40 3 примера : sin⁡2α+sin8α/cos2α-cos8α довести тождество: (ctg(π/2-α)*(sin⁡(3π/2 -α)+sin⁡(π+α)) / (ctg(π+α)*(cos⁡(2π+α)-sin⁡(2π-α) ) = -tg^2 α 16cos π/9 cos 2π/9 cos 4π/9=2

Julia1965 Julia1965    1   14.09.2019 13:10    0

Ответы
Tapoocek Tapoocek  07.10.2020 14:19
\frac{ctg(\frac{\pi}{2}-a)*[sin(\frac{3\pi}{2} -a)+sin(\pi+a)]}{ctg(\pi+a)*[cos(2\pi+a)-sin(2\pi-a)]}=
\frac{tg(a)*[-cos(a)-sin(a)]}{ctg(a)*[cos(a)+sin(a)]}=- \frac{tg(a)}{ctg(a)} =\\=
-tg(a): \frac{1}{tg(a)} =-tg^2(a)
------------------------------------
****sin(2a)=2sin(a)cos(a),\ \ \ -\ \textgreater \ \ \ \ cos(a)= \frac{sin(2a)}{2sin(a)} *****

16*cos(\frac{\pi}{9})*cos(\frac{2\pi}{9})*cos(\frac{4\pi}{9})=\\=
16*cos(20^0)*cos(40^0)*cos(80^0)=\\=
16*\frac{sin(40^0)}{2sin(20^0)}*\frac{sin(80^0)}{2sin(40^0)}*\frac{sin(160^0)}{2sin(80^0)}=2*\frac{sin(160^0)}{sin(20^0)}=\\=
2*\frac{sin(180^0-20^0)}{sin(20^0)}=2*\frac{sin(20^0)}{sin(20^0)}=2*1=2
----------------------
\frac{sin(2a)+sin(8a)}{cos(2a)-cos(8a)} = \frac{2sin( \frac{2a+8a}{2} )*cos(\frac{2a-8a}{2})}{2sin( \frac{2a+8a}{2})*sin( \frac{8a-2a}{2})} = \frac{cos(3a)}{sin(3a)} =ctg(3a)
ПОКАЗАТЬ ОТВЕТЫ
Другие вопросы по теме Алгебра