If `y=sinnx+cosnx` (x,n real0 then `-sqrt(2)leylesqrt(2)`.
a. True b. False

Answer

To solve the problem, we need to analyze the expression \( y = \sin(nx) + \cos(nx) \) and prove that \( -\sqrt{2} \leq y \leq \sqrt{2} \).

### Step-by-Step Solution:

1. **Start with the expression**:
\[ y = \sin(nx) + \cos(nx) \]

2. **Square both sides**:
\[ y^2 = (\sin(nx) + \cos(nx))^2 \]

3. **Expand the squared expression**:
\[ y^2 = \sin^2(nx) + \cos^2(nx) + 2\sin(nx)\cos(nx) \]

4. **Use the Pythagorean identity**:
\[ \sin^2(nx) + \cos^2(nx) = 1 \]
Therefore, we can rewrite \( y^2 \):
\[ y^2 = 1 + 2\sin(nx)\cos(nx) \]

5. **Use the double angle identity**:
\[ 2\sin(nx)\cos(nx) = \sin(2nx) \]
Thus, we have:
\[ y^2 = 1 + \sin(2nx) \]

6. **Determine the range of \( \sin(2nx) \)**:
The sine function oscillates between -1 and 1:
\[ -1 \leq \sin(2nx) \leq 1 \]

7. **Substitute the range of \( \sin(2nx) \) into the equation for \( y^2 \)**:
\[ 1 - 1 \leq y^2 \leq 1 + 1 \]
This simplifies to:
\[ 0 \leq y^2 \leq 2 \]

8. **Take the square root**:
Since \( y^2 \) is non-negative, we can take the square root:
\[ -\sqrt{2} \leq y \leq \sqrt{2} \]

9. **Conclusion**:
We have shown that:
\[ -\sqrt{2} \leq y \leq \sqrt{2} \]
Therefore, the statement is **True**.