Question

Express each of the following as a sum of two trigonometric ratios:sin5xcosx; cos7xsin5x ;cos9xcos3x ;sin3xsinx

Answer 1

To express each of the given trigonometric products as a sum of two trigonometric ratios, we will use the product-to-sum formulas. The formulas we will use are:

1. \(\sin A \cos B = \frac{1}{2} [\sin(A + B) + \sin(A - B)]\)
2. \(\cos A \sin B = \frac{1}{2} [\sin(B + A) - \sin(B - A)]\)
3. \(\cos A \cos B = \frac{1}{2} [\cos(A + B) + \cos(A - B)]\)
4. \(\sin A \sin B = \frac{1}{2} [\cos(A - B) - \cos(A + B)]\)

Now, let's apply these formulas to each case:

1. **For \( \sin 5x \cos x \)**:
\[
\sin 5x \cos x = \frac{1}{2} [\sin(5x + x) + \sin(5x - x)] = \frac{1}{2} [\sin 6x + \sin 4x]
\]

2. **For \( \cos 7x \sin 5x \)**:
\[
\cos 7x \sin 5x = \frac{1}{2} [\sin(5x + 7x) - \sin(5x - 7x)] = \frac{1}{2} [\sin 12x - \sin 2x]
\]

3. **For \( \cos 9x \cos 3x \)**:
\[
\cos 9x \cos 3x = \frac{1}{2} [\cos(9x + 3x) + \cos(9x - 3x)] = \frac{1}{2} [\cos 12x + \cos 6x]
\]

4. **For \( \sin 3x \sin x \)**:
\[
\sin 3x \sin x = \frac{1}{2} [\cos(3x - x) - \cos(3x + x)] = \frac{1}{2} [\cos 2x - \cos 4x]
\]

Putting it all together, we have:

1. \( \sin 5x \cos x = \frac{1}{2} [\sin 6x + \sin 4x] \)
2. \( \cos 7x \sin 5x = \frac{1}{2} [\sin 12x - \sin 2x] \)
3. \( \cos 9x \cos 3x = \frac{1}{2} [\cos 12x + \cos 6x] \)
4. \( \sin 3x \sin x = \frac{1}{2} [\cos 2x - \cos 4x] \)