Question

Решите алгебру плиз!
100 баллов

Answer 1

$1)\ \ tga=\sqrt{15}\\\\1+tg^2a=\dfrac{1}{\cos^2a}\ \ \to \ \ \ cos^2a=\dfrac{1}{1+tg^2a}=\dfrac{1}{1+15}=\dfrac{1}{16}\\\\\\0^\circ <a<90^\circ \ \ \ \to \ \ \ cosa>0\ \ ,\ \ cosa=\dfrac{1}{4}$

$2)\ \ sina=-\dfrac{1}{3}\ \ ,\ \ cos\beta =-\dfrac{2}{3}\\\\\\cos^2a=1-sin^2a=1-\dfrac{1}{9}=\dfrac{8}{9}\\\\\pi <a<\dfrac{3\pi }{2}\ \ \to \ \ cosa<0\ \ ,\ \ cosa=-\dfrac{2\sqrt2}{3}\\\\\\sin^2\beta =1-cos^2\beta =1-\dfrac{4}{9}=\dfrac{5}{9}\\\\\pi <\beta <\dfrac{3\pi }{2}\ \ \to \ \ sin\beta <0\ \ ,\ \ sin\beta =-\dfrac{\sqrt5}{3}\\\\\\sin(a+\beta )=sina\cdot cos\beta +cosa\cdot sin\beta =\dfrac{1}{3}\cdot \dfrac{2}{3}+\dfrac{2\sqrt2}{3}\cdot \dfrac{\sqrt5}{3}=\dfrac{2+2\sqrt{10}}{9}$

$3)\ \ \dfrac{(sina+cosa)^2-1}{tg(\frac{\pi}{2}-a)-sina\cdot cosa}-2\, tg^2a=\dfrac{sin^2a+cos^2a+2sina\cdot cosa-1}{ctga-sina\cdot cosa}-2\, tg^2a=\\\\\\=\dfrac{2\, sina\cdot cosa}{\dfrac{cosa}{sina}-sina\cdot cosa}-2\, tg^2a=\dfrac{2\, sin^2a\cdot cosa}{cosa-sin^2a\cdot cosa}-2\, tg^2a=\\\\\\=\dfrac{2\, sin^2a\cdot cosa}{cosa\cdot (1-sin^2a)}-2\, tg^2a=\dfrac{2sin^2a}{cos^2a}-2tg^2a=2tg^2a-2tg^2a=0$

$4)\ \ \dfrac{1+sin(\frac{3\pi}{2}+2a)-sin^2x}{sin2x\cdot sin(\frac{\pi}{2}+x)+sinx\cdot cos(\pi -2x)}=\dfrac{1-cos2x-sin^2x}{sin2x\cdot cosx-sinx\cdot cos2x}=\\\\\\=\dfrac{2sin^2x-sin^2x}{sin(2x-x)}=\dfrac{sin^2x}{sinx}=sinx\\\\\\\\\star \ \ sin^2x=\dfrac{1-cos2x}{2}\ \ \Rightarrow \ \ \ 1-cos2x=2sin^2x\ \ \star$