2 Use your calculator to evaluate each of the following: $$ 1.2 .1 \quad \cos (\hat{\mathrm{~A}}-\hat{\mathrm{B}}) \quad $$ if $$ \hat{\mathrm{A}}=60^{\circ} $$ and $$ \hat{\mathrm{B}}=30^{\circ} $$ $$ 1.2 .2 \quad \cos \hat{\mathrm{~A}}-\cos \hat{\mathrm{B}} $$ if $$ \hat{\mathrm{A}}=110^{\circ} $$ and $$ \hat{\mathrm{B}}=50^{\circ} $$ $$ 1.2 .3 \quad \cos \hat{\mathrm{~A}} \cos \hat{\mathrm{~B}}+\sin \hat{\mathrm{A}} \sin \hat{\mathrm{B}} $$ if $$ \hat{\mathrm{A}}=2250^{\circ} $$ and $$ \hat{\mathrm{B}}=135^{\circ} $$

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**Problem 1.2.1**

We need to evaluate  
$$
\cos (\hat{A}-\hat{B})
$$  
with $$\hat{A}=60^\circ$$ and $$\hat{B}=30^\circ$$.

**Step 1.** Compute the difference of the angles:  
$$
60^\circ - 30^\circ = 30^\circ
$$

**Step 2.** Evaluate the cosine of $$30^\circ$$:  
$$
\cos 30^\circ = \frac{\sqrt{3}}{2}
$$

Thus,  
$$
\cos (60^\circ-30^\circ) = \frac{\sqrt{3}}{2}.
$$

---

**Problem 1.2.2**

We need to evaluate  
$$
\cos \hat{A} - \cos \hat{B}
$$  
with $$\hat{A}=110^\circ$$ and $$\hat{B}=50^\circ$$.

**Step 1.** Evaluate each cosine using a calculator (or known approximations):  
$$
\cos 110^\circ \approx -0.3420,
$$
$$
\cos 50^\circ \approx 0.6428.
$$

**Step 2.** Subtract the two values:  
$$
\cos 110^\circ - \cos 50^\circ \approx -0.3420 - 0.6428 \approx -0.9848.
$$

Thus,  
$$
\cos 110^\circ - \cos 50^\circ \approx -0.9848.
$$

---

**Problem 1.2.3**

We need to evaluate  
$$
\cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B}
$$  
with $$\hat{A}=2250^\circ$$ and $$\hat{B}=135^\circ$$.

**Step 1.** Recognize that the expression uses the cosine difference identity:  
$$
\cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B} = \cos(\hat{A} - \hat{B}).
$$

**Step 2.** Compute the difference of the angles:  
$$
\hat{A} - \hat{B} = 2250^\circ - 135^\circ = 2115^\circ.
$$

**Step 3.** Reduce $$2115^\circ$$ modulo $$360^\circ$$:  
$$
2115^\circ - 360^\circ \times 5 = 2115^\circ - 1800^\circ = 315^\circ,
$$
since $$2115^\circ = 5 \times 360^\circ + 315^\circ$$.

**Step 4.** Evaluate $$\cos 315^\circ$$. Note that  
$$
\cos 315^\circ = \cos(360^\circ - 45^\circ) = \cos 45^\circ = \frac{\sqrt{2}}{2}.
$$

Thus,  
$$
\cos \hat{A} \cos \hat{B} + \sin \hat{A} \sin \hat{B} = \frac{\sqrt{2}}{2}.
$$

Alternatively, reducing $$\hat{A}$$ first gives:  
$$
2250^\circ \mod 360^\circ = 2250^\circ - 6 \times 360^\circ = 2250^\circ - 2160^\circ = 90^\circ,
$$
so  
$$
\cos 90^\circ \cos 135^\circ + \sin 90^\circ \sin 135^\circ = 0 \cdot \cos 135^\circ + 1 \cdot \sin 135^\circ = \sin 135^\circ,
$$
and since  
$$
\sin 135^\circ = \sin(180^\circ-45^\circ) = \sin 45^\circ = \frac{\sqrt{2}}{2},
$$
we obtain the same result.

---

**Final Answers:**

1. $$\cos (60^\circ-30^\circ)=\frac{\sqrt{3}}{2}$$  
2. $$\cos 110^\circ-\cos 50^\circ\approx-0.9848$$  
3. $$\cos 2250^\circ\cos135^\circ+\sin2250^\circ\sin135^\circ=\frac{\sqrt{2}}{2}$$
**Solutions:** 1. $$ \cos(60^\circ - 30^\circ) = \frac{\sqrt{3}}{2} $$ 2. $$ \cos 110^\circ - \cos 50^\circ \approx -0.9848 $$ 3. $$ \cos 2250^\circ \cos 135^\circ + \sin 2250^\circ \sin 135^\circ = \frac{\sqrt{2}}{2} $$

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