Tornenork Faluate, what using a calculator (a) $$ an 315^{\circ}-2 \cos 60^{\circ}+\sin 210 $$ (b) $$ \frac{ an 150^{\circ}}{ an 240^{\circ}}-\frac{\sin 300^{\circ}}{\sin 120^{\circ}} $$ (c) $$ \frac{ an 225^{\circ}-\sin x}{\sin x+\sin 270^{\circ}} $$ (b) $$ \sin 570^{\circ}+\cos 240^{\circ}- an 135^{\circ} $$ (c) $$ an 120^{\circ} \cdot \cos 210^{\circ}-\sin ^{2} 315^{\circ} $$

Question

Tornenork Faluate, what using a calculator (a) $$ 	an 315^{\circ}-2 \cos 60^{\circ}+\sin 210 $$ (b) $$ \frac{	an 150^{\circ}}{	an 240^{\circ}}-\frac{\sin 300^{\circ}}{\sin 120^{\circ}} $$ (c) $$ \frac{	an 225^{\circ}-\sin x}{\sin x+\sin 270^{\circ}} $$ (b) $$ \sin 570^{\circ}+\cos 240^{\circ}-	an 135^{\circ} $$ (c) $$ 	an 120^{\circ} \cdot \cos 210^{\circ}-\sin ^{2} 315^{\circ} $$

UpStudy AI · Accepted Answer

**(a)** Evaluate 
$$
\tan 315^{\circ} - 2\cos 60^{\circ} + \sin 210^{\circ}.
$$

1. Compute $$ \tan 315^{\circ} $$.  
   Since $$ 315^{\circ} = 360^{\circ} - 45^{\circ} $$, we have  
   $$
   \tan 315^{\circ} = -\tan 45^{\circ} = -1.
   $$
2. Compute $$ 2\cos 60^{\circ} $$.  
   We know $$ \cos 60^{\circ} = \frac{1}{2} $$, so  
   $$
   2\cos 60^{\circ} = 2\left(\frac{1}{2}\right) = 1.
   $$
3. Compute $$ \sin 210^{\circ} $$.  
   Since $$ 210^{\circ} = 180^{\circ} + 30^{\circ} $$,  
   $$
   \sin 210^{\circ} = -\sin 30^{\circ} = -\frac{1}{2}.
   $$
4. Combine the results:
   $$
   -1 - 1 - \frac{1}{2} = -2 - \frac{1}{2} = -\frac{5}{2}.
   $$

---

**(b)** Evaluate 
$$
\frac{\tan 150^{\circ}}{\tan 240^{\circ}} - \frac{\sin 300^{\circ}}{\sin 120^{\circ}}.
$$

1. Compute $$ \tan 150^{\circ} $$ and $$ \tan 240^{\circ} $$:
   - $$ 150^{\circ} = 180^{\circ} - 30^{\circ} $$ so  
     $$
     \tan 150^{\circ} = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}.
     $$
   - $$ 240^{\circ} = 180^{\circ} + 60^{\circ} $$ so  
     $$
     \tan 240^{\circ} = \tan 60^{\circ} = \sqrt{3}.
     $$
   - Thus,  
     $$
     \frac{\tan 150^{\circ}}{\tan 240^{\circ}} = \frac{-\frac{1}{\sqrt{3}}}{\sqrt{3}} = -\frac{1}{3}.
     $$

2. Compute $$ \sin 300^{\circ} $$ and $$ \sin 120^{\circ} $$:
   - $$ 300^{\circ} = 360^{\circ} - 60^{\circ} $$ so  
     $$
     \sin 300^{\circ} = -\sin 60^{\circ} = -\frac{\sqrt{3}}{2}.
     $$
   - $$ 120^{\circ} = 180^{\circ} - 60^{\circ} $$ so  
     $$
     \sin 120^{\circ} = \sin 60^{\circ} = \frac{\sqrt{3}}{2}.
     $$
   - Thus,  
     $$
     \frac{\sin 300^{\circ}}{\sin 120^{\circ}} = \frac{-\frac{\sqrt{3}}{2}}{\frac{\sqrt{3}}{2}} = -1.
     $$

3. Combine the parts:  
   $$
   -\frac{1}{3} - (-1) = -\frac{1}{3} + 1 = \frac{2}{3}.
   $$

---

**(c)** Evaluate 
$$
\frac{\tan 225^{\circ} - \sin x}{\sin x + \sin 270^{\circ}}.
$$

1. Compute $$ \tan 225^{\circ} $$:  
   $$ 225^{\circ} = 180^{\circ} + 45^{\circ} $$ gives  
   $$
   \tan 225^{\circ} = \tan 45^{\circ} = 1.
   $$

2. Note that $$ \sin 270^{\circ} = -1 $$. Thus, the expression becomes  
   $$
   \frac{1 - \sin x}{\sin x - 1}.
   $$
   Observe that  
   $$
   1 - \sin x = -(\sin x - 1).
   $$
   Therefore,  
   $$
   \frac{1 - \sin x}{\sin x - 1} = -1,
   $$
   provided that $$ \sin x \neq 1 $$.

---

**(d)** Evaluate 
$$
\sin 570^{\circ} + \cos 240^{\circ} - \tan 135^{\circ}.
$$

1. Compute $$ \sin 570^{\circ} $$:  
   $$ 570^{\circ} = 360^{\circ} + 210^{\circ} $$ so  
   $$
   \sin 570^{\circ} = \sin 210^{\circ} = -\frac{1}{2}.
   $$
2. Compute $$ \cos 240^{\circ} $$:  
   $$ 240^{\circ} = 180^{\circ} + 60^{\circ} $$ so  
   $$
   \cos 240^{\circ} = -\cos 60^{\circ} = -\frac{1}{2}.
   $$
3. Compute $$ \tan 135^{\circ} $$:  
   $$ 135^{\circ} = 180^{\circ} - 45^{\circ} $$ so  
   $$
   \tan 135^{\circ} = -\tan 45^{\circ} = -1.
   $$
4. Combine the parts:
   $$
   -\frac{1}{2} - \frac{1}{2} - (-1) = -1 + 1 = 0.
   $$

---

**(e)** Evaluate 
$$
\tan 120^{\circ} \cdot \cos 210^{\circ} - \sin^{2} 315^{\circ}.
$$

1. Compute $$ \tan 120^{\circ} $$:  
   $$ 120^{\circ} = 180^{\circ} - 60^{\circ} $$ gives  
   $$
   \tan 120^{\circ} = -\tan 60^{\circ} = -\sqrt{3}.
   $$
2. Compute $$ \cos 210^{\circ} $$:  
   $$ 210^{\circ} = 180^{\circ} + 30^{\circ} $$ so  
   $$
   \cos 210^{\circ} = -\cos 30^{\circ} = -\frac{\sqrt{3}}{2}.
   $$
3. Multiply:
   $$
   \tan 120^{\circ} \cdot \cos 210^{\circ} = (-\sqrt{3}) \cdot \left(-\frac{\sqrt{3}}{2}\right) = \frac{3}{2}.
   $$
4. Compute $$ \sin 315^{\circ} $$:  
   $$ 315^{\circ} = 360^{\circ} - 45^{\circ} $$ so  
   $$
   \sin 315^{\circ} = -\sin 45^{\circ} = -\frac{\sqrt{2}}{2}.
   $$
   Then,  
   $$
   \sin^{2} 315^{\circ} = \left(-\frac{\sqrt{2}}{2}\right)^2 = \frac{1}{2}.
   $$
5. Subtract:
   $$
   \frac{3}{2} - \frac{1}{2} = 1.
   $$

---

Final Answers:  
$$
\text{(a)}\ -\frac{5}{2},\quad \text{(b)}\ \frac{2}{3},\quad \text{(c)}\ -1\ (\sin x\neq 1),\quad \text{(d)}\ 0,\quad \text{(e)}\ 1.
$$
**(a)** $$ -\frac{5}{2} $$ **(b)** $$ \frac{2}{3} $$ **(c)** $$ -1 $$ (when $$ \sin x \neq 1 $$) **(d)** $$ 0 $$ **(e)** $$ 1 $$

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