Simplify the following expressions without using a calculator. $$ 6.1 \quad \frac{\sin 210^{\circ} \cos 300^{\circ} an 240^{\circ}}{\cos 120^{\circ} an 150^{\circ} \sin 330^{\circ}} $$ $$ 6.2 \quad[\sin (- heta)+\cos (360- heta)]\left[\cos (90- heta)+\frac{\sin heta}{ an heta} ight] $$ $$ 6.3 \quad $$ If $$ \quad an x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} $$ and $$ m^{2}+\frac{1}{m^{2}}=1 $$ Calculate the value of $$ x $$ without the use of a calculator.

Question

Simplify the following expressions without using a calculator. $$ 6.1 \quad \frac{\sin 210^{\circ} \cos 300^{\circ} 	an 240^{\circ}}{\cos 120^{\circ} 	an 150^{\circ} \sin 330^{\circ}} $$ $$ 6.2 \quad[\sin (-	heta)+\cos (360-	heta)]\left[\cos (90-	heta)+\frac{\sin 	heta}{	an 	heta}ight] $$ $$ 6.3 \quad $$ If $$ \quad 	an x=m+\frac{1}{m}, 90^{\circ} \leq x \leq 270^{\circ} $$ and $$ m^{2}+\frac{1}{m^{2}}=1 $$ Calculate the value of $$ x $$ without the use of a calculator.

UpStudy AI · Accepted Answer

Let's simplify each expression step by step.

### 6.1
We need to simplify the expression:
$$
\frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}}
$$

First, we find the values of the trigonometric functions involved:

- $$\sin 210^{\circ} = -\frac{1}{2}$$
- $$\cos 300^{\circ} = \frac{1}{2}$$
- $$\tan 240^{\circ} = \frac{\sin 240^{\circ}}{\cos 240^{\circ}} = \frac{-\frac{\sqrt{3}}{2}}{-\frac{1}{2}} = \sqrt{3}$$

Now, substituting these values into the numerator:
$$
\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ} = \left(-\frac{1}{2}\right) \left(\frac{1}{2}\right) \left(\sqrt{3}\right) = -\frac{\sqrt{3}}{4}
$$

Next, we calculate the denominator:
- $$\cos 120^{\circ} = -\frac{1}{2}$$
- $$\tan 150^{\circ} = \frac{\sin 150^{\circ}}{\cos 150^{\circ}} = \frac{\frac{1}{2}}{-\frac{\sqrt{3}}{2}} = -\frac{1}{\sqrt{3}}$$
- $$\sin 330^{\circ} = -\frac{1}{2}$$

Now substituting these values into the denominator:
$$
\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ} = \left(-\frac{1}{2}\right) \left(-\frac{1}{\sqrt{3}}\right) \left(-\frac{1}{2}\right) = -\frac{1}{4\sqrt{3}}
$$

Now we can substitute the numerator and denominator back into the expression:
$$
\frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}} = \frac{\sqrt{3}}{4} \cdot \frac{4\sqrt{3}}{1} = 3
$$

Thus, the simplified expression is:
$$
\boxed{3}
$$

### 6.2
We need to simplify the expression:
$$
[\sin (-\theta)+\cos (360-\theta)]\left[\cos (90-\theta)+\frac{\sin \theta}{\tan \theta}\right]
$$

First, simplify the first part:
- $$\sin(-\theta) = -\sin(\theta)$$
- $$\cos(360^{\circ} - \theta) = \cos(\theta)$$

So, the first part becomes:
$$
-\sin(\theta) + \cos(\theta)
$$

Now simplify the second part:
- $$\cos(90^{\circ} - \theta) = \sin(\theta)$$
- $$\frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta$$

Thus, the second part becomes:
$$
\sin(\theta) + \cos(\theta)
$$

Now we can combine both parts:
$$
(-\sin(\theta) + \cos(\theta))(\sin(\theta) + \cos(\theta))
$$

Using the distributive property:
$$
-\sin^2(\theta) - \sin(\theta)\cos(\theta) + \cos(\theta)\sin(\theta) + \cos^2(\theta) = -\sin^2(\theta) + \cos^2(\theta)
$$

Using the identity $$\cos^2(\theta) - \sin^2(\theta) = \cos(2\theta)$$:
$$
\cos(2\theta)
$$

Thus, the simplified expression is:
$$
\boxed{\cos(2\theta)}
$$

### 6.3
Given:
$$
\tan x = m + \frac{1}{m}, \quad 90^{\circ} \leq x \leq 270^{\circ} \quad \text{and} \quad m^{2} + \frac{1}{m^{2}} = 1
$$

From the equation $$m^2 + \frac{1}{m^2} = 1$$, we can derive:
$$
(m + \frac{1}{m})^2 = m^2 + 2 + \frac{1}{m^2} \implies (m + \frac{1}{m})^2 = 1 + 2 = 3
$$
Thus,
$$
m + \frac{1}{m} = \pm \sqrt{3}
$$

Now, we have two cases:
1. $$m + \frac{1}{m} = \sqrt{3}$$
2. $$m + \frac{1}{m} = -\sqrt{3}$$

For the first case:
$$
\tan x = \sqrt{3} \implies x = 60^{\circ} + k \cdot 180^{\circ}
$$
In the interval $$90^{\circ} \leq x \leq 270^{\circ}$$, we have:
$$
x = 60^{\circ} + 180^{\circ} = 240^{\circ}
$$

For the second case:
$$
\tan x = -\sqrt{3} \implies x = 120^{\circ} + k \cdot 180^{\circ}
$$
In the interval $$90^{\circ} \leq x \leq 270^{\circ}$$, we have:
$$
x = 120^{\circ} \quad \text{and} \quad x = 120^{\circ} + 180^{\circ} = 300^{\circ} \quad (\text{not in the interval})
$$

Thus, the only valid solutions are:
$$
x = 240^{\circ} \quad \text{and} \quad x = 120^{\circ}
$$

Since both values are valid, we can conclude:
$$
\boxed{240^{\circ}} \quad \text{or} \quad \boxed{120^{\circ}}
$$
$$ 6.1 \quad 3 $$ $$ 6.2 \quad \cos(2\theta) $$ $$ 6.3 \quad 240^{\circ} $$ or $$ 120^{\circ} $$

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