In right triangle ABC, the measure of angle C is 90^\circ. If \sin(A) - \cos(A) = \frac{1}… — SAT

In right triangle $ABC$ , the measure of angle $C$ is $90^\circ$ . If $\sin(A) - \cos(A) = \frac{1}{5}$ , what is the value of $\sin(A)\sin(B)$ ?

\frac{12}{25}

By the co-function identity for complementary angles in a right triangle,

\sin(B) = \cos(A)

, which means the expression

\sin(A)\sin(B)

is equivalent to

\sin(A)\cos(A)

. Squaring the equation

\sin(A) - \cos(A) = \frac{1}{5}

gives

\sin^2(A) - 2\sin(A)\cos(A) + \cos^2(A) = \frac{1}{25}

. Substituting the Pythagorean identity

\sin^2(A) + \cos^2(A) = 1

simplifies this to

1 - 2\sin(A)\cos(A) = \frac{1}{25}

. Solving for

\sin(A)\cos(A)

yields

2\sin(A)\cos(A) = \frac{24}{25}

, or

\sin(A)\cos(A) = \frac{12}{25}

Express the target expression in terms of a single angle using complementary relationships.

Since angle

C = 90^\circ

, angles

A

and

B

are complementary, so

\sin(B) = \cos(A)

. Thus,

\sin(A)\sin(B) = \sin(A)\cos(A)

In any right triangle, the sine of one acute angle equals the cosine of the other acute angle.

Square both sides of the given equation

\sin(A) - \cos(A) = \frac{1}{5}

(\sin(A) - \cos(A))^2 = \left(\frac{1}{5}\right)^2 \implies \sin^2(A) - 2\sin(A)\cos(A) + \cos^2(A) = \frac{1}{25}

Squaring the difference allows us to introduce the product term

\sin(A)\cos(A)

and the squared terms.

Apply the Pythagorean identity to simplify the equation.

Since

\sin^2(A) + \cos^2(A) = 1

, the equation becomes

1 - 2\sin(A)\cos(A) = \frac{1}{25}

The sum of the squares of sine and cosine of the same angle is always equal to 1.

Solve for the product

\sin(A)\cos(A)

2\sin(A)\cos(A) = 1 - \frac{1}{25} = \frac{24}{25} \implies \sin(A)\cos(A) = \frac{12}{25}

Subtracting

1/25

from

1

and dividing the resulting fraction by

2

isolates the target product.

Pythagorean identity and co-function relationships in right triangles

Question