Question

In a right triangle \( ABC \), right angled at \( B \), prove that \( \sin A = \cos (90^\circ - A) \) and \( \cos A = \sin (90^\circ - A) \).

Hint:

Remember: In a right triangle, the two acute angles are complementary. So, sine of one acute angle is equal to cosine of the other, and cosine of one acute angle is equal to sine of the other.

Answer 1

Step 1: Write the given condition of the triangle.}
In \( \triangle ABC \), it is given that \( \angle B = 90^\circ \). Since the sum of all angles of a triangle is \( 180^\circ \), we get \[ \angle A + \angle C + 90^\circ = 180^\circ \] So, \[ \angle A + \angle C = 90^\circ \] Hence, \[ \angle C = 90^\circ - \angle A \]
Step 2: Write \( \sin A \) in terms of sides.}
In a right triangle, \[ \sin A = \frac{\text{Perpendicular}}{\text{Hypotenuse}} = \frac{BC}{AC} \]
Step 3: Write \( \cos (90^\circ - A) \) using angle \( C \).}
Since \[ C = 90^\circ - A \] we have \[ \cos (90^\circ - A) = \cos C \] Now, \[ \cos C = \frac{\text{Adjacent side to } C}{\text{Hypotenuse}} = \frac{BC}{AC} \]
Step 4: Compare both values.}
From above, \[ \sin A = \frac{BC}{AC} \] and \[ \cos (90^\circ - A) = \frac{BC}{AC} \] Therefore, \[ \sin A = \cos (90^\circ - A) \]
Step 5: Prove the second identity.}
Now, \[ \cos A = \frac{\text{Base}}{\text{Hypotenuse}} = \frac{AB}{AC} \] Also, since \( C = 90^\circ - A \), \[ \sin (90^\circ - A) = \sin C \] and \[ \sin C = \frac{\text{Perpendicular to } C}{\text{Hypotenuse}} = \frac{AB}{AC} \] Thus, \[ \cos A = \sin (90^\circ - A) \]
Step 6: Write the conclusion.}
Hence proved, \[ \boxed{\sin A = \cos (90^\circ - A)} \] and \[ \boxed{\cos A = \sin (90^\circ - A)} \]