5. Without using a calculator: a) Evaluate $$ \sin ^{2} 40^{\circ}+\cos ^{2} 40^{\circ} $$ b) Write $$ \cos 80^{\circ} $$ in terms of $$ \cos 40^{\circ} $$ and $$ \sin 40^{\circ} $$

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Let's solve each part step by step.

### Part a: Evaluate $$ \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} $$

We can use the Pythagorean identity for sine and cosine, which states that:

$$
\sin^{2} \theta + \cos^{2} \theta = 1
$$

In this case, we have $$ \theta = 40^{\circ} $$. Therefore, we can directly apply the identity:

$$
\sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1
$$

So, the answer for part a is:

$$
\sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1
$$

### Part b: Write $$ \cos 80^{\circ} $$ in terms of $$ \cos 40^{\circ} $$ and $$ \sin 40^{\circ} $$

To express $$ \cos 80^{\circ} $$ in terms of $$ \cos 40^{\circ} $$ and $$ \sin 40^{\circ} $$, we can use the cosine of a double angle formula:

$$
\cos(2\theta) = \cos^{2} \theta - \sin^{2} \theta
$$

Here, we can let $$ \theta = 40^{\circ} $$. Thus, we have:

$$
\cos 80^{\circ} = \cos(2 \times 40^{\circ}) = \cos^{2} 40^{\circ} - \sin^{2} 40^{\circ}
$$

We can also use the identity $$ \cos^{2} \theta = 1 - \sin^{2} \theta $$ to express $$ \cos^{2} 40^{\circ} $$:

$$
\cos^{2} 40^{\circ} = 1 - \sin^{2} 40^{\circ}
$$

Substituting this into the equation gives:

$$
\cos 80^{\circ} = (1 - \sin^{2} 40^{\circ}) - \sin^{2} 40^{\circ}
$$

This simplifies to:

$$
\cos 80^{\circ} = 1 - 2\sin^{2} 40^{\circ}
$$

Alternatively, we can express it in terms of $$ \cos 40^{\circ} $$:

Using the identity $$ \sin^{2} \theta = 1 - \cos^{2} \theta $$:

$$
\cos 80^{\circ} = \cos^{2} 40^{\circ} - (1 - \cos^{2} 40^{\circ}) = 2\cos^{2} 40^{\circ} - 1
$$

Thus, we can express $$ \cos 80^{\circ} $$ in terms of $$ \cos 40^{\circ} $$ and $$ \sin 40^{\circ} $$ as:

$$
\cos 80^{\circ} = 2\cos^{2} 40^{\circ} - 1
$$

In summary, the answers are:
- a) $$ \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1 $$
- b) $$ \cos 80^{\circ} = 2\cos^{2} 40^{\circ} - 1 $$ or $$ \cos 80^{\circ} = 1 - 2\sin^{2} 40^{\circ} $$
a) $$ \sin^{2} 40^{\circ} + \cos^{2} 40^{\circ} = 1 $$ b) $$ \cos 80^{\circ} = 2\cos^{2} 40^{\circ} - 1 $$

5. Without using a calculator: a) Evaluate \( \sin ^{2} 40^{\circ}+\cos ^{2} 40^{\circ} \) b) Write \( \cos 80^{\circ} \) in terms of \( \cos 40^{\circ} \) and \( \sin 40^{\circ} \)

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