To prove that \(\sin(x) + \cos(x) > \sin(2x)\) for \(0^\circ < x < 90^\circ\), we'll use trigonometric identities and analysis.

### Step-by-Step Proof:

#### 1. Express \(\sin(2x)\) in terms of \(\sin(x)\) and \(\cos(x)\):
We know the double-angle identity for sine:
\[
\sin(2x) = 2\sin(x)\cos(x)
\]
So, the inequality we need to prove becomes:
\[
\sin(x) + \cos(x) > 2\sin(x)\cos(x)
\]

#### 2. Rearrange the inequality:
Move all terms to one side:
\[
\sin(x) + \cos(x) - 2\sin(x)\cos(x) > 0
\]
Factor the left-hand side:
\[
(\sin(x) + \cos(x))(1 - \sin(x)\cos(x)) > 0
\]
We now need to show that \((\sin(x) + \cos(x))(1 - \sin(x)\cos(x)) > 0\) for \(0^\circ < x < 90^\circ\).

#### 3. Analyze each factor:
- \(\sin(x) + \cos(x)\) is always positive for \(0^\circ < x < 90^\circ\), because both \(\sin(x)\) and \(\cos(x)\) are positive in this interval.
  
- Now, consider \(1 - \sin(x)\cos(x)\). Since \(0^\circ < x < 90^\circ\), both \(\sin(x)\) and \(\cos(x)\) are less than or equal to 1, so:
  \[
  \sin(x)\cos(x) \leq \frac{1}{2}
  \]
  The maximum value of \(\sin(x)\cos(x)\) occurs at \(x = 45^\circ\), where \(\sin(45^\circ) = \cos(45^\circ) = \frac{\sqrt{2}}{2}\), giving:
  \[
  \sin(45^\circ)\cos(45^\circ) = \frac{1}{2}
  \]
  Therefore, \(1 - \sin(x)\cos(x)\) is always positive and reaches its minimum value of \(0.5\) at \(x = 45^\circ\). Hence, for all \(0^\circ < x < 90^\circ\), we have:
  \[
  1 - \sin(x)\cos(x) > 0
  \]

#### 4. Conclusion:
Both \(\sin(x) + \cos(x)\) and \(1 - \sin(x)\cos(x)\) are positive for \(0^\circ < x < 90^\circ\). Therefore, the product of these two expressions is positive, meaning:
\[
(\sin(x) + \cos(x))(1 - \sin(x)\cos(x)) > 0
\]
Thus, \(\sin(x) + \cos(x) > \sin(2x)\) for all \(0^\circ < x < 90^\circ\), as required.

---

Would you like further details or have any questions? Here are 5 related questions to expand on this:

1. What are the values of \(\sin(x) + \cos(x)\) and \(\sin(2x)\) at specific points like \(x = 30^\circ\) or \(x = 60^\circ\)?
2. Can you graph the functions \(\sin(x) + \cos(x)\) and \(\sin(2x)\) to visually confirm the inequality?
3. How would the inequality change if the domain were different, such as \(90^\circ < x < 180^\circ\)?
4. Can you prove \(\sin(x) + \cos(x) \leq \sqrt{2}\) for all \(x\)?
5. How can you extend this inequality to other trigonometric identities like \(\cos(2x)\)?

**Tip:** Always look for trigonometric identities when dealing with inequalities or expressions involving multiple angles!

Prove that sinx + cosx > sin2x, 0 < x < 90°

Prove the inequality sin(x) + cos(x) > sin(2x) for the interval 0° < x < 90° using trigonometric identities and analysis. This detailed step-by-step solution demonstrates the application of the double-angle identity and explores the positivity of trigonometric expressions in the first quadrant. Suitable for students in Grades 10-12.

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