Mathematics - Trigonometric Ratios & Identities Question with Solution | TestHub

$A={\sin }^{4}x+{\cos }^{4}x=({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x=1-2(\sin x\cos x{)}^2=1-2{\left(\frac{\sin 2x}{2}\right)}^2=1-\frac{1}{2}{\sin }^{2}2x$. This implies ${\sin }^{2}2x=2(1-A)$.$B={\sin }^{8}x+{\cos }^{8}x=({\sin }^{4}x+{\cos }^{4}x{)}^2-2{\sin }^{4}x{\cos }^{4}x=A^2-2(\sin x\cos x{)}^4=A^2-2{\left(\frac{\sin 2x}{2}\right)}^4=A^2-\frac{2}{16}{\sin }^{4}2x=A^2-\frac{1}{8}{\sin }^{4}2x$. Substitute ${\sin }^{2}2x=2(1-A)$ into the expression for $B$:$B=A^2-\frac{1}{8}(2(1-A){)}^2=A^2-\frac{1}{8}\cdot 4(1-A{)}^2=A^2-\frac{1}{2}(1-A{)}^2$.$B=A^2-\frac{1}{2}(1-2A+A^2)=A^2-\frac{1}{2}+A-\frac{1}{2}{A}^2=\frac{1}{2}{A}^2+A-\frac{1}{2}$. Multiplying by 2, we get $2B=A^2+2A-1$.