please-help-me-to-show-that-tan-2-pi-8-2tan-pi-8-1-0-

Question Number 75131 by mathocean1 last updated on 07/Dec/19

please help me to show that

\tan^{2} (\frac{π}{8}) + 2 \tan (\frac{π}{8}) - 1 = 0

Answered by mr W last updated on 07/Dec/19

\tan \frac{π}{4} = 1

\Rightarrow \tan 2 \times \frac{π}{8} = 1

\Rightarrow \frac{2 \tan \frac{π}{8}}{1 - \tan^{2} \frac{π}{8}} = 1

\Rightarrow 2 \tan \frac{π}{8} = 1 - \tan^{2} \frac{π}{8}

\Rightarrow \tan^{2} \frac{π}{8} + 2 \tan \frac{π}{8} - 1 = 0

Commented by mathocean1 last updated on 07/Dec/19

thank you sir \dots can you help me to

deduct \tan \frac{π}{8} ?

Commented by peter frank last updated on 07/Dec/19

\tan 2 θ = \frac{2 \tan θ}{1 - \tan^{2} θ}

b u t θ = \frac{π}{8}

\tan 2 \frac{π}{8} = \frac{2 \tan \frac{π}{8}}{1 - \tan^{2} \frac{π}{8}}

\tan \frac{π}{4} = \frac{2 \tan \frac{π}{8}}{1 - \tan^{2} \frac{π}{8}}

Commented by MJS last updated on 07/Dec/19

right idea, just put θ = \frac{α}{2}

\tan α = \frac{2 \tan \frac{α}{2}}{1 - \tan^{2} \frac{α}{2}}

now solve for \tan \frac{α}{2}

\tan \frac{α}{2} = - \frac{1 \pm \sqrt{1 + \tan^{2} α}}{\tan α}

now α = \frac{π}{4} \Rightarrow \tan α = 1

\Rightarrow \tan \frac{π}{8} = - 1 + \sqrt{2}

knowing \tan \frac{π}{8} > 0 we cannot use the

second solution - 1 - \sqrt{2}

Facebook Tweet Pin