Question

In \( \triangle ABC \), \( \angle B = 90^\circ \), \( BC = 6 \, \text{cm} \), and \( AC = 10 \, \text{cm} \). What is the value of \( (\sin 2A - \cos 2A) \)?

• A. \( \frac{17}{25} \)
• B. \( \frac{9}{25} \)
• C. \( \frac{2}{5} \)
• D. \( \frac{1}{5} \)

Hint:

For double angle identities, remember that \( \sin 2A = 2 \sin A \cos A \) and \( \cos 2A = \cos^2 A - \sin^2 A \). These can simplify the computation for angle-related questions.

Answer 1

The Correct Option is A

We are given that \( \triangle ABC \) is a right-angled triangle with \( \angle B = 90^\circ \), \( BC = 6 \, \text{cm} \), and \( AC = 10 \, \text{cm} \). To find \( \sin 2A - \cos 2A \), we first need to find \( \sin A \) and \( \cos A \).
Using Pythagoras' theorem: \[ AB^2 = AC^2 - BC^2 = 10^2 - 6^2 = 100 - 36 = 64 \quad \Rightarrow \quad AB = 8 \, \text{cm}. \] Now, we have: \[ \sin A = \frac{BC}{AC} = \frac{6}{10} = 0.6, \quad \cos A = \frac{AB}{AC} = \frac{8}{10} = 0.8. \] Next, use the double angle formulas: \[ \sin 2A = 2 \sin A \cos A = 2 \times 0.6 \times 0.8 = 0.96, \] \[ \cos 2A = \cos^2 A - \sin^2 A = 0.8^2 - 0.6^2 = 0.64 - 0.36 = 0.28. \] Thus, \[ \sin 2A - \cos 2A = 0.96 - 0.28 = 0.68 = \frac{17}{25}. \]