Answer
The value of \( x \) is \( 120^{\circ} \) or \( 240^{\circ} \).
Solution
Let's solve each problem step by step.
---
## **6.1**
\[
\frac{\sin 210^{\circ} \cos 300^{\circ} \tan 240^{\circ}}{\cos 120^{\circ} \tan 150^{\circ} \sin 330^{\circ}}
\]
**Step-by-Step Solution:**
1. **Evaluate each trigonometric function:**
- \(\sin 210^{\circ}\):
\(210^{\circ} = 180^{\circ} + 30^{\circ}\), so \(\sin 210^{\circ} = -\sin 30^{\circ} = -\frac{1}{2}\).
- \(\cos 300^{\circ}\):
\(300^{\circ} = 360^{\circ} - 60^{\circ}\), so \(\cos 300^{\circ} = \cos 60^{\circ} = \frac{1}{2}\).
- \(\tan 240^{\circ}\):
\(240^{\circ} = 180^{\circ} + 60^{\circ}\), so \(\tan 240^{\circ} = \tan 60^{\circ} = \sqrt{3}\).
- \(\cos 120^{\circ}\):
\(120^{\circ} = 180^{\circ} - 60^{\circ}\), so \(\cos 120^{\circ} = -\cos 60^{\circ} = -\frac{1}{2}\).
- \(\tan 150^{\circ}\):
\(150^{\circ} = 180^{\circ} - 30^{\circ}\), so \(\tan 150^{\circ} = -\tan 30^{\circ} = -\frac{1}{\sqrt{3}}\).
- \(\sin 330^{\circ}\):
\(330^{\circ} = 360^{\circ} - 30^{\circ}\), so \(\sin 330^{\circ} = -\sin 30^{\circ} = -\frac{1}{2}\).
2. **Substitute the values into the expression:**
\[
\frac{(-\frac{1}{2}) \cdot (\frac{1}{2}) \cdot \sqrt{3}}{(-\frac{1}{2}) \cdot (-\frac{1}{\sqrt{3}}) \cdot (-\frac{1}{2})}
\]
3. **Simplify the numerator and the denominator:**
- **Numerator:**
\[
-\frac{1}{2} \times \frac{1}{2} \times \sqrt{3} = -\frac{\sqrt{3}}{4}
\]
- **Denominator:**
\[
-\frac{1}{2} \times -\frac{1}{\sqrt{3}} \times -\frac{1}{2} = -\frac{1}{4\sqrt{3}}
\]
4. **Compute the overall fraction:**
\[
\frac{-\frac{\sqrt{3}}{4}}{-\frac{1}{4\sqrt{3}}} = \frac{\sqrt{3}}{1} \times \sqrt{3} = 3
\]
**Answer:**
\[
\boxed{3}
\]
---
## **6.2**
\[
[\sin (-\theta) + \cos (360^{\circ} - \theta)] \left[ \cos (90^{\circ} - \theta) + \frac{\sin \theta}{\tan \theta} \right]
\]
**Step-by-Step Solution:**
1. **Simplify each trigonometric function:**
- \(\sin(-\theta) = -\sin \theta\)
- \(\cos(360^{\circ} - \theta) = \cos \theta\)
- \(\cos(90^{\circ} - \theta) = \sin \theta\)
- \(\frac{\sin \theta}{\tan \theta}\):
\[
\tan \theta = \frac{\sin \theta}{\cos \theta} \implies \frac{\sin \theta}{\tan \theta} = \frac{\sin \theta}{\frac{\sin \theta}{\cos \theta}} = \cos \theta
\]
2. **Substitute the simplified forms back into the expression:**
\[
[-\sin \theta + \cos \theta] \left[ \sin \theta + \cos \theta \right]
\]
3. **Recognize the pattern:**
This is of the form \((a + b)(a - b) = a^2 - b^2\).
4. **Apply the identity:**
\[
(\cos \theta - \sin \theta)(\cos \theta + \sin \theta) = \cos^2 \theta - \sin^2 \theta
\]
5. **Use the double-angle identity:**
\[
\cos^2 \theta - \sin^2 \theta = \cos(2\theta)
\]
**Answer:**
\[
\boxed{\cos 2\theta}
\]
---
## **Calculating \(x\) Given:**
\[
\tan x = m + \frac{1}{m}, \quad 90^{\circ} \leq x \leq 270^{\circ}
\]
\[
m^2 + \frac{1}{m^2} = 1
\]
**Step-by-Step Solution:**
1. **Given:**
\[
m^2 + \frac{1}{m^2} = 1
\]
2. **Let’s find \(\tan x\):**
\[
\tan x = m + \frac{1}{m}
\]
Square both sides to relate with the given equation:
\[
(\tan x)^2 = \left(m + \frac{1}{m}\right)^2 = m^2 + 2 + \frac{1}{m^2}
\]
Substitute the given \(m^2 + \frac{1}{m^2} = 1\):
\[
(\tan x)^2 = 1 + 2 = 3 \implies \tan x = \pm\sqrt{3}
\]
3. **Determine the possible values for \(x\) in the given interval:**
- The interval \(90^{\circ} \leq x \leq 270^{\circ}\) spans the **2nd** and **3rd** quadrants.
- **In the 2nd Quadrant (90° to 180°):**
\(\tan x\) is **negative**. Thus, \(\tan x = -\sqrt{3}\).
\[
x = 180^{\circ} - 60^{\circ} = 120^{\circ}
\]
- **In the 3rd Quadrant (180° to 270°):**
\(\tan x\) is **positive**. Thus, \(\tan x = \sqrt{3}\).
\[
x = 180^{\circ} + 60^{\circ} = 240^{\circ}
\]
**Answer:**
\[
\boxed{120^{\circ} \text{ or } 240^{\circ}}
\]
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