JEE Main · 2019 · 12 Jan 2019 · Shift 2 · Trigonometric Ratios and Identities

The left side is a sum of four non-negative terms ${\sin }^4\alpha$, $4{\cos }^4\beta$, $1$ and $1$, so by the AM-GM inequality $\frac{{\sin }^4\alpha +4{\cos }^4\beta +1+1}{4}\ge {\left({\sin }^4\alpha \cdot 4{\cos }^4\beta \cdot 1\cdot 1\right)}^{1/4}=\sqrt{2}\sin \alpha \cos \beta$. This rearranges to ${\sin }^4\alpha +4{\cos }^4\beta +2\ge 4\sqrt{2}\sin \alpha \cos \beta$, and the given equation is precisely the equality case. Equality in AM-GM demands all four terms be equal, so ${\sin }^4\alpha =4{\cos }^4\beta =1$, giving $\sin \alpha =1$ and $\cos \beta =\frac{1}{\sqrt{2}}$ (the positive value keeps the right side equal to $4$). With $\alpha ,\beta \in [0,\pi ]$ this forces $\alpha =\frac{\pi }{2}$ and $\beta =\frac{\pi }{4}$, hence $\sin \beta =\frac{1}{\sqrt{2}}$. Now convert the required difference into a product: $\cos (\alpha +\beta )-\cos (\alpha -\beta )=-2\sin \alpha \sin \beta =-2\cdot 1\cdot \frac{1}{\sqrt{2}}=-\sqrt{2}$.