(a) Prove the following: (1) $$ (\cos heta+\sin heta)^{2}=1+\sin 2 heta $$ (3) $$ \cos ^{4} \alpha-\sin ^{4} \alpha=\cos 2 \alpha $$ (5) $$ \frac{\sin heta+\sin 2 heta}{1+\cos heta+\cos 2 heta}= an heta $$

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(a) Prove the following: (1) $$ (\cos 	heta+\sin 	heta)^{2}=1+\sin 2 	heta $$ (3) $$ \cos ^{4} \alpha-\sin ^{4} \alpha=\cos 2 \alpha $$ (5) $$ \frac{\sin 	heta+\sin 2 	heta}{1+\cos 	heta+\cos 2 	heta}=	an 	heta $$

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### (1) Proof of 
$$
(\cos \theta+\sin \theta)^{2}=1+\sin 2 \theta
$$

- Expand the left-hand side:
  $$
  (\cos \theta+\sin \theta)^2 = \cos^2 \theta + 2\sin\theta\cos\theta + \sin^2 \theta.
  $$
- Recognize that:
  $$
  \cos^2 \theta + \sin^2 \theta = 1
  $$
  and
  $$
  2\sin\theta\cos\theta = \sin 2\theta.
  $$
- So, we have:
  $$
  (\cos \theta+\sin \theta)^2 = 1 + \sin 2\theta.
  $$

---

### (3) Proof of 
$$
\cos ^{4} \alpha - \sin ^{4} \alpha = \cos 2 \alpha
$$

- Write the left-hand side as a difference of squares:
  $$
  \cos ^{4} \alpha - \sin ^{4} \alpha = (\cos^2 \alpha)^2 - (\sin^2 \alpha)^2.
  $$
- Factor using the formula $$a^2 - b^2 = (a - b)(a + b)$$:
  $$
  (\cos^2 \alpha - \sin^2 \alpha)(\cos^2 \alpha + \sin^2 \alpha).
  $$
- Since
  $$
  \cos^2 \alpha + \sin^2 \alpha = 1,
  $$
  it becomes:
  $$
  \cos ^{4} \alpha - \sin ^{4} \alpha = \cos^2 \alpha - \sin^2 \alpha.
  $$
- Recognize that:
  $$
  \cos^2 \alpha - \sin^2 \alpha = \cos 2 \alpha.
  $$
- Therefore:
  $$
  \cos ^{4} \alpha - \sin ^{4} \alpha = \cos 2 \alpha.
  $$

---

### (5) Proof of 
$$
\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=\tan \theta
$$

- **Step 1:** Express $$\sin 2\theta$$ and $$\cos 2\theta$$ using double angle formulas.

Recall that:
  $$
  \sin 2\theta = 2\sin \theta \cos \theta \quad \text{and} \quad \cos 2\theta = 2\cos^2 \theta - 1.
  $$

- **Step 2:** Simplify the numerator:

$$
  \sin \theta + \sin 2\theta = \sin \theta + 2\sin \theta \cos \theta = \sin \theta (1+2\cos \theta).
  $$

- **Step 3:** Simplify the denominator:

$$
  1+\cos \theta+\cos 2\theta = 1+\cos \theta+(2\cos^2 \theta-1) = \cos \theta + 2\cos^2 \theta.
  $$
  
  Factor out $$\cos \theta$$:
  
  $$
  \cos \theta + 2\cos^2 \theta = \cos \theta(1+2\cos \theta).
  $$

- **Step 4:** Substitute back into the fraction:

$$
  \frac{\sin \theta(1+2\cos \theta)}{\cos \theta(1+2\cos \theta)}.
  $$
  
  Cancel the common factor $$(1+2\cos \theta)$$ (provided $$(1+2\cos \theta) \neq 0$$) to obtain:

$$
  \frac{\sin \theta}{\cos \theta} = \tan \theta.
  $$

Thus, the identity is verified:

$$
\frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta} = \tan \theta.
$$
$$ (\cos \theta + \sin \theta)^2 = 1 + \sin 2\theta $$ $$ \cos^4 \alpha - \sin^4 \alpha = \cos 2\alpha $$ $$ \frac{\sin \theta + \sin 2\theta}{1 + \cos \theta + \cos 2\theta} = \tan \theta $$

(a) Prove the following: (1) \( (\cos \theta+\sin \theta)^{2}=1+\sin 2 \theta \) (3) \( \cos ^{4} \alpha-\sin ^{4} \alpha=\cos 2 \alpha \) (5) \( \frac{\sin \theta+\sin 2 \theta}{1+\cos \theta+\cos 2 \theta}=\tan \theta \)

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