1. **Problem statement:** Simplify the 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) $\arcsin(\cos(\frac{17\pi}{5}))$, $\arccos(\sin(\frac{17\pi}{5}))$, $\arctan(\frac{1}{\tan(\frac{17\pi}{5})})$ 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 $\arcsin(\sin x)$ and $\arctan(\tan x)$, the output is the unique angle in the principal range equivalent to $x$ modulo $2\pi$. - For $\arccos(\cos x)$, the output is the unique angle in $[0, \pi]$ equivalent to $x$ modulo $2\pi$. 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 $\frac{2\pi}{5} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, it equals $\frac{2\pi}{5}$. 4. **Simplify (ii):** - $\frac{17\pi}{5} = 3\pi + \frac{2\pi}{5}$ (since $17/5=3.4$) - Reduce modulo $2\pi$: $3\pi + \frac{2\pi}{5} - 2\pi = \pi + \frac{2\pi}{5} = \frac{7\pi}{5} \approx 4.398$. - For $\arcsin(\sin(\frac{17\pi}{5}))$, find equivalent angle in $[-\frac{\pi}{2}, \frac{\pi}{2}]$: - $\sin(\frac{17\pi}{5}) = \sin(\frac{7\pi}{5}) = -\sin(\frac{3\pi}{5})$ - $\arcsin(\sin(\frac{17\pi}{5})) = -\frac{3\pi}{5}$ - For $\arccos(\cos(\frac{17\pi}{5}))$, since $\cos(\frac{17\pi}{5})=\cos(\frac{7\pi}{5})$, and $\frac{7\pi}{5} > \pi$, use $2\pi - \frac{7\pi}{5} = \frac{3\pi}{5}$ - So $\arccos(\cos(\frac{17\pi}{5})) = \frac{3\pi}{5}$ - For $\arctan(\tan(\frac{17\pi}{5}))$, reduce $\frac{17\pi}{5}$ modulo $\pi$ (period of tan is $\pi$): - $\frac{17\pi}{5} - 3\pi = \frac{2\pi}{5}$ - Since $\frac{2\pi}{5} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, $\arctan(\tan(\frac{17\pi}{5})) = \frac{2\pi}{5}$ 5. **Simplify (iii):** - $-\frac{17\pi}{5} = -3\pi - \frac{2\pi}{5}$ - For $\arcsin(\sin(-\frac{17\pi}{5}))$: - $\sin(-x) = -\sin x$, so $\sin(-\frac{17\pi}{5}) = -\sin(\frac{17\pi}{5}) = -\sin(\frac{7\pi}{5}) = \sin(\frac{3\pi}{5})$ - Since $\frac{3\pi}{5} \in [-\frac{\pi}{2}, \frac{\pi}{2}]$? No, $\frac{3\pi}{5} > \frac{\pi}{2}$, so use $\pi - \frac{3\pi}{5} = \frac{2\pi}{5}$ but with sign consideration, the principal value is $\arcsin(\sin(-\frac{17\pi}{5})) = \frac{2\pi}{5}$ - For $\arccos(\cos(-\frac{17\pi}{5}))$: - $\cos$ is even, so $\cos(-\frac{17\pi}{5}) = \cos(\frac{17\pi}{5}) = \cos(\frac{7\pi}{5})$ - As before, $\arccos(\cos(\frac{7\pi}{5})) = \frac{3\pi}{5}$ - For $\arctan(\tan(-\frac{17\pi}{5}))$: - $\tan$ is odd, so $\tan(-\frac{17\pi}{5}) = -\tan(\frac{17\pi}{5})$ - $\arctan(\tan(-\frac{17\pi}{5})) = -\arctan(\tan(\frac{17\pi}{5})) = -\frac{2\pi}{5}$ 6. **Simplify (iv):** - $\arcsin(\cos(\frac{17\pi}{5}))$: - $\cos(\frac{17\pi}{5}) = \cos(\frac{7\pi}{5}) = -\cos(\frac{3\pi}{5})$ - $\cos(\frac{3\pi}{5}) = \cos(108^\circ) \approx -0.3090$, so $\cos(\frac{7\pi}{5}) \approx -(-0.3090) = 0.3090$ - $\arcsin(0.3090) \approx 0.314$ (which is $\frac{\pi}{10}$) - $\arccos(\sin(\frac{17\pi}{5}))$: - $\sin(\frac{17\pi}{5}) = \sin(\frac{7\pi}{5}) = -\sin(\frac{3\pi}{5}) \approx -0.9511$ - $\arccos(-0.9511) \approx 2.827$ (which is $\pi - \arcsin(0.9511) = \pi - \frac{3\pi}{10} = \frac{7\pi}{10}$) - $\arctan(\frac{1}{\tan(\frac{17\pi}{5})})$: - $\frac{1}{\tan x} = \cot x = \tan(\frac{\pi}{2} - x)$ - So $\arctan(\frac{1}{\tan(\frac{17\pi}{5})}) = \arctan(\tan(\frac{\pi}{2} - \frac{17\pi}{5}))$ - Simplify inside: $\frac{\pi}{2} - \frac{17\pi}{5} = \frac{5\pi}{10} - \frac{34\pi}{10} = -\frac{29\pi}{10}$ - Reduce modulo $\pi$: $-\frac{29\pi}{10} + 3\pi = -\frac{29\pi}{10} + \frac{30\pi}{10} = \frac{\pi}{10}$ - Since $\frac{\pi}{10} \in (-\frac{\pi}{2}, \frac{\pi}{2})$, the value is $\frac{\pi}{10}$ **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) $\arcsin(\sin(\frac{17\pi}{5})) = -\frac{3\pi}{5}$, $\arccos(\cos(\frac{17\pi}{5})) = \frac{3\pi}{5}$, $\arctan(\tan(\frac{17\pi}{5})) = \frac{2\pi}{5}$ (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) $\arcsin(\cos(\frac{17\pi}{5})) = \frac{\pi}{10}$, $\arccos(\sin(\frac{17\pi}{5})) = \frac{7\pi}{10}$, $\arctan(\frac{1}{\tan(\frac{17\pi}{5})}) = \frac{\pi}{10}$