Question 78 of 155MediumMCQ
If (sin−1x)2−(cos−1x)2=a, 0<x<1, a=0, then the value of 2x2−1 is Solutions · 1
Official solution
\[(\sin^{-1}x+\cos^{-1}x)(\sin^{-1}x-\cos^{-1}x)=a\ \Rightarrow\ \frac{\pi}{2}(\sin^{-1}x-\cos^{-1}x)=a\ \Rightarrow\ \sin^{-1}x-\cos^{-1}x=\frac{2a}{\pi}.\]
Combined with sin−1x+cos−1x=2π: cos−1x=4π−πa, so 2cos−1x=2π−π2a, giving
\[\cos^{-1}(2x^2-1)=2\cos^{-1}x=\frac{\pi}{2}-\frac{2a}{\pi}\ \Rightarrow\ 2x^2-1=\cos\left(\frac{\pi}{2}-\frac{2a}{\pi}\right)=\sin\left(\frac{2a}{\pi}\right).\] Answer the question to unlock