1. **Problem statement:** Simplify the following expressions involving inverse trigonometric functions: (i) $\arcsin(\sin(\frac{2\pi}{5}))$, $\arccos(\cos(\frac{2\pi}{5}))$, $\arctan(\tan(\frac{2\pi}{5}))$ (ii) $\arcsin(\sin(\frac{17\pi}{5}))$, $\arccos(\cos(\frac{17\pi}{5}))$, $\arctan(\tan(\frac{17\pi}{5}))$ (iii) $\arcsin(\sin(-\frac{17\pi}{5}))$, $\arccos(\cos(-\frac{17\pi}{5}))$, $\arctan(\tan(-\frac{17\pi}{5}))$ (iv) $\arcsin(\cos(\frac{17\pi}{5}))$, $\arccos(\sin(\frac{17\pi}{5}))$, $\arctan\left(\frac{1}{\tan(\frac{17\pi}{5})}\right)$ 2. **Key formulas and ranges:** - $\arcsin(x)$ returns values in $[-\frac{\pi}{2}, \frac{\pi}{2}]$ - $\arccos(x)$ returns values in $[0, \pi]$ - $\arctan(x)$ returns values in $(-\frac{\pi}{2}, \frac{\pi}{2})$ For simplification, use periodicity and principal value ranges: - $\sin(\theta)$ is $2\pi$ periodic - $\cos(\theta)$ is $2\pi$ periodic - $\tan(\theta)$ is $\pi$ periodic 3. **Simplify (i):** - $\frac{2\pi}{5} \approx 1.2566$ lies in $[0, \pi]$ and also in $[-\frac{\pi}{2}, \frac{\pi}{2}]$? No, since $\frac{\pi}{2} \approx 1.5708$, $\frac{2\pi}{5} < \frac{\pi}{2}$, so: - $\arcsin(\sin(\frac{2\pi}{5})) = \frac{2\pi}{5}$ - $\arccos(\cos(\frac{2\pi}{5})) = \frac{2\pi}{5}$ - For $\arctan(\tan(\frac{2\pi}{5}))$, since $\tan$ is $\pi$ periodic and $\frac{2\pi}{5} < \frac{\pi}{2}$, the principal value is $\frac{2\pi}{5}$ 4. **Simplify (ii):** - $\frac{17\pi}{5} = 3\pi + \frac{2\pi}{5}$ (since $3\pi = \frac{15\pi}{5}$) - Use periodicity: - $\sin(\frac{17\pi}{5}) = \sin(3\pi + \frac{2\pi}{5}) = -\sin(\frac{2\pi}{5})$ - $\arcsin(\sin(\frac{17\pi}{5})) = \arcsin(-\sin(\frac{2\pi}{5})) = -\arcsin(\sin(\frac{2\pi}{5})) = -\frac{2\pi}{5}$ (since $-\frac{2\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$) - $\cos(\frac{17\pi}{5}) = \cos(3\pi + \frac{2\pi}{5}) = -\cos(\frac{2\pi}{5})$ - $\arccos(\cos(\frac{17\pi}{5})) = \arccos(-\cos(\frac{2\pi}{5})) = \pi - \frac{2\pi}{5} = \frac{3\pi}{5}$ - For $\arctan(\tan(\frac{17\pi}{5}))$, since $\tan$ is $\pi$ periodic: - $\frac{17\pi}{5} = 3\pi + \frac{2\pi}{5}$ - $\tan(\frac{17\pi}{5}) = \tan(\frac{2\pi}{5})$ - Principal value of $\arctan(\tan(\frac{17\pi}{5}))$ is $\frac{2\pi}{5}$ (since $\frac{2\pi}{5} < \frac{\pi}{2}$) 5. **Simplify (iii):** - $-\frac{17\pi}{5} = -3\pi - \frac{2\pi}{5}$ - $\sin(-\frac{17\pi}{5}) = -\sin(\frac{17\pi}{5}) = -(-\sin(\frac{2\pi}{5})) = \sin(\frac{2\pi}{5})$ - $\arcsin(\sin(-\frac{17\pi}{5})) = \arcsin(\sin(\frac{2\pi}{5})) = \frac{2\pi}{5}$ - $\cos(-\frac{17\pi}{5}) = \cos(\frac{17\pi}{5}) = -\cos(\frac{2\pi}{5})$ - $\arccos(\cos(-\frac{17\pi}{5})) = \arccos(-\cos(\frac{2\pi}{5})) = \frac{3\pi}{5}$ - $\tan(-\frac{17\pi}{5}) = -\tan(\frac{17\pi}{5}) = -\tan(\frac{2\pi}{5})$ - $\arctan(\tan(-\frac{17\pi}{5})) = \arctan(-\tan(\frac{2\pi}{5})) = -\frac{2\pi}{5}$ 6. **Simplify (iv):** - $\arcsin(\cos(\frac{17\pi}{5}))$: - $\cos(\frac{17\pi}{5}) = -\cos(\frac{2\pi}{5})$ - Numerically, $\cos(\frac{2\pi}{5}) \approx 0.3090$, so $\cos(\frac{17\pi}{5}) \approx -0.3090$ - $\arcsin(-0.3090) \approx -0.3142$ (since $\arcsin$ range is $[-\frac{\pi}{2}, \frac{\pi}{2}]$) - $\arccos(\sin(\frac{17\pi}{5}))$: - $\sin(\frac{17\pi}{5}) = -\sin(\frac{2\pi}{5}) \approx -0.9511$ - $\arccos(-0.9511) \approx 2.823$ (in $[0, \pi]$) - $\arctan\left(\frac{1}{\tan(\frac{17\pi}{5})}\right)$: - $\tan(\frac{17\pi}{5}) = \tan(\frac{2\pi}{5}) \approx 1.3764$ - $\frac{1}{1.3764} \approx 0.7265$ - $\arctan(0.7265) \approx 0.628$ (in $(-\frac{\pi}{2}, \frac{\pi}{2})$) **Final answers:** (i) $\arcsin(\sin(\frac{2\pi}{5})) = \frac{2\pi}{5}$, $\arccos(\cos(\frac{2\pi}{5})) = \frac{2\pi}{5}$, $\arctan(\tan(\frac{2\pi}{5})) = \frac{2\pi}{5}$ (ii) $\arcsin(\sin(\frac{17\pi}{5})) = -\frac{2\pi}{5}$, $\arccos(\cos(\frac{17\pi}{5})) = \frac{3\pi}{5}$, $\arctan(\tan(\frac{17\pi}{5})) = \frac{2\pi}{5}$ (iii) $\arcsin(\sin(-\frac{17\pi}{5})) = \frac{2\pi}{5}$, $\arccos(\cos(-\frac{17\pi}{5})) = \frac{3\pi}{5}$, $\arctan(\tan(-\frac{17\pi}{5})) = -\frac{2\pi}{5}$ (iv) $\arcsin(\cos(\frac{17\pi}{5})) \approx -0.3142$, $\arccos(\sin(\frac{17\pi}{5})) \approx 2.823$, $\arctan\left(\frac{1}{\tan(\frac{17\pi}{5})}\right) \approx 0.628$