Given that \cos x - \sin x = \frac{1}{\sqrt{5}} for an acute angle x, what is the exact va… — JAMB UTME

Given that $\cos x - \sin x = \frac{1}{\sqrt{5}}$ for an acute angle $x$ , what is the exact value of $\cos^3 x - \sin^3 x$ ?

\frac{7\sqrt{5}}{25}

Squaring

\cos x - \sin x = \frac{1}{\sqrt{5}}

yields

1 - 2\sin x \cos x = \frac{1}{5}

, which gives

\sin x \cos x = \frac{2}{5}

. Using the difference of cubes factorization,

\cos^3 x - \sin^3 x = (\cos x - \sin x)(1 + \sin x \cos x) = \frac{1}{\sqrt{5}} \left(1 + \frac{2}{5}\right) = \frac{7}{5\sqrt{5}}

. Rationalizing the denominator produces

\frac{7\sqrt{5}}{25}

Square both sides of the given equation to find the product

\sin x \cos x

(\cos x - \sin x)^2 = \left(\frac{1}{\sqrt{5}}\right)^2 \implies \cos^2 x - 2\sin x \cos x + \sin^2 x = \frac{1}{5}

Squaring allows us to use the Pythagorean identity

\sin^2 x + \cos^2 x = 1

to isolate

\sin x \cos x

Simplify using

\cos^2 x + \sin^2 x = 1

to solve for

\sin x \cos x

1 - 2\sin x \cos x = \frac{1}{5} \implies 2\sin x \cos x = 1 - \frac{1}{5} = \frac{4}{5} \implies \sin x \cos x = \frac{2}{5}

Finding the product of

\sin x

and

\cos x

is necessary for the algebraic expansion of the difference of cubes.

Apply the difference of cubes algebraic identity to

\cos^3 x - \sin^3 x

\cos^3 x - \sin^3 x = (\cos x - \sin x)(\cos^2 x + \sin x \cos x + \sin^2 x) = (\cos x - \sin x)(1 + \sin x \cos x)

The identity

a^3 - b^3 = (a - b)(a^2 + ab + b^2)

breaks the target expression down into known terms.

Substitute the known values into the expression and rationalize the denominator.

\left(\frac{1}{\sqrt{5}}\right) \left(1 + \frac{2}{5}\right) = \frac{1}{\sqrt{5}} \times \frac{7}{5} = \frac{7}{5\sqrt{5}} = \frac{7\sqrt{5}}{25}

Evaluating the product and rationalizing

\frac{7}{5\sqrt{5}}

yields the final exact surd form.

Basic Trigonometric Ratios, Special Angles, and Identities

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