(b) $$ \frac{\sin \left(180^{\circ}-x ight)-\sin (-x)}{\sin \left(360^{\circ}+x ight)} $$ (d) $$ \frac{\cos \left(720^{\circ}+ heta ight)-\cos (- heta)}{ an \left(180^{\circ}- heta ight)} $$

Question

(b) $$ \frac{\sin \left(180^{\circ}-xight)-\sin (-x)}{\sin \left(360^{\circ}+xight)} $$ (d) $$ \frac{\cos \left(720^{\circ}+	hetaight)-\cos (-	heta)}{	an \left(180^{\circ}-	hetaight)} $$

UpStudy AI · Accepted Answer

**(b) Simplify**
  
1. Rewrite the numerator using trigonometric identities:
   - $$ \sin(180^\circ - x) = \sin x $$
   - $$ \sin(-x) = -\sin x $$
     
   Thus, the numerator becomes:
   $$
   \sin x - (-\sin x) = \sin x + \sin x = 2\sin x
   $$
  
2. Rewrite the denominator using the periodicity of sine:
   $$
   \sin(360^\circ + x) = \sin x
   $$
  
3. The expression becomes:
   $$
   \frac{2\sin x}{\sin x}
   $$
  
4. Cancel $$\sin x$$ (assuming $$\sin x \neq 0$$):
   $$
   2
   $$
  
**(b) Answer:** $$2$$

---

**(d) Simplify**
  
1. Rewrite the numerator using trigonometric identities:
   - Periodicity of cosine gives:
     $$
     \cos(720^\circ + \theta) = \cos \theta
     $$
   - Cosine is an even function:
     $$
     \cos(-\theta) = \cos \theta
     $$
     
   Thus, the numerator becomes:
   $$
   \cos \theta - \cos \theta = 0
   $$
  
2. Rewrite the denominator using the tangent identity:
   $$
   \tan(180^\circ - \theta) = -\tan\theta
   $$
  
3. The expression becomes:
   $$
   \frac{0}{-\tan \theta}
   $$
  
4. Since the numerator is $$0$$ (assuming $$\tan\theta \neq 0$$ so the denominator is defined), the entire expression is $$0$$.

**(d) Answer:** $$0$$
**(b) Simplify:** The expression simplifies to $$2$$. **(d) Simplify:** The expression simplifies to $$0$$.

(b)
(d)

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(b) (d)