(b) $$ \frac{\sin A}{\sin B}-\frac{\cos A}{\cos B}=\frac{2 \sin (A-B)}{\sin 2 B} $$

Question

UpStudy AI · Accepted Answer

1. Start with the left-hand side (LHS):
   $$
   \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B}.
   $$

2. Write the difference as a single fraction by finding a common denominator $$\sin B \cos B$$:
   $$
   \frac{\sin A \cos B - \cos A \sin B}{\sin B\cos B}.
   $$

3. Recognize that the numerator $$\sin A \cos B - \cos A \sin B$$ is the sine of a difference:
   $$
   \sin A \cos B - \cos A \sin B = \sin (A - B).
   $$

4. Thus, the expression becomes:
   $$
   \frac{\sin (A - B)}{\sin B\cos B}.
   $$

5. Recall the double-angle formula for sine:
   $$
   \sin 2B = 2\sin B \cos B.
   $$

6. Solve for $$\sin B\cos B$$:
   $$
   \sin B\cos B = \frac{\sin 2B}{2}.
   $$

7. Substitute $$\sin B\cos B = \frac{\sin 2B}{2}$$ into the expression:
   $$
   \frac{\sin (A - B)}{\frac{\sin 2B}{2}} = \frac{2\sin (A - B)}{\sin 2B}.
   $$

8. This shows that:
   $$
   \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B} = \frac{2\sin (A-B)}{\sin 2B}.
   $$
$$ \frac{\sin A}{\sin B} - \frac{\cos A}{\cos B} = \frac{2\sin (A - B)}{\sin 2B} $$

(b) \( \frac{\sin A}{\sin B}-\frac{\cos A}{\cos B}=\frac{2 \sin (A-B)}{\sin 2 B} \)

Upstudy AI Solution

Answer

Solution

Bonus Knowledge

Related Questions

Latest Trigonometry Questions