Cho sin + cos = 1/3, tính sin - cos

To solve for \(\sin x - \cos x\) given that \(\sin x + \cos x = \frac{1}{3}\), we can use the following steps:

1. **Square both sides of the given equation:**
\[
(\sin x + \cos x)^2 = \left(\frac{1}{3}\right)^2
\]
\[
\sin^2 x + 2\sin x \cos x + \cos^2 x = \frac{1}{9}
\]

2. **Use the Pythagorean identity:**
\[
\sin^2 x + \cos^2 x = 1
\]
Substitute this into the equation:
\[
1 + 2\sin x \cos x = \frac{1}{9}
\]

3. **Solve for \(2\sin x \cos x\):**
\[
2\sin x \cos x = \frac{1}{9} - 1
\]
\[
2\sin x \cos x = \frac{1}{9} - \frac{9}{9}
\]
\[
2\sin x \cos x = -\frac{8}{9}
\]
\[
\sin x \cos x = -\frac{4}{9}
\]

4. **Express \(\sin x - \cos x\) in terms of \(\sin x + \cos x\):**
\[
(\sin x - \cos x)^2 = \sin^2 x - 2\sin x \cos x + \cos^2 x
\]
Using the Pythagorean identity again:
\[
(\sin x - \cos x)^2 = 1 - 2\sin x \cos x
\]
Substitute \(\sin x \cos x = -\frac{4}{9}\):
\[
(\sin x - \cos x)^2 = 1 - 2\left(-\frac{4}{9}\right)
\]
\[
(\sin x - \cos x)^2 = 1 + \frac{8}{9}
\]
\[
(\sin x - \cos x)^2 = \frac{9}{9} + \frac{8}{9}
\]
\[
(\sin x - \cos x)^2 = \frac{17}{9}
\]

5. **Take the square root of both sides:**
\[
\sin x - \cos x = \pm \sqrt{\frac{17}{9}}
\]
\[
\sin x - \cos x = \pm \frac{\sqrt{17}}{3}
\]

Therefore, \(\sin x - \cos x = \pm \frac{\sqrt{17}}{3}\).