Question

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

Answer 1

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

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

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

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

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