If sin θ cos θ = k, where \(0 \le \theta \le \frac{\pi }{2}\) , then which one of the following is correct?
0 ≤ k ≤ 0.5 only
The question asks for the correct range of the expression \(k = \sin \theta \cos \theta\) given that the angle \(\theta\) is restricted to the interval \(0 \le \theta \le \frac{\pi }{2}\).
Finding the Range of sin θ cos θ
We are given the expression \(k = \sin \theta \cos \theta\). To find its range, we can use a standard trigonometric identity that relates the product of sine and cosine to a single sine function.
Recall the double angle identity for sine:
\(\sin(2\theta) = 2 \sin \theta \cos \theta\)
We can rearrange this identity to express \(\sin \theta \cos \theta\) in terms of \(\sin(2\theta)\):
\(\sin \theta \cos \theta = \frac{1}{2} \sin(2\theta)\)
So, the expression for \(k\) can be written as:
\(k = \frac{1}{2} \sin(2\theta)\)
Determining the Range based on θ
The given range for \(\theta\) is \(0 \le \theta \le \frac{\pi }{2}\).
To find the range of \(k\), we first need to find the range of \(2\theta\). We can multiply the inequality by 2:
\(2 \times 0 \le 2 \times \theta \le 2 \times \frac{\pi }{2}\)
\(0 \le 2\theta \le \pi\)
Now we need to determine the range of \(\sin(2\theta)\) when \(2\theta\) is in the interval \([0, \pi]\). The sine function starts at 0 at \(0\), increases to its maximum value of 1 at \(\frac{\pi}{2}\), and then decreases back to 0 at \(\pi\). Therefore, for \(0 \le 2\theta \le \pi\), the range of \(\sin(2\theta)\) is:
\(0 \le \sin(2\theta) \le 1\)
Finding the Range of k
Since \(k = \frac{1}{2} \sin(2\theta)\), we can find the range of \(k\) by multiplying the inequality for \(\sin(2\theta)\) by \(\frac{1}{2}\):
\(\frac{1}{2} \times 0 \le \frac{1}{2} \sin(2\theta) \le \frac{1}{2} \times 1\)
\(0 \le k \le 0.5\)
Let's check the values of \(k\) at the boundaries of the \(\theta\) range:
- When \(\theta = 0\), \(k = \sin(0) \cos(0) = 0 \times 1 = 0\).
- When \(\theta = \frac{\pi}{2}\), \(k = \sin(\frac{\pi}{2}) \cos(\frac{\pi}{2}) = 1 \times 0 = 0\).
The maximum value occurs when \(2\theta = \frac{\pi}{2}\), which means \(\theta = \frac{\pi}{4}\).
- When \(\theta = \frac{\pi}{4}\), \(k = \sin(\frac{\pi}{4}) \cos(\frac{\pi}{4}) = \frac{1}{\sqrt{2}} \times \frac{1}{\sqrt{2}} = \frac{1}{2} = 0.5\).
So, the minimum value of \(k\) is 0 and the maximum value is 0.5. The range of \(k\) is indeed \(0 \le k \le 0.5\).
Comparing with Options
Let's compare our derived range \(0 \le k \le 0.5\) with the given options:
- Option 1: \(0 \le k \le 1\) - This range is too wide.
- Option 2: \(0 \le k \le 0.5\) only - This matches our calculated range exactly.
- Option 3: \(0.5 \le k \le 1\) only - This range is incorrect as it starts from the maximum value.
- Option 4: \(0 < k < 1\) - This range excludes the endpoints 0 and 0.5, which are included in the range of k.
The correct range for \(k\) is \(0 \le k \le 0.5\).
Revision Table: Trigonometric Identities and Range
| Concept | Description | Example/Formula |
|---|---|---|
| Double Angle Identity for Sine | Relates the sine of twice an angle to the product of sine and cosine of the angle. | \(\sin(2\theta) = 2 \sin \theta \cos \theta\) |
| Range of Sine Function | The values that \(\sin(x)\) can take for a given range of \(x\). | For \(0 \le x \le \pi\), the range of \(\sin(x)\) is \([0, 1]\). |
| Range of \(k\) | The set of possible values for \(k = \frac{1}{2} \sin(2\theta)\) when \(0 \le \theta \le \frac{\pi}{2}\). | \(0 \le k \le 0.5\) |
Additional Information: Understanding Range of Functions
The range of a function is the set of all possible output values it can produce given its domain (the allowed input values). In this problem, the domain of the function \(k(\theta) = \sin \theta \cos \theta\) is restricted to \(0 \le \theta \le \frac{\pi}{2}\).
By transforming the expression for \(k\) into \(k = \frac{1}{2} \sin(2\theta)\), we made it easier to determine the range because the range of the standard sine function \(\sin(x)\) for different intervals of \(x\) is well-known.
For \(x\) in the interval \([0, \pi]\), the sine function starts at 0, goes up to a maximum of 1 at \(x=\frac{\pi}{2}\), and comes back down to 0 at \(x=\pi\). So, its range is \([0, 1]\).
Since \(k\) is half of \(\sin(2\theta)\), the range of \(k\) is half the range of \(\sin(2\theta)\), which is \([\frac{1}{2} \times 0, \frac{1}{2} \times 1]\) or \([0, 0.5]\).
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