Question

Тригонометрия срочно!!! 55б

Answer 1

Ответ:

$\star \ \ tg\alpha =-2\ \ \Rightarrow \ \ \dfrac{sin\alpha }{cos\alpha }=-2\ \ ,\ \ sin\alpha =-2\, cos\alpha \ \ \star \\\\\star \ \ 1+tg^2\alpha =\dfrac{1}{cos^2\alpha }\ \ \Rightarrow \ \ cos^2a=\dfrac{1}{1+tg^2\alpha }=\dfrac{1}{1+4}=\dfrac{1}{5}\ \ \star$

$1+5sin2\alpha -3\, (cos2\alpha )^{-1}=1+5\cdot 2\, sin\alpha \cdot cos\alpha -\dfrac{3}{cos2\alpha }=\\\\\\=1+10\cdot (-2\, cos\alpha )\cdot cos\alpha -\dfrac{3}{cos^2\alpha -sin^2\alpha }=\\\\\\=1-20cos^2\alpha -\dfrac{3}{cos^2\alpha -(-2\, cos\alpha )^2}=1-20\cdot \dfrac{1}{5}+\dfrac{3}{3\, cos^2\alpha }=\\\\\\=1-4+5=2$