тригонометрия

[Гост]

Опростете израза $cotg\alpha-tg\alpha-2tg2\alpha-4tg4\alpha$
Отговорът е $8cotg8\alpha$
$\frac{3-4cos2\alpha +cos4\alpha}{3+4cos2\alpha +cos4\alpha}$
Отговорът е $tg^{4}\alpha$

[Nathi123]

[tex]A= \frac{3-4cos2\alpha+cos4\alpha}{3+4cos2\alpha+cos4\alpha} = \frac{1+cos4\alpha+2(1-2cos2\alpha)}{1+cos4\alpha+2(1+2cos2\alpha)}[/tex]
[tex]\Leftrightarrow A=\frac{2cos^{2}2\alpha+2-4cos2\alpha}{2cos^{2}2\alpha+2+4cos2\alpha}=\frac{2cos2\alpha(cos2\alpha-1)+2(1-cos2\alpha)}{2cos2\alpha(cos2\alpha+1)+2(1+cos2\alpha)}\Leftrightarrow[/tex]
[tex]A=\frac{-4cos2 \alpha sin^{2}\alpha+4sin^{2}\alpha}{4cos2\alpha cos^{2}2\alpha+4cos^{2}\alpha}=\frac{4sin^{2}\alpha(1-cos2\alpha)}{4cos^{2}\alpha(1+cos2\alpha)}=tg^{2}\alpha\frac{2sin^{2}\alpha}{2cos^{2}\alpha}=tg^{4}\alpha[/tex].

[Nathi123]

[tex]B= cotg\alpha - tg\alpha -2tg2\alpha -4tg4\alpha = \frac{cos\alpha}{sin\alpha}-\frac{sin\alpha}{cos\alpha} -2tg2\alpha -4tg4\alpha \Leftrightarrow[/tex]
[tex]B=\frac{cos^{2}\alpha-sin^{2}\alpha}{sin\alpha.cos\alpha}-2tg2\alpha -4tg4\alpha = 2cotg2\alpha-2tg2\alpha -4tg4\alpha = 4cotg4\alpha - 4tg4\alpha = 8tg8\alpha[/tex].
Т.е 3 пъти използваме,че [tex]cotg\alpha - tg\alpha = 2cotg2\alpha[/tex].