Do not use a calculator whev answering Parts, 5.1 Apply the compound angle capansion $$ \cos (x+y)=\cos x \cos y-\sin x \sin y $$ to cos $$ 2 x $$ and siniplify your answer $$ \begin{array}{l} \cos 2 x=\cos (x+x) \end{array} $$ (2) 5.2 Complete the square identity: $$ \sin ^{2} x+\cos ^{2} x=\ldots \ldots $$ (I) 53 Use the identity in 5.2 and rewrito $$ \cos ^{2} x-\sin ^{2} x $$ in ierns of sin $$ x $$ only: $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ (2) 5.4 Use the identity in 5.2 and rewrite $$ \cos ^{2} x-\sin ^{2} x $$ in terms of cos $$ x $$ only: $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ $$ \qquad $$ 5.5 Simplify each expression to a singlo term: $$ 55.1 \cos ^{2} 19-\sin ^{2} 19= $$ $$ \qquad $$ $$ 5.2 \cos ^{2} 19-\sin ^{2} 19= $$ $$ \qquad $$ $$ 553 \quad 2 \cos ^{2} 50^{\circ}-1= $$ $$ \qquad $$ $$ 554 \quad 1-2 \sin ^{2} 50^{\circ}= $$. $$ \qquad $$ $$ 555 \quad 6 \cos ^{2} \frac{\theta}{2}-3= $$ $$ \qquad $$ 56 It is given that $$ \cos 32^{\circ}=t $$. Without using a calculator, express cos 16 in terms of i $$ \qquad $$

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Let's solve each part step by step.

### Part 5.1
We need to apply the compound angle expansion for cosine:
$$
\cos(x+y) = \cos x \cos y - \sin x \sin y
$$
to $$\cos(2x)$$. Here, we can set $$x = x$$ and $$y = x$$:
$$
\cos(2x) = \cos(x+x) = \cos x \cos x - \sin x \sin x
$$
This simplifies to:
$$
\cos(2x) = \cos^2 x - \sin^2 x
$$

### Part 5.2
The complete square identity is:
$$
\sin^2 x + \cos^2 x = 1
$$

### Part 5.3
We want to rewrite $$\cos^2 x - \sin^2 x$$ in terms of $$\sin x$$ only. We can use the identity from Part 5.2:
$$
\cos^2 x = 1 - \sin^2 x
$$
Substituting this into $$\cos^2 x - \sin^2 x$$:
$$
\cos^2 x - \sin^2 x = (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x
$$

### Part 5.4
Now, we rewrite $$\cos^2 x - \sin^2 x$$ in terms of $$\cos x$$ only. From the identity in Part 5.2:
$$
\sin^2 x = 1 - \cos^2 x
$$
Substituting this into $$\cos^2 x - \sin^2 x$$:
$$
\cos^2 x - \sin^2 x = \cos^2 x - (1 - \cos^2 x) = \cos^2 x - 1 + \cos^2 x = 2\cos^2 x - 1
$$

### Part 5.5
Now we simplify each expression to a single term:

#### 5.5.1
$$
\cos^2 19 - \sin^2 19 = \cos(2 \cdot 19) = \cos(38)
$$

#### 5.5.2
Using the same identity:
$$
\cos^2 19 - \sin^2 19 = \cos(2 \cdot 19) = \cos(38)
$$

#### 5.5.3
$$
2\cos^2 50^\circ - 1 = \cos(2 \cdot 50^\circ) = \cos(100^\circ)
$$

#### 5.5.4
$$
1 - 2\sin^2 50^\circ = \cos(2 \cdot 50^\circ) = \cos(100^\circ)
$$

#### 5.5.5
$$
6\cos^2 \frac{\theta}{2} - 3 = 3(2\cos^2 \frac{\theta}{2} - 1) = 3\cos(\theta)
$$

### Part 5.6
Given that $$\cos 32^\circ = t$$, we want to express $$\cos 16^\circ$$ in terms of $$t$$. We can use the half-angle formula:
$$
\cos(16^\circ) = \cos\left(\frac{32^\circ}{2}\right) = \sqrt{\frac{1 + \cos 32^\circ}{2}} = \sqrt{\frac{1 + t}{2}}
$$

Thus, the final answers are:
- 5.1: $$\cos(2x) = \cos^2 x - \sin^2 x$$
- 5.2: $$\sin^2 x + \cos^2 x = 1$$
- 5.3: $$\cos^2 x - \sin^2 x = 1 - 2\sin^2 x$$
- 5.4: $$\cos^2 x - \sin^2 x = 2\cos^2 x - 1$$
- 5.5.1: $$\cos(38)$$
- 5.5.2: $$\cos(38)$$
- 5.5.3: $$\cos(100^\circ)$$
- 5.5.4: $$\cos(100^\circ)$$
- 5.5.5: $$3\cos(\theta)$$
- 5.6: $$\cos 16^\circ = \sqrt{\frac{1 + t}{2}}$$
Here are the simplified answers for each part: 1. **5.1**: $$\cos(2x) = \cos^2 x - \sin^2 x$$ 2. **5.2**: $$\sin^2 x + \cos^2 x = 1$$ 3. **5.3**: $$\cos^2 x - \sin^2 x = 1 - 2\sin^2 x$$ 4. **5.4**: $$\cos^2 x - \sin^2 x = 2\cos^2 x - 1$$ 5. **5.5.1**: $$\cos(38)$$ 6. **5.5.2**: $$\cos(38)$$ 7. **5.5.3**: $$\cos(100^\circ)$$ 8. **5.5.4**: $$\cos(100^\circ)$$ 9. **5.5.5**: $$3\cos(\theta)$$ 10. **5.6**: $$\cos 16^\circ = \sqrt{\frac{1 + t}{2}}$$

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