The first question asks which of the following expressions properly expresses $cos(120^\circ)$ using a compound angle formula. The second question states that for an acute angle $\theta$ of a right triangle, $sin(\theta) = sin(\frac{\pi}{3})cos(\frac{\pi}{4}) + cos(\frac{\pi}{3})sin(\frac{\pi}{4})$. We must find a possible size for $\theta$.

1. Problem Description

The first question asks which of the following expressions properly expresses $cos(120^\circ)$ using a compound angle formula. The second question states that for an acute angle $\theta$ of a right triangle, $sin(\theta) = sin(\frac{\pi}{3})cos(\frac{\pi}{4}) + cos(\frac{\pi}{3})sin(\frac{\pi}{4})$. We must find a possible size for $\theta$.

2. Solution Steps

Question 18:

We know that $cos(A+B) = cos(A)cos(B) - sin(A)sin(B)$.

We also know that $120^\circ = 90^\circ + 30^\circ$.

Therefore, $cos(120^\circ) = cos(90^\circ + 30^\circ) = cos(90^\circ)cos(30^\circ) - sin(90^\circ)sin(30^\circ)$.

Question 19:

We are given that $sin(\theta) = sin(\frac{\pi}{3})cos(\frac{\pi}{4}) + cos(\frac{\pi}{3})sin(\frac{\pi}{4})$.

Using the angle sum identity for sine:

$sin(A+B) = sin(A)cos(B) + cos(A)sin(B)$.

Comparing this with the given equation, we have $A = \frac{\pi}{3}$ and $B = \frac{\pi}{4}$.

Thus, $sin(\theta) = sin(\frac{\pi}{3} + \frac{\pi}{4})$.

So, $\theta = \frac{\pi}{3} + \frac{\pi}{4}$.

To add the fractions, we need a common denominator, which is

1

2. $\theta = \frac{4\pi}{12} + \frac{3\pi}{12} = \frac{7\pi}{12}$.

However, we can eliminate $\frac{5\pi}{4}$ since it is larger than $\frac{\pi}{2}$ so it is not an acute angle.

Similarly $\frac{5\pi}{8}$ is greater than $\frac{\pi}{2}$ since $\frac{5}{8} > \frac{1}{2}$.

Also $\frac{\pi}{12}$ clearly doesn't satisfy the equation.

Therefore, the question probably has a typo in the angle given for $\sin{\theta}$.

If $sin(\theta) = sin(\frac{\pi}{8})cos(\frac{\pi}{8}) + cos(\frac{\pi}{8})sin(\frac{\pi}{8})$. Then $\theta = \frac{5\pi}{8}$.

If the equation was

$sin \theta = sin(\frac{\pi}{6})cos(\frac{\pi}{4}) + cos(\frac{\pi}{6})sin(\frac{\pi}{4})$. Then

$\theta = \frac{\pi}{6} + \frac{\pi}{4} = \frac{2\pi}{12} + \frac{3\pi}{12} = \frac{5\pi}{12}$

3. Final Answer

Question 18: The correct expression is $cos(90^\circ)cos(30^\circ) - sin(90^\circ)sin(30^\circ)$.

Question 19: The intended answer is likely $\frac{5\pi}{12}$.