We have an answer to "Calculate tan⁡9∘−tan⁡27∘−tan⁡63∘+tan⁡81∘ ?"

Kienastsrx

Answered question

2022-03-16

Calculate

{\tan 9}^{\circ} - {\tan 27}^{\circ} - {\tan 63}^{\circ} + {\tan 81}^{\circ}

Answer & Explanation

Quentin Olsen

Beginner2022-03-17Added 5 answers

You'd better to know

\sin 2 θ = 2 \sin θ \cos θ

\sin α - \sin β = 2 \cos \frac{α + β}{2} \sin \frac{α - β}{2}

Then we have

{\tan 9}^{\circ} - {\tan 27}^{\circ} - {\tan 63}^{\circ} + {\tan 81}^{\circ} = \frac{{\sin 9}^{\circ}}{{\cos 9}^{\circ}} - \frac{{\sin 27}^{\circ}}{{\cos 27}^{\circ}} - \frac{{\cos 27}^{\circ}}{{\sin 27}^{\circ}} + \frac{{\cos 9}^{\circ}}{{\sin 9}^{\circ}}

= \frac{1}{{\sin 9}^{\circ} {\cos 9}^{\circ}} - \frac{1}{{\sin 27}^{\circ} {\cos 27}^{\circ}}

= \frac{2}{{\sin 18}^{\circ}} - \frac{2}{{\sin 54}^{\circ}}

= \frac{2 ({\sin 54}^{\circ} - {\sin 18}^{\circ})}{{\sin 18}^{\circ} {\sin 54}^{\circ}}

= 2 \frac{2 {\cos 36}^{\circ} {\sin 18}^{\circ}}{{\sin 18}^{\circ} {\cos 36}^{\circ}}

= 4

bicicletagyp

Beginner2022-03-18Added 4 answers

${\tan 9}^{\circ} + {\tan 81}^{\circ} = \frac{{\sin 9}^{\circ}}{{\cos 9}^{\circ}} + \frac{{\sin 81}^{\circ}}{{\cos 81}^{\circ}} = \frac{\sin (9^{\circ} + 81^{\circ})}{}} {{\cos 9}^{\circ} {\cos 81}^{\circ}}$ Similarly:
$= \frac{1}{{\cos 9}^{\circ} {\cos 81}^{\circ}} = \frac{1 {\cos 9}^{\circ}}{{\sin 9}^{\circ}} = \frac{2}{{\sin 18}^{\circ}}$
${\tan 27}^{\circ} + {\tan 63}^{\circ} = \frac{1}{{\cos 27}^{\circ} {\cos 63}^{\circ}} = \frac{1}{{\cos 27}^{\circ} {\sin 27}^{\circ}}$
$= \frac{2}{{\sin 54}^{\circ}}$
Now let $x = \sin 18 °$ , then from ${\cos 36}^{\circ} = {\sin 54}^{\circ}$ we have:
$1 - 2 x^{2} = 3 x - 4 x^{3} \to 4 x^{3} - 2 x^{2} - 3 x + 1 = 0$ , sind $x =\neq 1 \to (x - 1) (4 x^{2} + 2 x - 1) = 0$
$x = \frac{\sqrt{5} - 1}{4} = {\sin 18}^{\circ}$ , and ${\sin 54}^{\circ} = 1 - 2 x^{2} = 1 - 2 {(\frac{\sqrt{5} - 1}{4})}^{2} = \frac{\sqrt{5} + 1}{4}$
Thus:
$S = \frac{2}{\frac{\sqrt{5} - 1}{4}} - \frac{2}{\frac{\sqrt{5} + 1}{4}} = 4$

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