Functions#12

Note that if $\tan \frac \theta 2 = x$ , then by half-angle tangent substitution we have:

$\begin{aligned} \cos^{-1} \left(\frac {1-x^2}{1+x^2}\right) & = \frac \pi 2 - 2 \sin^{-1} \left(\frac {2x}{1+x^2}\right) \\ \theta & = \frac \pi 2 - 2\theta \\ \implies \theta & = \frac \pi 6 \\ \sin^{-1} \left(\frac {2x}{1+x^2}\right) & = \frac \pi 6 \\ \frac {2x}{1+x^2} & = \frac 12 \\ x^2 - 4x + 1 & = 0 \\ \implies x & = 2 \pm \sqrt 3 \end{aligned}$

$\begin{aligned} \implies \lambda & = (2+\sqrt 3)^2 + (2-\sqrt 3)^2 = 14 \\ \lambda - 8 & = 14-8 = \boxed{6}\end{aligned}$

$\cos^{-1}(\frac{1 - x^2}{1 + x^2}) = \frac{\pi}{2} - 2\sin^{-1}(\frac{2x}{1 + x^2})$ $\ldots$ (given)

$\frac{1 - x^2}{1 + x^2} = \cos(\frac{\pi}{2} - 2\sin^{-1}(\frac{2x}{1 + x^2}))$ $\ldots$ ( $\cos^{-1}y = x \implies y = \cos x$ )

$\frac{1 - x^2}{1 + x^2} = \sin(2\sin^{-1}(\frac{2x}{1 + x^2}))$ $\ldots$ ( $\cos(\frac{\pi}{2} - x) = \sin x$ )

$\frac{1 - x^2}{1 + x^2} = 2 \sin(\sin^{-1}(\frac{2x}{1 + x^2})) \cos(\sin^{-1}(\frac{2x}{1 + x^2}))$ $\ldots$ ( $\sin(2x) = 2 \sin x \cos x$ )

$\frac{1 - x^2}{1 + x^2} = 2 \cdot \frac{2x}{1 + x^2} \cdot \frac{\sqrt{(1 + x^2)^2 - (2x)^2}}{1 + x^2}$ $\ldots$ ( $\sin^{-1}(\sin x) = x$ , $\sin^{-1}(\cos \frac{b}{c}) = \frac{\sqrt{c^2 - b^2}}{c}$ )

$\frac{1 - x^2}{1 + x^2} = 2 \cdot \frac{2x}{1 + x^2} \cdot \frac{1 - x^2}{1 + x^2}$ $\ldots$ (difference of squares, simplify)

$1 = 2 \cdot \frac{2x}{1 + x^2}$ , $x \neq \pm 1$ $\ldots$ (divide by $\frac{1 - x^2}{1 + x^2}$ )

$x^2 - 4x + 1 = 0$ , $x \neq \pm 1$ $\ldots$ (rearrange)

$x = 2 \pm \sqrt{3}$ $\ldots$ (quadratic equation)

Therefore, $\lambda - 8 = (2 + \sqrt{3})^2 + (2 - \sqrt{3})^2 - 8 = 14 - 8 = \boxed{6}$ .