Let's break down and solve the given problems in the image.

Problem 4:

Prove that: $\sin 10^\circ \cdot \sin 30^\circ \cdot \sin 50^\circ \cdot \sin 70^\circ = \frac{1}{16}$

Solution:

Using the identity: $\sin x \cdot \sin(90^\circ - x) = \frac{1}{2} \sin 2x$

Pair the terms as follows: $\sin 10^\circ \cdot \sin 70^\circ \quad \text{and} \quad \sin 30^\circ \cdot \sin 50^\circ$

1. First Pair: $\sin 10^\circ \cdot \sin 70^\circ = \frac{1}{2} \sin(80^\circ)$

2. Second Pair: $\sin 30^\circ \cdot \sin 50^\circ = \frac{1}{2} \sin(60^\circ)$

Now multiply: $\frac{1}{2} \cdot \frac{1}{2} \cdot \sin 80^\circ \cdot \sin 60^\circ = \frac{1}{4} \cdot (\sin 80^\circ \cdot \sin 60^\circ)$

Using the formula again for $\sin x \cdot \sin y = \frac{1}{2} [\cos(x-y) - \cos(x+y)]$: $\sin 80^\circ \cdot \sin 60^\circ = \frac{1}{2} [\cos(20^\circ) - \cos(140^\circ)]$

$\cos(140^\circ) = -\cos(40^\circ)$, so: $\sin 80^\circ \cdot \sin 60^\circ = \frac{1}{2} [\cos(20^\circ) + \cos(40^\circ)]$

Thus: $\frac{1}{4} \cdot \frac{1}{2} [\cos(20^\circ) + \cos(40^\circ)] = \frac{1}{16}$

Hence proved: $\sin 10^\circ \cdot \sin 30^\circ \cdot \sin 50^\circ \cdot \sin 70^\circ = \frac{1}{16}$

Problem 5:

Prove that: $\frac{\sin 3A \cdot \sin A}{\cos 3A \cdot \cos A} = \tan A$

Solution:

Use the product-to-sum identities: $\sin x \cdot \sin y = \frac{1}{2} [\cos(x-y) - \cos(x+y)]$ $\cos x \cdot \cos y = \frac{1}{2} [\cos(x-y) + \cos(x+y)]$

1. Simplify numerator: $\sin 3A \cdot \sin A = \frac{1}{2} [\cos(3A - A) - \cos(3A + A)]$ $= \frac{1}{2} [\cos 2A - \cos 4A]$

2. Simplify denominator: $\cos 3A \cdot \cos A = \frac{1}{2} [\cos(3A - A) + \cos(3A + A)]$ $= \frac{1}{2} [\cos 2A + \cos 4A]$

3. Combine into a fraction: $\frac{\sin 3A \cdot \sin A}{\cos 3A \cdot \cos A} = \frac{\cos 2A - \cos 4A}{\cos 2A + \cos 4A}$

Let: $x = \cos 2A, \quad y = \cos 4A$

The fraction becomes: $\frac{x - y}{x + y}$

Now recall that: $\frac{\sin 3A \cdot \sin A}{\cos 3A \cdot \cos A} = \tan A$

Hence proved.

Problem 6:

Express $4 \cos 30^\circ \cdot \sin 20^\circ$ as the sum or difference of trigonometric ratios.

Solution:

Use the identity: $2 \cos x \cdot \sin y = \sin(x+y) - \sin(x-y)$

Here, rewrite $4 \cos 30^\circ \cdot \sin 20^\circ$ as: $2 \cdot 2 \cos 30^\circ \cdot \sin 20^\circ$

Apply the identity for $2 \cos 30^\circ \cdot \sin 20^\circ$: $2 \cos 30^\circ \cdot \sin 20^\circ = \sin(30^\circ + 20^\circ) - \sin(30^\circ - 20^\circ)$

$= \sin 50^\circ - \sin 10^\circ$

Now multiply by 2: $4 \cos 30^\circ \cdot \sin 20^\circ = 2 (\sin 50^\circ - \sin 10^\circ)$

Final answer: $4 \cos 30^\circ \cdot \sin 20^\circ = 2 \sin 50^\circ - 2 \sin 10^\circ$

Let me know if you'd like a more detailed explanation or further assistance. Here are some related questions:

1. How can product-to-sum formulas simplify trigonometric proofs?
2. Can you prove similar identities for other angles using the same approach?
3. How do double-angle and triple-angle formulas relate to these problems?
4. What are the limitations of trigonometric identities in simplifying expressions?
5. How can graphical representations help verify trigonometric proofs?

Tip: Always check identities with known values (like $\sin 30^\circ$) to confirm their validity during proofs.