JEE Main 2022 — Mathematics Trigonometry

Let $\frac{{\mathrm{sin}}^{-1}x}{\alpha }=\frac{{\mathrm{cos}}^{-1}x}{\beta }=k$αsin−1x​=βcos−1x​=k

i.e. ${\mathrm{sin}}^{-1}x=k\alpha$sin−1x=kα and ${\mathrm{cos}}^{-1}x=k\beta$cos−1x=kβ

$$\Rightarrow {\mathrm{sin}}^{-1}x+{\mathrm{cos}}^{-1}x=k(\alpha +\beta )$$⇒sin−1x+cos−1x=k(α+β)

i.e. $k(\alpha +\beta )=\frac{\pi }{2}$k(α+β)=2π​

$$\Rightarrow k=\frac{\pi }{2(\alpha +\beta )}$$⇒k=2(α+β)π​

Now $\frac{2\pi \alpha }{\alpha +\beta }=4k\alpha =4{\mathrm{sin}}^{-1}x$α+β2πα​=4kα=4sin−1x

So $\mathrm{sin}(\frac{2\pi \alpha }{\alpha +\beta })=\mathrm{sin}(4{\mathrm{sin}}^{-1}x)$sin(α+β2πα​)=sin(4sin−1x)

$$=2\mathrm{sin}(2{\mathrm{sin}}^{-1}x)\mathrm{cos}(2{\mathrm{sin}}^{-1}x)$$=2sin(2sin−1x)cos(2sin−1x)

$$=2\mathrm{sin}({\mathrm{sin}}^{-1}2x\sqrt{1-{x}^{2}})\mathrm{cos}({\mathrm{sin}}^{-1}2x\sqrt{1-{x}^{2}})$$=2sin(sin−12x1−x2​)cos(sin−12x1−x2​)

$$=2\mathrm{sin}({\mathrm{sin}}^{-1}2x\sqrt{1-{x}^{2}})\mathrm{cos}({\mathrm{cos}}^{-1}(1-2{x}^{2}))$$=2sin(sin−12x1−x2​)cos(cos−1(1−2x2))

$$=4x\sqrt{1-{x}^{2}}(1-2{x}^{2})$$=4x1−x2​(1−2x2)