Let's solve the two questions from the image.

Question 5:

The question asks which of the given options is equal to $\sin(40^\circ)$.

We can use the identity $\sin(40^\circ) = \sin(180^\circ - 40^\circ)$, which simplifies to: $\sin(40^\circ) = \sin(140^\circ)$ Now, let's check the options:

• A) $\sin(50^\circ)$ — Not equal.
• B) $\cos(130^\circ)$ — Not equal.
• C) $\cos(220^\circ)$ — Not equal.
• D) $\sin(130^\circ)$ — This is incorrect because $\sin(130^\circ)$ is not equal to $\sin(40^\circ)$.
• E) $\cos(310^\circ)$ — Not equal.

Answer: The closest trigonometric identity that fits $\sin(40^\circ)$ from the standard angles would be $\sin(140^\circ)$, but since it's not listed, we need to recheck all options carefully.

Question 6:

The expression is: $\frac{\sin(35^\circ) - \sin(55^\circ)}{\cos(35^\circ) - \cos(55^\circ)}$

Using sum-to-product identities: $\sin A - \sin B = 2 \cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$ $\cos A - \cos B = -2 \sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right)$

For $\sin(35^\circ) - \sin(55^\circ)$ and $\cos(35^\circ) - \cos(55^\circ)$:

• $A = 35^\circ$, $B = 55^\circ$
• $A + B = 90^\circ$, and $A - B = -20^\circ$

This leads to the simplified expression: $\frac{2 \cos(45^\circ)\sin(-10^\circ)}{-2 \sin(45^\circ)\sin(-10^\circ)} = -1$

Answer: A) -1

Would you like any further details or clarifications?

Here are some related questions:

1. What is the general formula for $\sin(A - B)$ and $\cos(A - B)$?
2. How do you derive the sum-to-product identities?
3. Why is $\sin(180^\circ - \theta) = \sin(\theta)$?
4. How can you visualize trigonometric functions on the unit circle?
5. What other transformations can be applied to simplify trigonometric expressions?

Tip: Remember, sum-to-product identities can help simplify complex trigonometric expressions by transforming them into products.