$$ [ \sin ( - heta ) + \cos ( 360 - heta ) ] [ \cos ( 90 - heta ) + \frac { \sin heta } { an heta } ] $$

Question

$$ [ \sin ( - 	heta ) + \cos ( 360 - 	heta ) ] [ \cos ( 90 - 	heta ) + \frac { \sin 	heta } { 	an 	heta } ] $$

UpStudy AI · Accepted Answer

Let's simplify the expression step by step using trigonometric identities.

**Given Expression:**
$$
[ \sin ( - \theta ) + \cos ( 360^\circ - \theta ) ] \times \left[ \cos ( 90^\circ - \theta ) + \frac{ \sin \theta }{ \tan \theta } \right]
$$

### Step 1: Simplify the First Bracket
$$
\sin(-\theta) + \cos(360^\circ - \theta)
$$

- **Using the Odd-Even Identities:**
  - $$\sin(-\theta) = -\sin\theta$$
  - $$\cos(360^\circ - \theta) = \cos\theta$$ (since cosine is periodic with period $$360^\circ$$ and even)

$$
\sin(-\theta) + \cos(360^\circ - \theta) = -\sin\theta + \cos\theta
$$

### Step 2: Simplify the Second Bracket
$$
\cos(90^\circ - \theta) + \frac{\sin\theta}{\tan\theta}
$$

- **Using Co-Function Identity:**
  - $$\cos(90^\circ - \theta) = \sin\theta$$

- **Simplify $$\frac{\sin\theta}{\tan\theta}$$:**
  - $$\tan\theta = \frac{\sin\theta}{\cos\theta}$$ 
  - Therefore, $$\frac{\sin\theta}{\tan\theta} = \frac{\sin\theta}{\frac{\sin\theta}{\cos\theta}} = \cos\theta$$

$$
\cos(90^\circ - \theta) + \frac{\sin\theta}{\tan\theta} = \sin\theta + \cos\theta
$$

### Step 3: Multiply the Simplified Brackets
$$
(-\sin\theta + \cos\theta)(\sin\theta + \cos\theta)
$$

- **Use the Distributive Property:**
$$
= -\sin^2\theta - \sin\theta \cos\theta + \sin\theta \cos\theta + \cos^2\theta
$$

- **Combine Like Terms:**
$$
= -\sin^2\theta + \cos^2\theta
$$

### Step 4: Use the Double Angle Identity
$$
\cos^2\theta - \sin^2\theta = \cos(2\theta)
$$

**Final Simplified Form:**
$$
\cos(2\theta)
$$

**Answer:**
After simplifying, the expression is equal to the cosine of twice θ. In symbols,

cos 2θ
The expression simplifies to $$ \cos(2\theta) $$.

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