Question

Matematica e prea grea ​

Answer 1

$\sqrt{}$Răspuns:

1. sin(x-y) = 1

2.cos(x+y) = $-\frac{1+\sqrt{2} }{3}$

3. tg(x-y) = $\frac{2\sqrt{2}+\sqrt{5} }{1+2\sqrt{10} }$

4. ctg(x+y) = $\frac{7}{9}$

Explicație pas cu pas:

1. sin(x-y) = ?

x, y ∈ ($\frac{\pi}2}$,$\pi$) => sin x > 0, cos x < 0

                    sin y > 0, cos y < 0

sin x = $\frac{2}{3}$

sin²x + cos²x = 1

($\frac{2}{3}$)² + cos²x = 1

$\frac{4}{9}$ + cos²x = 1 | - $\frac{4}{9}$

cos²x = $\frac{5}{9}$ | $\sqrt{}$

cos x = ± $\frac{\sqrt{5}}{3}$

dar cos x < 0 => cos x = - $\frac{\sqrt{5}}{3}$

cos y = $\frac{2}{3}$

sin²y + cos²y = 1

sin²y + $(-\frac{2}{3}) ^{2}$= 1

sin²y + $\frac{4}{9}$ = 1 | - $\frac{4}{9}$

sin²y = $\frac{5}{9}$ |  $\sqrt{}$

sin y = ± $\frac{\sqrt{5}}{3}$

dar sin y > 0 = > sin y = $\frac{\sqrt{5}}{3}$

sin(x-y) = sin x · cos y - sin y · cos x

sin(x-y) = $\frac{2}{3}.\frac{2}{3} - \frac{\sqrt{5}}{3}.\frac{-\sqrt{5}}{3}$

sin(x-y) = $\frac{4}{9} + \frac{5}{9}$

sin(x-y) = $\frac{9}{9} = 1$

2. cos(x+y) = ?

x ∈ $(\pi, \frac{3\pi }{2})$ => sin x < 0, cos x < 0

y ∈ $(\frac{3\pi }{2}, 2\pi)$ => sin y < 0, cos y > 0

sin x = $-\frac{1}{\sqrt{3}}$

cos²x = 1 - sin²x

cos²x = $1-\frac{1}{3}$

cos²x = $\frac{2}{3}$ | $\sqrt{}$

cos x = ± $\frac{\sqrt{6}}{3}$

dar cos x < 0 => cos x = $-\frac{\sqrt{6} }{3}$

tg y = -$\sqrt{5}$ => sin y = $-\sqrt{5} cos y$

cos²y = 1 - sin²y

cos²y = 1 - 5cos²y | + 5cos²y

6cos²y = 1 | :6

cos²y = $\frac{1}{6}$ | $\sqrt{}$

cos y = $\frac{\sqrt{6}}{6}$ => sin y = $-\frac{\sqrt{30}}{6}$

cos(x+y) = cos x · cos y - sin x · sin y

cos(x+y) = $-\frac{\sqrt{6} }{3} . \frac{\sqrt{6} }{6} - \frac{-\sqrt{3} }{3} . \frac{-\sqrt{6} }{3}$

cos(x+y) = $-\frac{6}{18} - \frac{3\sqrt{2} }{9}$

cos(x+y) = $-\frac{3}{9} - \frac{3\sqrt{2} }{9}$

cos(x+y) = $-\frac{1+\sqrt{2} }{3}$

3. tg(x-y) = ?

x ∈ $(0, \frac{\pi}{2} )$ => sin x , cos x > 0

cos x = $\frac{1}{3}$

sin²x = 1 - cos²x

sin²x = 1 - $\frac{1}{9}$

sin²x = $\frac{8}{9}$ | $\sqrt{}$

sin x = $\frac{2\sqrt{2} }{3}$

tg x = $\frac{sinx}{cosx} = \frac{\frac{2\sqrt{2} }{3}}{\frac{1}{3} }$ = 2$\sqrt{2}$

tg(x-y) = $\frac{tgx-tgy}{1+tgx*tgy}$ = $\frac{2\sqrt{2}+\sqrt{5} }{1 + 2\sqrt{2}*\sqrt{5} }$ = $\frac{2\sqrt{2}+\sqrt{5} }{1+2\sqrt{10} }$

4. ctg(x+y) = ?

tg x = -3

ctg x = $tgx^{-1}$ = $-\frac{1}{3}$

ctg y = $-\frac{2}{3}$

ctg(x+y) = $\frac{ctg x*ctgy -1 }{ctgx + ctgy}$

ctg(x+y) = $\frac{-\frac{1}{3}*-\frac{2}{3}-1 }{-\frac{1}{3}- \frac{2}{3} }$ = $\frac{\frac{2}{9}-1}{-\frac{3}{3} } = \frac{\frac{2}{9} - \frac{9}{9} }{-1} = \frac{7}{9}$