[tex]sina=+\frac{3}{4} \\ \frac{\pi} {2} \ \textless \ a\ \textless \ \pi[/tex] Надо найти cosa, tga,ctga, tg2a. Created by 321NIYAZ123. algebra-ru

tex] Надо найти cosa, tga,ctga, tg2a...

321NIYAZ123 @321NIYAZ123

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[tex]sina=+\frac{3}{4} \\ \frac{\pi} {2} \ \textless \ a\ \textless \ \pi[/tex]
Надо найти cosa, tga,ctga, tg2a

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Artem112

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Так как $\dfrac{\pi} {2}$ , то угол а - второй четверти, где косинус, тангенс и котангенс отрицательные.

$\sin^2a+\cos^2a=1\\\cos^2a=1-\sin^2a\\\cos a=-\sqrt{1-\sin^2a} \\\cos a=-\sqrt{1-\left(\dfrac{3}{4}\right)^2 } =-\sqrt{1-\dfrac{9}{16} } =-\sqrt{\dfrac{7}{16} } =-\dfrac{\sqrt{7}} {4}$

$\mathrm{tg}a=\dfrac{\sin a}{\cos a}=\dfrac{3} {4}:\left(-\dfrac{\sqrt{7}} {4} \right)=\dfrac{3} {4}\cdot\left(-\dfrac{4} {\sqrt{7}} \right)=-\dfrac{3} {\sqrt{7}} =-\dfrac{3\sqrt{7}} {7}$

$\mathrm{ctg}a=\dfrac{1}{\mathrm{tg}a} =1:\left(-\dfrac{3} {\sqrt{7}}\right) =-\dfrac{\sqrt{7}} {3}$

$\mathrm{tg}2a=\dfrac{2\mathrm{tg}a}{1-\mathrm{tg}^2a} =\dfrac{2\cdot\left(-\dfrac{3} {\sqrt{7}}\right)}{1-\left(-\dfrac{3} {\sqrt{7}}\right)^2} =\dfrac{-\dfrac{6} {\sqrt{7}}}{1-\dfrac{9} {7}} =\dfrac{-\dfrac{6} {\sqrt{7}}}{-\dfrac{2} {7}} =\\\\=\dfrac{6}{\sqrt{7}}:\dfrac{2} {7}=\dfrac{6}{\sqrt{7}}\cdot\dfrac{7} {2}=3\sqrt{7}$

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