Question

40 баловвв!!!! 3 примера Упростить: 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

Answer 1

$1)\; \; \frac{sin2a+sin8a}{cos2a-cos8a}=\frac{2\cdot sin5a\cdot cos3a}{-2\cdot sin5a\cdot sin(-3a)}=\frac{cos3a}{sin3a}=ctg3a\\\\2)\; \; \frac{ctg(\frac{\pi}{2}-a)\cdot (sin(\frac{3\pi}{2}-a)+sin(\pi +a))}{ctg(\pi +a)\cdot (cos(2\pi +a)-sin(2\pi -a))}=\frac{tga\cdot (-cosa-sina)}{ctga\cdot (cosa+sina)}=\\\\= \frac{-tga\cdot (cosa+sina)}{\frac{1}{tga}\cdot (cosa+sina)} =-tg^2a$

$3)\; \; 16\cdot cos \frac{\pi }{9}\cdot cos\frac{2\pi }{9}\cdot cos\frac{4\pi }{9} =\\\\= \frac{8}{sin\frac{\pi}{9}}\cdot (\underbrace {2\, sin \frac{\pi }{9}\cdot cos\frac{\pi }{9}}_{sin\frac{2\pi }{9}})\cdot cos\frac{2\pi }{9}\cdot cos\frac{4\pi }{9}=\\\\=\frac{4}{sin\frac{\pi}{9}}\cdot (\underbrace {2\, sin\frac{2\pi }{9}\cdot cos\frac{2\pi }{9}}_{sin\frac{4\pi}{9}})\cdot cos\frac{4\pi }{9}=\\\\=\frac{2}{sin\frac{\pi}{9}}\cdot (\underbrace{2sin \frac{4\pi }{9}\cdot cos\frac{4\pi }{9}}_{sin\frac{8\pi }{9}})=\frac{2sin\frac{8\pi}{9}}{sin\frac{\pi}{9}} = \frac{2sin(\pi-\frac{\pi}{9})}{sin\frac{\pi}{9}}=\\\\=\frac{2sin\frac{\pi}{9}}{sin\frac{\pi}{9}} =2$

Answer 2

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

$16*cos(\frac{\pi}{9})*cos(\frac{2\pi}{9})*cos(\frac{4\pi}{9})=\\= 16*cos(20^0)*cos(40^0)*cos(80^0)=\\= 16*\frac{sin(40^0)}{2sin(20^0)}*\frac{sin(80^0)}{2sin(40^0)}*\frac{sin(160^0)}{2sin(80^0)}=2*\frac{sin(160^0)}{sin(20^0)}=\\= 2*\frac{sin(180^0-20^0)}{sin(20^0)}=2*\frac{sin(20^0)}{sin(20^0)}=2*1=2$
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$\frac{sin(2a)+sin(8a)}{cos(2a)-cos(8a)} = \frac{2sin( \frac{2a+8a}{2} )*cos(\frac{2a-8a}{2})}{2sin( \frac{2a+8a}{2})*sin( \frac{8a-2a}{2})} = \frac{cos(3a)}{sin(3a)} =ctg(3a)$