1. **Convert degrees to radians:** Use the formula $\text{radians} = \text{degrees} \times \frac{\pi}{180}$. 2. **Convert each angle:** - 23.1: $40^\circ = 40 \times \frac{\pi}{180} = \frac{2\pi}{9}$ (not matching given options) - 23.2: $120^\circ = 120 \times \frac{\pi}{180} = \frac{2\pi}{3}$ - 23.3: $-240^\circ = -240 \times \frac{\pi}{180} = -\frac{4\pi}{3}$ - 23.4: $-20^\circ = -20 \times \frac{\pi}{180} = -\frac{\pi}{9}$ 3. **Evaluate trigonometric expressions (25):** - 25.1: $\tan\left(\frac{4\pi}{3}\right) - \sin\left(\frac{13\pi}{6}\right) + \cos\left(\frac{11\pi}{3}\right)$ - $\tan\left(\frac{4\pi}{3}\right) = \tan\left(\pi + \frac{\pi}{3}\right) = \tan\left(\frac{\pi}{3}\right) = \sqrt{3}$ - $\sin\left(\frac{13\pi}{6}\right) = \sin\left(2\pi + \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$ - $\cos\left(\frac{11\pi}{3}\right) = \cos\left(2\pi + \frac{5\pi}{3}\right) = \cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}$ - Sum: $\sqrt{3} - \frac{1}{2} + \frac{1}{2} = \sqrt{3}$ - 25.2: $\sin\left(\frac{3\pi}{2}\right) + \tan\left(\frac{17\pi}{4}\right) - 2 \cos\left(-\frac{3\pi}{4}\right)$ - $\sin\left(\frac{3\pi}{2}\right) = -1$ - $\tan\left(\frac{17\pi}{4}\right) = \tan\left(4\pi + \frac{\pi}{4}\right) = \tan\left(\frac{\pi}{4}\right) = 1$ - $\cos\left(-\frac{3\pi}{4}\right) = \cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$ - Expression: $-1 + 1 - 2 \times \left(-\frac{\sqrt{2}}{2}\right) = 0 + \sqrt{2} = \sqrt{2}$ - 25.3: $2 \sin\left(\frac{11\pi}{3}\right) + \tan\left(-\frac{2\pi}{3}\right) + \cos(5\pi)$ - $\sin\left(\frac{11\pi}{3}\right) = \sin\left(2\pi + \frac{5\pi}{3}\right) = \sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}$ - $\tan\left(-\frac{2\pi}{3}\right) = -\tan\left(\frac{2\pi}{3}\right) = -(-\sqrt{3}) = \sqrt{3}$ - $\cos(5\pi) = \cos(\pi) = -1$ - Sum: $2 \times \left(-\frac{\sqrt{3}}{2}\right) + \sqrt{3} - 1 = -\sqrt{3} + \sqrt{3} - 1 = -1$ 4. **Compare values (26):** - 26.1: $\sin\left(\frac{2\pi}{3}\right) = \sin\left(\frac{5\pi}{6}\right) = \frac{\sqrt{3}}{2}$, so $=$ - 26.2: $\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, $\cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2}$, so $=$ - 26.3: $\tan\left(-\frac{2\pi}{3}\right) = \sqrt{3}$, $\tan\left(\frac{\pi}{3}\right) = \sqrt{3}$, so $=$ - 26.4: $\sin\left(\frac{13\pi}{6}\right) = \sin\left(2\pi + \frac{\pi}{6}\right) = \frac{1}{2}$, $\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}$, so $<$ 5. **Problem 27:** Find $\alpha \in [-\pi, 2\pi[$ such that $|\sin \alpha| = |\sin \frac{5\pi}{6}| = \frac{1}{2}$. 6. **Evaluate expressions (28):** - 28.1: $2 \cos(\pi - \alpha) + \sin(6\pi - \alpha) + \cos(-\alpha)$ - Use identities: $\cos(\pi - \alpha) = -\cos \alpha$, $\sin(6\pi - \alpha) = -\sin \alpha$, $\cos(-\alpha) = \cos \alpha$ - Expression: $2(-\cos \alpha) - \sin \alpha + \cos \alpha = -2\cos \alpha - \sin \alpha + \cos \alpha = -\cos \alpha - \sin \alpha$ - 28.2: $\tan(\pi + \alpha) + \frac{\sin(-\alpha)}{\cos(\pi + \alpha)}$ - $\tan(\pi + \alpha) = \tan \alpha$, $\sin(-\alpha) = -\sin \alpha$, $\cos(\pi + \alpha) = -\cos \alpha$ - Expression: $\tan \alpha + \frac{-\sin \alpha}{-\cos \alpha} = \tan \alpha + \frac{\sin \alpha}{\cos \alpha} = \tan \alpha + \tan \alpha = 2 \tan \alpha$ - 28.3: $\sin(7\pi + \alpha) - \tan(5\pi - \alpha) - \sin(-\alpha)$ - $\sin(7\pi + \alpha) = -\sin \alpha$, $\tan(5\pi - \alpha) = -\tan \alpha$, $\sin(-\alpha) = -\sin \alpha$ - Expression: $-\sin \alpha - (-\tan \alpha) - (-\sin \alpha) = -\sin \alpha + \tan \alpha + \sin \alpha = \tan \alpha$ - 28.4: $\sin^2(\pi - \alpha) + \cos^2(-\alpha) - \sin(5\pi + \alpha)$ - $\sin(\pi - \alpha) = \sin \alpha$, $\cos(-\alpha) = \cos \alpha$, $\sin(5\pi + \alpha) = -\sin \alpha$ - Expression: $\sin^2 \alpha + \cos^2 \alpha - (-\sin \alpha) = 1 + \sin \alpha$ 7. **Problem 29:** Given $\alpha \in [-\frac{\pi}{2}, \frac{\pi}{2}]$ with $\sin \alpha = -\frac{3}{5}$, find exact value of $-2 \tan(-\alpha) \times \sin(\alpha - \pi)$. - $\tan(-\alpha) = -\tan \alpha$. - $\sin(\alpha - \pi) = -\sin \alpha$. - $\tan \alpha = \frac{\sin \alpha}{\cos \alpha} = \frac{-3/5}{\sqrt{1 - (-3/5)^2}} = \frac{-3/5}{4/5} = -\frac{3}{4}$. - So $\tan(-\alpha) = -(-\frac{3}{4}) = \frac{3}{4}$. - $\sin(\alpha - \pi) = -(-\frac{3}{5}) = \frac{3}{5}$. - Expression: $-2 \times \frac{3}{4} \times \frac{3}{5} = -2 \times \frac{9}{20} = -\frac{18}{20} = -\frac{9}{10}$. 8. **Problem 30:** Given unit circle with angles $\alpha, \beta$. - 30.1: $\sin(\pi - \beta) + \sin(-\alpha) < 0$? - $\sin(\pi - \beta) = \sin \beta$, $\sin(-\alpha) = -\sin \alpha$. - Expression: $\sin \beta - \sin \alpha < 0$ depends on $\alpha, \beta$. - 30.2: $\tan(\pi - \alpha) \times \tan(-\beta) > 0$? - $\tan(\pi - \alpha) = -\tan \alpha$, $\tan(-\beta) = -\tan \beta$. - Product: $(-\tan \alpha)(-\tan \beta) = \tan \alpha \tan \beta > 0$ if $\tan \alpha$ and $\tan \beta$ have same sign. - 30.3: $\cos(\alpha - \pi) \times \sin(\pi + \beta) > 0$? - $\cos(\alpha - \pi) = -\cos \alpha$, $\sin(\pi + \beta) = -\sin \beta$. - Product: $(-\cos \alpha)(-\sin \beta) = \cos \alpha \sin \beta > 0$ depends on signs. - 30.4: $\cos(\pi - \alpha) + \cos(-\beta) > 0$? - $\cos(\pi - \alpha) = -\cos \alpha$, $\cos(-\beta) = \cos \beta$. - Sum: $-\cos \alpha + \cos \beta > 0$ depends. - 30.5: $\tan(\pi + \alpha) - \tan(2\pi + \beta) < 0$? - $\tan(\pi + \alpha) = \tan \alpha$, $\tan(2\pi + \beta) = \tan \beta$. - Expression: $\tan \alpha - \tan \beta < 0$ depends. 9. **Problem 31:** - 31.1: $\cos(5\pi + \alpha) < 0$ and $\sin(-\alpha) > 0$ - $\cos(5\pi + \alpha) = \cos(\pi + \alpha) = -\cos \alpha < 0 \Rightarrow \cos \alpha > 0$ - $\sin(-\alpha) = -\sin \alpha > 0 \Rightarrow \sin \alpha < 0$ - So $\cos \alpha > 0$, $\sin \alpha < 0$ means $\alpha$ in 4th quadrant. - 31.2: $\cos(-\alpha) < 0$ and $\tan(\pi + \alpha) > 0$ - $\cos(-\alpha) = \cos \alpha < 0$ - $\tan(\pi + \alpha) = \tan \alpha > 0$ - So $\cos \alpha < 0$, $\tan \alpha > 0$ means $\alpha$ in 3rd quadrant. Final answers are detailed above for each problem.