Вычислить tg(α-30°),если cosα=1/√2 и 0<α<π/2

Question

Aratius

5 год назад

ОТВЕТЫ

Катерина

Nov 7, 2020

$\mathrm{tg}(a - 30^\circ) =$

$= \frac{\sin(a - 30^\circ)}{\cos(a - 30^\circ)} =$

$= \frac{\sin(a)\cos(30^\circ) - \cos(a)\sin(30^\circ)}{\cos(a)\cos(30^\circ)+\sin(a)\sin(30^\circ)} =$

$= \frac{\sin(a)\cdot\frac{\sqrt{3}}{2} - \cos(a)\cdot\frac{1}{2}}{\cos(a)\cdot\frac{\sqrt{3}}{2}+\sin(a)\cdot\frac{1}{2}} =$

$= \frac{\sin(a)\cdot\sqrt{3} - \cos(a)}{\cos(a)\cdot\sqrt{3} + \sin(a)} = V$

$\cos(a) = \frac{1}{\sqrt{2}}$

$\sin(a) = \pm\sqrt{1 - \cos^2(a)} = \pm\sqrt{1 - (\frac{1}{\sqrt{2}})^2} =$

$= \pm\sqrt{1 - \frac{1}{2}} = \pm\sqrt{\frac{1}{2}} = \pm\frac{1}{\sqrt{2}}$

Но т.к. $0$ , то $\sin(a) > 0$ поэтому

$\sin(a) = \frac{1}{\sqrt{2}}$ , а $\cos(a) = \frac{1}{\sqrt{2}}$

$V = \frac{\frac{1}{\sqrt{2}}\cdot\sqrt{3} - \frac{1}{\sqrt{2}} }{\frac{1}{\sqrt{2}}\cdot\sqrt{3} + \frac{1}{\sqrt{2}} } =$

$= \frac{\sqrt{3}-1}{\sqrt{3}+1} = \frac{(\sqrt{3}-1)^2}{(\sqrt{3}+1)\cdot(\sqrt{3}-1)} = \frac{3 - 2\cdot\sqrt{3} + 1}{3 - 1} =$

$= \frac{4 - 2\cdot\sqrt{3}}{2} = 2 - \sqrt{3}$

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