Question-161668

Question Number 161668 by mnjuly1970 last updated on 21/Dec/21

Answered by mr W last updated on 21/Dec/21

\frac{\sin^{2} x}{1 - \sin^{2} x} = 1 + \sin^{2} x

\sin^{4} x + \sin^{2} x - 1 = 0

\sin^{2} x = \frac{- 1 + \sqrt{5}}{2} = ϕ

4 \sin^{2} \frac{x}{2} (1 - \sin^{2} \frac{x}{2}) = ϕ

\sin^{4} \frac{x}{2} - \sin^{2} \frac{x}{2} + \frac{ϕ}{4} = 0

\sin^{2} \frac{x}{2} = \frac{1 \pm \sqrt{1 - ϕ}}{2}

{(\sin^{2} \frac{x}{2})}_{m a x} = \frac{1 + \sqrt{1 - ϕ}}{2}

= \frac{1 + \sqrt{\frac{3 - \sqrt{5}}{2}}}{2} = \frac{1}{2} + \frac{\sqrt{6 - 2 \sqrt{5}}}{4} = \frac{1}{2} + \frac{\sqrt{5} - 1}{4} = \frac{\sqrt{5} + 1}{2}

\frac{\sqrt{6 - 2 \sqrt{5}}}{4} = \frac{\sqrt{6 - \sqrt{20}}}{4}

\Rightarrow a n s w e r (3) i s c o r r e c t :

Commented by mnjuly1970 last updated on 21/Dec/21

g e a t e f u l s i r W \dots v e r y e x c e l l e n t