Question

Вычислите. Алгебра 9 класс.​

Answer 1

Объяснение:

3) а)

$\tg \beta = - 4 <=> \frac{ \sin \beta }{ \cos \beta } = - 4 \\ \sin \beta = - 4 \cos \beta \\ 1 = \sin^{2} \beta +\cos^{2} \beta = \\ = 16\cos^{2} \beta +\cos^{2} \beta = \\ = 17\cos^{2} \beta$

б) Подставим (-4 cos b) вместо sin b

$\frac{2 \sin \beta \cos \beta + 3}{4 { \cos}^{2} \beta + { \sin}^{2} \beta } = \frac{2 (- 4\cos \beta) \cos \beta + 3}{4 { \cos}^{2} \beta + ( { - 4\cos \beta })^{2}} = \\ \frac{3- 8\cos^{2} \beta }{4 { \cos}^{2} \beta + 16\cos ^{2} \beta } = \\ \frac{3 \cdot( 17\cos^{2} \beta ) - 8\cos^{2} \beta }{ 20\cos ^{2} \beta } = \\ = \frac{51\cos^{2} \beta - 8\cos^{2} \beta }{ 20\cos ^{2} \beta } = \\ = \frac{43 \: \: \cancel{\cos^{2} \beta} }{ 20 \: \: \cancel{\cos^{2} \beta} } = \frac{43}{20} = 2 \frac{3}{20} = 2.15$

4)

$\tg \beta = 3 < = > \frac{ \sin \beta }{ \cos \beta } = 3 \\ \sin \beta = 3 \cos \beta$

$\frac{{\cos}^{2} \beta + 2}{{ \cos}^{2} \beta + 3 \sin \beta \cos \beta } = \\ = \frac{\cos^{2} \beta + 2\sin^{2} \beta + 2\cos^{2} \beta}{{ \cos}^{2} \beta + 3 \sin \beta \cos \beta } = \\ = \frac{3\cos^{2} \beta + 2\sin^{2} \beta }{{ \cos}^{2} \beta + 3 \sin \beta \cos \beta } =$

Подставим (3 cos b) вместо sin b

$\frac{3\cos^{2} \beta + 2\sin^{2} \beta }{{ \cos}^{2} \beta + 3 \sin \beta \cos \beta } = \\ \: = \frac{3\cos^{2} \beta + 2 \cdot(3\cos \beta)^{2} }{{ \cos}^{2} \beta + 3 \cdot (3\cos \beta)\cos \beta } = \\ = \frac{3\cos^{2} \beta + 2 \cdot 9\cos^{2} \beta}{{ \cos}^{2} \beta + 9\cos^{2} \beta} = \\ = \frac{(3 + 18) {\cos}^{2} \beta }{(1+ 9) {\cos}^{2} \beta } = \frac{21 \:\cancel{\cos^{2} \beta}}{10\cancel{\cos^{2} \beta} } = \frac{21}{10} = 2.1$