}{\cos{{\left({{\cos}^{{-{1}}}{y}}\right)}}}={y}$$
On simplifying,
$$\displaystyle{\cos{{\left({{\sin}^{{-{1}}}{x}}\right)}}}{\cos{{\left({{\cos}^{{-{1}}}{y}}\right)}}}{\sin{{\left({{\sin}^{{-{1}}}{x}}\right)}}}{\cos{{\left({{\cos}^{{-{1}}}{y}}\right)}}}$$
On applying the Pythagorean identities as shown below,
$$\displaystyle{\cos{{\left({{\sin}^{{-{1}}}{x}}\right)}}}{\quad\text{and}\quad}{\sin{{\left({{\cos}^{{-{1}}}{y}}\right)}}}$$
$$\displaystyle{\cos{{\left({{\sin}^{{-{1}}}{x}}\right)}}}{y}+{x}{\sin{{\left({{\cos}^{{-{1}}}{y}}\right)}}}=\sqrt{{{1}-{{\sin}^{{2}}{\left({{\sin}^{{-{1}}}{x}}\right)}}}}{y}+{x}\sqrt{{{1}-{{\cos}^{{2}}{\left({{\cos}^{{-{1}}}{y}}\right)}}}}$$
On simplification the equation obtained is,
$$\displaystyle\sqrt{{{1}-{{\sin}^{{2}}{\left({{\sin}^{{-{1}}}{x}}\right)}}}}{y}+{x}\sqrt{{{1}-{{\cos}^{{2}}{\left({{\cos}^{{-{1}}}{y}}\right)}}}}=\sqrt{{{1}-{x}^{{2}}}}{y}+{x}\sqrt{{{1}-{y}^{{2}}}}$$
$$\displaystyle={y}\sqrt{{{1}-{x}^{{2}}}}+{x}\sqrt{{{1}-{y}^{{2}}}}$$
Hence,the simplified expression is $$\displaystyle{y}\sqrt{{{1}-{x}^{{2}}}}+{x}\sqrt{{{1}-{y}^{{2}}}}$$