Mathematics - Inverse Trigonometric Functions Question with Solution | TestHub

We use the properties ${\sin }^{-1}(\sin x)=x-2n\pi$, ${\cos }^{-1}(\cos x)=x-2n\pi$ or $2n\pi -x$, and ${\tan }^{-1}(\tan x)=x-n\pi$, such that the result lies in the principal value branch. For ${\sin }^{-1}(\sin 5)$: Since $3\pi /2<5<2\pi$, we have $\sin 5<0$. The principal value branch is $[-\pi /2,\pi /2]$. We choose $y=5-2\pi \approx 5-6.28=-1.28$, which is in $[-\pi /2,\pi /2]$. So ${\sin }^{-1}(\sin 5)=5-2\pi$. For ${\cos }^{-1}(\cos 10)$: Since $3\pi <10<4\pi$, we have $\cos 10>0$. The principal value branch is $[0,\pi ]$. We choose $y=4\pi -10\approx 12.56-10=2.56$, which is in $[0,\pi ]$. So ${\cos }^{-1}(\cos 10)=4\pi -10$. For ${\tan }^{-1}(\tan 15)$: Since $4\pi <15<5\pi$, we have $\tan 15<0$. The principal value branch is $(-\pi /2,\pi /2)$. We choose $y=15-5\pi \approx 15-15.7=-0.7$, which is in $(-\pi /2,\pi /2)$. So ${\tan }^{-1}(\tan 15)=15-5\pi$. Summing these values: $(5-2\pi )+(4\pi -10)+(15-5\pi )=(5-10+15)+(-2\pi +4\pi -5\pi )=10-3\pi$.