Question

Даю 40 балов!!!
1)Доведіть тотожність
sin^3a-cos a+sin a•cos^3a=0,5 sin2a
2)Обчислити
sin^2a-3cos^2a/2sin^2a+cos^2a
Якщо tga=3

Answer 1

Sin³α * Cosα + Sinα * Cos³α = 0,5Sin2α

Sin³α * Cosα + Sinα * Cos³α = SinαCosα(Sin²α + Cos²α) = SinαCosα * 1=

= SinαCosα = 0,5 * 2SinαCosα = 0,5Sin2α

0,5Sin2α = 0,5Sin2α

$2)\frac{Sin^{2}\alpha-3Cos^{2}\alpha}{2Sin^{2}\alpha+Cos^{2}\alpha} =\frac{\frac{Sin^{2}\alpha}{Cos^{2}\alpha}-\frac{-3Cos^{2}\alpha}{Cos^{2}\alpha}}{\frac{2Sin^{2}\alpha}{Cos^{2}\alpha}+\frac{Cos^{2}\alpha}{Cos^{2}\alpha}}=\frac{tg^{2}\alpha-3}{2tg^{2}\alpha+1}$

$tg\alpha=3\Rightarrow \frac{3^{2}-3}{2*3^{2} +1} =\frac{6}{19}$

Answer 2

$1)\; \; sin^3a\cdot cosa+sina\cdot cos^3a=sina\cdot cosa\cdot (sin^2a+cos^2a)=\\\\=sina\cdot cosa\cdot 1=\frac{1}{2}\cdot 2\, sina\cdot cosa=\frac{1}{2}\cdot sin2a$

$2)\; \; tga=3\\\\\frac{sin^2a-3\, cos^2a}{2\, sin^2a+cos^2a}=\frac{cos^2a\cdot (tga^2a-3)}{cos^2a\cdot (2tg^2a+1)}=\frac{tg^2a-3}{tg^2a+1}=\frac{3^2-3}{2\cdot 3^2+1}=\frac{6}{19}$