Consider the following statements: 1. The equation 2 sin 2 θ - cos θ + 4 = 0 is possible for all θ 2. tan θ + cot θ cannot be less than 2, where 0 < θ < \(\frac{\pi}{{2}}\) Which of the above statements is / are correct?
2 only
Analyzing Trigonometry Statements
Let's carefully examine each statement provided in the question about trigonometric equations and inequalities.
Analyzing Statement 1: Equation Possibility
The first statement is: The equation \(2 \sin 2 \theta - \cos \theta + 4 = 0\) is possible for all \(\theta\).
We need to determine if this equation can hold true for any value of \(\theta\). Let's rearrange the equation:
\[2 \sin 2 \theta - \cos \theta = -4\]
Now, let's consider the possible values of the terms on the left side of the equation.
- The range of \(\sin x\) is \([-1, 1]\). Therefore, the range of \(\sin 2 \theta\) is \([-1, 1]\).
- The range of \(2 \sin 2 \theta\) is \([2 \times -1, 2 \times 1]\), which is \([-2, 2]\).
- The range of \(\cos \theta\) is \([-1, 1]\). Therefore, the range of \(-\cos \theta\) is \([-1 \times 1, -1 \times -1]\), which is \([-1, 1]\).
The expression on the left side is the sum of \(2 \sin 2 \theta\) and \(-\cos \theta\). To find the maximum possible value of this sum, we take the maximum possible value of each term:
Maximum value of \(2 \sin 2 \theta - \cos \theta \le\) (Maximum value of \(2 \sin 2 \theta\)) + (Maximum value of \(-\cos \theta\))
Maximum value of \(2 \sin 2 \theta - \cos \theta \le 2 + 1 = 3\).
So, the expression \(2 \sin 2 \theta - \cos \theta\) can take values up to a maximum of 3. It can never be equal to -4.
Since the left side of the equation \(2 \sin 2 \theta - \cos \theta = -4\) can never be equal to -4, the equation \(2 \sin 2 \theta - \cos \theta + 4 = 0\) is not possible for any value of \(\theta\), let alone for all \(\theta\).
Thus, statement 1 is incorrect.
Analyzing Statement 2: Trigonometry Inequality
The second statement is: \(\tan \theta + \cot \theta\) cannot be less than 2, where \(0 < \theta < \frac{\pi}{{2}}\).
The condition \(0 < \theta < \frac{\pi}{{2}}\) means \(\theta\) is in the first quadrant. In the first quadrant, both \(\tan \theta\) and \(\cot \theta\) are positive values.
We can use the Arithmetic Mean-Geometric Mean (AM-GM) inequality. For any two non-negative numbers \(a\) and \(b\), the AM-GM inequality states that \(\frac{{a+b}}{{2}} \ge \sqrt{{ab}}\).
Let \(a = \tan \theta\) and \(b = \cot \theta\). Since \(\theta\) is in the first quadrant, \(\tan \theta > 0\) and \(\cot \theta > 0\). Applying the AM-GM inequality:
\[\frac{{\tan \theta + \cot \theta}}{{2}} \ge \sqrt{{\tan \theta \cdot \cot \theta}}\]
We know that \(\tan \theta \cdot \cot \theta = \tan \theta \cdot \frac{{1}}{{\tan \theta}} = 1\), provided \(\tan \theta \ne 0\), which is true for \(0 < \theta < \frac{\pi}{{2}}\).
Substituting this into the inequality:
\[\frac{{\tan \theta + \cot \theta}}{{2}} \ge \sqrt{{1}}\]
\[\frac{{\tan \theta + \cot \theta}}{{2}} \ge 1\]
Multiplying both sides by 2:
\[\tan \theta + \cot \theta \ge 2\]
The equality \(\tan \theta + \cot \theta = 2\) holds when \(\tan \theta = \cot \theta\), which happens when \(\tan \theta = 1\) (since \(\theta\) is in the first quadrant), i.e., when \(\theta = \frac{{\pi}}{{4}}\).
For all other values of \(\theta\) in the range \(0 < \theta < \frac{\pi}{{2}}\), \(\tan \theta + \cot \theta > 2\).
Therefore, \(\tan \theta + \cot \theta\) is always greater than or equal to 2 for \(0 < \theta < \frac{\pi}{{2}}\). This means \(\tan \theta + \cot \theta\) cannot be less than 2.
Thus, statement 2 is correct.
Conclusion
Based on our analysis:
- Statement 1 is incorrect.
- Statement 2 is correct.
| Statement | Correctness |
|---|---|
| 1. \(2 \sin 2 \theta - \cos \theta + 4 = 0\) is possible for all \(\theta\) | Incorrect |
| 2. \(\tan \theta + \cot \theta\) cannot be less than 2, where \(0 < \theta < \frac{\pi}{{2}}\) | Correct |
The option that states only statement 2 is correct is the right answer.
Revision Table: Key Trigonometry Concepts
| Concept | Description / Formula |
|---|---|
| Range of \(\sin x\) | \([-1, 1]\) |
| Range of \(\cos x\) | \([-1, 1]\) |
| \(\sin 2 \theta\) identity | \(2 \sin \theta \cos \theta\) |
| \(\cot \theta\) definition | \(\frac{{1}}{{\tan \theta}}\) (for \(\tan \theta \ne 0\)) |
| AM-GM Inequality | For \(a, b \ge 0\), \(\frac{{a+b}}{{2}} \ge \sqrt{{ab}}\). Equality holds when \(a=b\). |
Additional Information: Proving Trigonometric Inequalities
Trigonometric inequalities can often be proven using various methods, including:
- Using basic trigonometric identities to simplify the expression.
- Applying standard inequalities like AM-GM, Cauchy-Schwarz, etc.
- Analyzing the range of trigonometric functions or combinations of functions.
- Using calculus (derivatives) to find minimum or maximum values of functions over a given interval.
In the case of \(\tan \theta + \cot \theta\) for \(\theta\) in the first quadrant, besides AM-GM, you could also use algebraic manipulation:
\[\tan \theta + \cot \theta = \tan \theta + \frac{{1}}{{\tan \theta}}\]
Let \(x = \tan \theta\). Since \(\theta\) is in the first quadrant, \(x > 0\). We want to show \(x + \frac{{1}}{{{x}}} \ge 2\). This is equivalent to \(x + \frac{{1}}{{{x}}} - 2 \ge 0\). Combine the terms: \(\frac{{x^2 + 1 - 2x}}{{x}} \ge 0\). The numerator is \(x^2 - 2x + 1 = (x-1)^2\). So we have \(\frac{{(x-1)^2}}{{x}} \ge 0\). Since \(x = \tan \theta > 0\) and \((x-1)^2 \ge 0\) for any real \(x\), the inequality \(\frac{{(x-1)^2}}{{x}} \ge 0\) is always true for \(x > 0\). Equality holds when \((x-1)^2 = 0\), i.e., \(x=1\), which means \(\tan \theta = 1\). This confirms that \(\tan \theta + \cot \theta \ge 2\) for \(0 < \theta < \frac{\pi}{{2}}\).
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