Is given that $$ p=\cos \alpha+\sin \alpha $$ and $$ q=\cos \alpha-\sin \alpha $$ 1 Determine the following trigonometric ratios in terms of $$ p $$ and/or $$ q $$ (a) $$ \cos 2 \alpha $$ (b) $$ an \alpha $$ (c) $$ \frac{1+\sin 2 a}{\cos 2 a} $$

Question

Is given that $$ p=\cos \alpha+\sin \alpha $$ and $$ q=\cos \alpha-\sin \alpha $$ 1 Determine the following trigonometric ratios in terms of $$ p $$ and/or $$ q $$ (a) $$ \cos 2 \alpha $$ (b) $$ 	an \alpha $$ (c) $$ \frac{1+\sin 2 a}{\cos 2 a} $$

UpStudy AI · Accepted Answer

To solve the problem, we start with the given expressions for $$ p $$ and $$ q $$:

$$
p = \cos \alpha + \sin \alpha
$$
$$
q = \cos \alpha - \sin \alpha
$$

We will derive the required trigonometric ratios step by step.

### (a) Determine $$ \cos 2\alpha $$

Using the double angle formula for cosine, we have:

$$
\cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha
$$

We can express $$ \cos^2 \alpha $$ and $$ \sin^2 \alpha $$ in terms of $$ p $$ and $$ q $$:

1. From $$ p $$ and $$ q $$, we can find $$ \cos^2 \alpha $$ and $$ \sin^2 \alpha $$:
   - Adding $$ p $$ and $$ q $$:
   $$
   p + q = (\cos \alpha + \sin \alpha) + (\cos \alpha - \sin \alpha) = 2\cos \alpha \implies \cos \alpha = \frac{p + q}{2}
   $$
   - Subtracting $$ q $$ from $$ p $$:
   $$
   p - q = (\cos \alpha + \sin \alpha) - (\cos \alpha - \sin \alpha) = 2\sin \alpha \implies \sin \alpha = \frac{p - q}{2}
   $$

2. Now, we can find $$ \cos^2 \alpha $$ and $$ \sin^2 \alpha $$:
   $$
   \cos^2 \alpha = \left(\frac{p + q}{2}\right)^2 = \frac{(p + q)^2}{4}
   $$
   $$
   \sin^2 \alpha = \left(\frac{p - q}{2}\right)^2 = \frac{(p - q)^2}{4}
   $$

3. Substitute these into the double angle formula:
   $$
   \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \frac{(p + q)^2}{4} - \frac{(p - q)^2}{4}
   $$

4. Simplifying this:
   $$
   \cos 2\alpha = \frac{(p + q)^2 - (p - q)^2}{4}
   $$
   $$
   = \frac{(p^2 + 2pq + q^2) - (p^2 - 2pq + q^2)}{4}
   $$
   $$
   = \frac{4pq}{4} = pq
   $$

Thus, we have:

$$
\cos 2\alpha = pq
$$

### (b) Determine $$ \tan \alpha $$

Using the definitions of sine and cosine, we have:

$$
\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{\frac{p - q}{2}}{\frac{p + q}{2}} = \frac{p - q}{p + q}
$$

Thus, we have:

$$
\tan \alpha = \frac{p - q}{p + q}
$$

### (c) Determine $$ \frac{1 + \sin 2\alpha}{\cos 2\alpha} $$

Using the double angle formulas, we know:

$$
\sin 2\alpha = 2\sin \alpha \cos \alpha
$$

Substituting the expressions for $$ \sin \alpha $$ and $$ \cos \alpha $$:

$$
\sin 2\alpha = 2 \left(\frac{p - q}{2}\right) \left(\frac{p + q}{2}\right) = \frac{(p - q)(p + q)}{2}
$$

Now, substituting into the expression:

$$
\frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{1 + \frac{(p - q)(p + q)}{2}}{pq}
$$

Simplifying this:

$$
= \frac{2 + (p - q)(p + q)}{2pq}
$$
$$
= \frac{2 + p^2 - q^2}{2pq}
$$

Thus, we have:

$$
\frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{2 + p^2 - q^2}{2pq}
$$

### Summary of Results

1. $$ \cos 2\alpha = pq $$
2. $$ \tan \alpha = \frac{p - q}{p + q} $$
3. $$ \frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{2 + p^2 - q^2}{2pq} $$
$$ \cos 2\alpha = pq $$, $$ \tan \alpha = \frac{p - q}{p + q} $$, and $$ \frac{1 + \sin 2\alpha}{\cos 2\alpha} = \frac{2 + p^2 - q^2}{2pq} $$.

Is given that and
1 Determine the following trigonometric ratios in terms of and/or
(a)
(b)
©

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Solution

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