Which one is true ?

Options

A. `sin1 gtsin2 gt sin3`
B. `sin1 ltsin2ltsin3`
C. `sin1lt sin3ltsin2`
D. `sin3lt sin1 ltsin2`

Explanation

To solve the problem of finding the relationship between \(\sin 1\), \(\sin 2\), and \(\sin 3\), we will use the sine difference formula and analyze the values step by step.

### Step 1: Use the Sine Difference Formula
We start by applying the sine difference formula:
\[ \sin c - \sin d = 2 \sin\left(\frac{c - d}{2}\right) \cos\left(\frac{c + d}{2}\right) \]

### Step 2: Calculate \(\sin 2 - \sin 1\)
Let \(c = 2\) and \(d = 1\):
\[ \sin 2 - \sin 1 = 2 \sin\left(\frac{2 - 1}{2}\right) \cos\left(\frac{2 + 1}{2}\right) \]
This simplifies to:
\[ \sin 2 - \sin 1 = 2 \sin\left(\frac{1}{2}\right) \cos\left(\frac{3}{2}\right) \]

### Step 3: Analyze the Values
- The value \(\frac{1}{2}\) radians is approximately \(0.5\), which is between \(0\) and \(\frac{\pi}{2}\). Therefore, \(\sin\left(\frac{1}{2}\right)\) is positive.
- The value \(\frac{3}{2}\) radians is approximately \(1.5\), which is also between \(0\) and \(\frac{\pi}{2}\). However, since \(\frac{3}{2}\) is greater than \(\frac{\pi}{2}\), \(\cos\left(\frac{3}{2}\right)\) is negative.

Thus, we have:
\[ \sin 2 - \sin 1 = 2 \cdot \text{(positive)} \cdot \text{(negative)} = \text{(negative)} \]
This means:
\[ \sin 2 < \sin 1 \]

### Step 4: Calculate \(\sin 3 - \sin 1\)
Now, let’s find \(\sin 3 - \sin 1\):
\[ \sin 3 - \sin 1 = 2 \sin\left(\frac{3 - 1}{2}\right) \cos\left(\frac{3 + 1}{2}\right) \]
This simplifies to:
\[ \sin 3 - \sin 1 = 2 \sin(1) \cos(2) \]

### Step 5: Analyze the Values Again
- The value \(1\) radians is between \(0\) and \(\frac{\pi}{2}\), so \(\sin(1)\) is positive.
- The value \(2\) radians is between \(\frac{\pi}{2}\) and \(\pi\), so \(\cos(2)\) is negative.

Thus, we have:
\[ \sin 3 - \sin 1 = 2 \cdot \text{(positive)} \cdot \text{(negative)} = \text{(negative)} \]
This means:
\[ \sin 3 < \sin 1 \]

### Step 6: Conclusion
From the above analysis, we have:
1. \(\sin 2 < \sin 1\)
2. \(\sin 3 < \sin 1\)

Combining these results, we can conclude:
\[ \sin 3 < \sin 1 < \sin 2 \]

### Final Answer
Thus, the relationship is:
\[ \sin 3 < \sin 1 < \sin 2 \]