JEE Main 2022 — Mathematics Trigonometry

Given $x=\mathrm{sin}(2{\mathrm{tan}}^{-1}\alpha )=\mathrm{sin}({\mathrm{tan}}^{-1}\frac{2\alpha }{1-{\alpha }^{2}})=\mathrm{sin}({\mathrm{sin}}^{-1}\frac{2\alpha }{1+{\alpha }^{2}})=\frac{2\alpha }{1+{\alpha }^{2}}\cdots (i)$x=sin(2tan−1α)=sin(tan−11−α22α​)=sin(sin−11+α22α​)=1+α22α​⋯(i)

and $y=\mathrm{sin}(\frac{1}{2}{\mathrm{tan}}^{-1}\frac{4}{3})=\mathrm{sin}({\mathrm{tan}}^{-1}\frac{1}{2})=\mathrm{sin}({\mathrm{sin}}^{-1}\frac{1}{\sqrt{5}})=\frac{1}{\sqrt{5}}...(\mathrm{ii})$y=sin(21​tan−134​)=sin(tan−121​)=sin(sin−15​1​)=5​1​...(ii)

Now, ${y}^{2}=1-x$y2=1−x

$\Rightarrow \frac{1}{5}=1-\frac{2\alpha }{1+{\alpha }^{2}}$⇒51​=1−1+α22α​ (from (i)&(\mathrm{ii}))

$$\Rightarrow 1+{\alpha }^{2}=5+5{\alpha }^{2}-10\alpha$$⇒1+α2=5+5α2−10α

$$\Rightarrow 2{\alpha }^{2}-5\alpha +2=0$$⇒2α2−5α+2=0

$$\therefore \alpha =2,\frac{1}{2}$$∴α=2,21​

$$\therefore \underset{\alpha \in S}{\sum }16{\alpha }^{3}=16\times {2}^{3}+16\times \frac{1}{{2}^{3}}=16(\frac{65}{8})$$∴α∈S∑​16α3=16×23+16×231​=16(865​)

$$=130$$=130