If sin β = 5/13 and 0° ≤ β ≤ 90°, than find without using tables or calculators cos β - 2 sin β / 2 tan β

Given $\sin \beta = \frac{5}{13}$ and $0^\circ \le \beta \le 90^\circ$. We use the Pythagorean identity $\sin^2 \beta + \cos^2 \beta = 1$.

$$\left(\frac{5}{13}\right)^2 + \cos^2 \beta = 1$$ $$\frac{25}{169} + \cos^2 \beta = 1$$ $$\cos^2 \beta = 1 - \frac{25}{169}$$ $$\cos^2 \beta = \frac{169 - 25}{169}$$ $$\cos^2 \beta = \frac{144}{169}$$

Since $0^\circ \le \beta \le 90^\circ$, $\cos \beta$ is positive.

$$\cos \beta = \sqrt{\frac{144}{169}}$$ $$\cos \beta = \frac{12}{13}$$

We use the identity $\tan \beta = \frac{\sin \beta}{\cos \beta}$.

$$\tan \beta = \frac{\frac{5}{13}}{\frac{12}{13}}$$ $$\tan \beta = \frac{5}{12}$$

The numerator is $\cos \beta - 2 \sin \beta$.

$$\cos \beta - 2 \sin \beta = \frac{12}{13} - 2\left(\frac{5}{13}\right)$$ $$= \frac{12}{13} - \frac{10}{13}$$ $$= \frac{2}{13}$$

The denominator is $2 \tan \beta$.

$$2 \tan \beta = 2\left(\frac{5}{12}\right)$$ $$= \frac{10}{12}$$ $$= \frac{5}{6}$$

Now, divide the numerator by the denominator.

$$\frac{\cos \beta - 2 \sin \beta}{2 \tan \beta} = \frac{\frac{2}{13}}{\frac{5}{6}}$$ $$= \frac{2}{13} \times \frac{6}{5}$$ $$= \frac{12}{65}$$

The value of the expression is $\frac{12}{65}$.