Statement I: The equation (sin−1x)3+(cos−1x)3−aπ3=0 has a solution for all a≥321.
Statement II: For any x∈R, sin−1x+cos−1x=2π and 0≤(sin−1x−4π)2≤169π2.
Solutions · 1
Official solution
Since sin−1x∈[−2π,2π], −43π≤sin−1x−4π≤4π, so 0≤(sin−1x−4π)2≤169π2 --- Statement II is true.
For Statement I: using (sin−1x)3+(cos−1x)3=(sin−1x+cos−1x)[(sin−1x+cos−1x)2−3sin−1xcos−1x],
\[=\frac{\pi}{2}\left[\frac{\pi^2}{4}-3\sin^{-1}x\left(\frac{\pi}{2}-\sin^{-1}x\right)\right]=a\pi^3\ \Rightarrow\ \left(\sin^{-1}x-\frac{\pi}{4}\right)^2=\frac{\pi^2}{48}(32a-1).\]
Substituting the bound 0≤(sin−1x−4π)2≤169π2 gives 0≤32a−1≤27⇒321≤a≤87, confirming a solution exists for a in this range --- Statement I is true (as keyed).
Both statements are true.