$$ \begin{array}{ll} ext { Gisen: } & \cos (A-B)=\cos A \cos B+\sin A \sin B \ 24.4 .1 & ext { Use the formain for } \cos (A-B) ext { to derive a formula for } \sin (A-B) \ 244.2 \quad ext { Without using a culculator, show that } \ & \sin \left(x+64^{\circ} ight) \cos \left(x+379^{\circ} ight)+\sin \left(x+19^{\circ} ight) \cos \left(x+244^{\circ} ight)=\frac{1}{\sqrt{2}} \ ext { for all values of } x .\end{array} $$

Question

$$ \begin{array}{ll}	ext { Gisen: } & \cos (A-B)=\cos A \cos B+\sin A \sin B \ 24.4 .1 & 	ext { Use the formain for } \cos (A-B) 	ext { to derive a formula for } \sin (A-B) \ 244.2 \quad 	ext { Without using a culculator, show that } \ & \sin \left(x+64^{\circ}ight) \cos \left(x+379^{\circ}ight)+\sin \left(x+19^{\circ}ight) \cos \left(x+244^{\circ}ight)=\frac{1}{\sqrt{2}} \ 	ext { for all values of } x .\end{array} $$

UpStudy AI · Accepted Answer

To solve the problem, we will follow the steps outlined in the prompt.

### Step 1: Derive a formula for $$ \sin(A - B) $$

We start with the known formula for $$ \cos(A - B) $$:

$$
\cos(A - B) = \cos A \cos B + \sin A \sin B
$$

We can use the identity for $$ \sin(A - B) $$ in terms of $$ \sin $$ and $$ \cos $$:

$$
\sin(A - B) = \sin A \cos B - \cos A \sin B
$$

To derive this, we can use the following relationships:

1. $$ \sin(A) = \cos\left(\frac{\pi}{2} - A\right) $$
2. $$ \cos(A) = \sin\left(\frac{\pi}{2} - A\right) $$

Using the cosine of a sum formula, we can express $$ \sin(A - B) $$ as follows:

$$
\sin(A - B) = \sin A \cos B - \cos A \sin B
$$

This is derived from the sine addition formula:

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

Thus, we have derived the formula for $$ \sin(A - B) $$:

$$
\sin(A - B) = \sin A \cos B - \cos A \sin B
$$

### Step 2: Show that

$$
\sin\left(x + 64^{\circ}\right) \cos\left(x + 379^{\circ}\right) + \sin\left(x + 19^{\circ}\right) \cos\left(x + 244^{\circ}\right) = \frac{1}{\sqrt{2}}
$$

We will simplify each term using the sine and cosine addition formulas.

1. **First term:**

$$
   \sin\left(x + 64^{\circ}\right) \cos\left(x + 379^{\circ}\right)
   $$

We can simplify $$ \cos\left(x + 379^{\circ}\right) $$:

$$
   379^{\circ} = 360^{\circ} + 19^{\circ} \implies \cos\left(x + 379^{\circ}\right) = \cos\left(x + 19^{\circ}\right)
   $$

Thus, the first term becomes:

$$
   \sin\left(x + 64^{\circ}\right) \cos\left(x + 19^{\circ}\right)
   $$

2. **Second term:**

$$
   \sin\left(x + 19^{\circ}\right) \cos\left(x + 244^{\circ}\right)
   $$

We can simplify $$ \cos\left(x + 244^{\circ}\right) $$:

$$
   244^{\circ} = 180^{\circ} + 64^{\circ} \implies \cos\left(x + 244^{\circ}\right) = -\sin\left(x + 64^{\circ}\right)
   $$

Thus, the second term becomes:

$$
   \sin\left(x + 19^{\circ}\right)(-\sin\left(x + 64^{\circ}\right))
   $$

Now, we can combine both terms:

$$
\sin\left(x + 64^{\circ}\right) \cos\left(x + 19^{\circ}\right) - \sin\left(x + 19^{\circ}\right) \sin\left(x + 64^{\circ}\right)
$$

Using the sine subtraction formula:

$$
\sin(A) \cos(B) - \sin(B) \sin(A) = \sin(A - B)
$$

Let $$ A = x + 64^{\circ} $$ and $$ B = x + 19^{\circ} $$:

$$
\sin\left((x + 64^{\circ}) - (x + 19^{\circ})\right) = \sin(64^{\circ} - 19^{\circ}) = \sin(45^{\circ}) = \frac{1}{\sqrt{2}}
$$

Thus, we have shown that:

$$
\sin\left(x + 64^{\circ}\right) \cos\left(x + 379^{\circ}\right) + \sin\left(x + 19^{\circ}\right) \cos\left(x + 244^{\circ}\right) = \frac{1}{\sqrt{2}}
$$

### Final Answer

1. The derived formula for $$ \sin(A - B) $$ is:

$$
\sin(A - B) = \sin A \cos B - \cos A \sin B
$$

2. The equation

$$
\sin\left(x + 64^{\circ}\right) \cos\left(x + 379^{\circ}\right) + \sin\left(x + 19^{\circ}\right) \cos\left(x + 244^{\circ}\right) = \frac{1}{\sqrt{2}}
$$

holds true for all values of $$ x $$.
The derived formula for $$ \sin(A - B) $$ is: $$ \sin(A - B) = \sin A \cos B - \cos A \sin B $$ Additionally, the equation $$ \sin\left(x + 64^{\circ}\right) \cos\left(x + 379^{\circ}\right) + \sin\left(x + 19^{\circ}\right) \cos\left(x + 244^{\circ}\right) = \frac{1}{\sqrt{2}} $$ holds true for all values of $$ x $$.

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