[tex]\sqrt{}[/tex]Răspuns:
1. sin(x-y) = 1
2.cos(x+y) = [tex]-\frac{1+\sqrt{2} }{3}[/tex]
3. tg(x-y) = [tex]\frac{2\sqrt{2}+\sqrt{5} }{1+2\sqrt{10} }[/tex]
4. ctg(x+y) = [tex]\frac{7}{9}[/tex]
Explicație pas cu pas:
1. sin(x-y) = ?
x, y ∈ ([tex]\frac{\pi}2}[/tex],[tex]\pi[/tex]) => sin x > 0, cos x < 0
sin y > 0, cos y < 0
sin x = [tex]\frac{2}{3}[/tex]
sin²x + cos²x = 1
([tex]\frac{2}{3}[/tex])² + cos²x = 1
[tex]\frac{4}{9}[/tex] + cos²x = 1 | - [tex]\frac{4}{9}[/tex]
cos²x = [tex]\frac{5}{9}[/tex] | [tex]\sqrt{}[/tex]
cos x = ± [tex]\frac{\sqrt{5}}{3}[/tex]
dar cos x < 0 => cos x = - [tex]\frac{\sqrt{5}}{3}[/tex]
cos y = [tex]\frac{2}{3}[/tex]
sin²y + cos²y = 1
sin²y + [tex](-\frac{2}{3}) ^{2}[/tex]= 1
sin²y + [tex]\frac{4}{9}[/tex] = 1 | - [tex]\frac{4}{9}[/tex]
sin²y = [tex]\frac{5}{9}[/tex] | [tex]\sqrt{}[/tex]
sin y = ± [tex]\frac{\sqrt{5}}{3}[/tex]
dar sin y > 0 = > sin y = [tex]\frac{\sqrt{5}}{3}[/tex]
sin(x-y) = sin x · cos y - sin y · cos x
sin(x-y) = [tex]\frac{2}{3}.\frac{2}{3} - \frac{\sqrt{5}}{3}.\frac{-\sqrt{5}}{3}[/tex]
sin(x-y) = [tex]\frac{4}{9} + \frac{5}{9}[/tex]
sin(x-y) = [tex]\frac{9}{9} = 1[/tex]
2. cos(x+y) = ?
x ∈ [tex](\pi, \frac{3\pi }{2})[/tex] => sin x < 0, cos x < 0
y ∈ [tex](\frac{3\pi }{2}, 2\pi)[/tex] => sin y < 0, cos y > 0
sin x = [tex]-\frac{1}{\sqrt{3}}[/tex]
cos²x = 1 - sin²x
cos²x = [tex]1-\frac{1}{3}[/tex]
cos²x = [tex]\frac{2}{3}[/tex] | [tex]\sqrt{}[/tex]
cos x = ± [tex]\frac{\sqrt{6}}{3}[/tex]
dar cos x < 0 => cos x = [tex]-\frac{\sqrt{6} }{3}[/tex]
tg y = -[tex]\sqrt{5}[/tex] => sin y = [tex]-\sqrt{5} cos y[/tex]
cos²y = 1 - sin²y
cos²y = 1 - 5cos²y | + 5cos²y
6cos²y = 1 | :6
cos²y = [tex]\frac{1}{6}[/tex] | [tex]\sqrt{}[/tex]
cos y = [tex]\frac{\sqrt{6}}{6}[/tex] => sin y = [tex]-\frac{\sqrt{30}}{6}[/tex]
cos(x+y) = cos x · cos y - sin x · sin y
cos(x+y) = [tex]-\frac{\sqrt{6} }{3} . \frac{\sqrt{6} }{6} - \frac{-\sqrt{3} }{3} . \frac{-\sqrt{6} }{3}[/tex]
cos(x+y) = [tex]-\frac{6}{18} - \frac{3\sqrt{2} }{9}[/tex]
cos(x+y) = [tex]-\frac{3}{9} - \frac{3\sqrt{2} }{9}[/tex]
cos(x+y) = [tex]-\frac{1+\sqrt{2} }{3}[/tex]
3. tg(x-y) = ?
x ∈ [tex](0, \frac{\pi}{2} )[/tex] => sin x , cos x > 0
cos x = [tex]\frac{1}{3}[/tex]
sin²x = 1 - cos²x
sin²x = 1 - [tex]\frac{1}{9}[/tex]
sin²x = [tex]\frac{8}{9}[/tex] | [tex]\sqrt{}[/tex]
sin x = [tex]\frac{2\sqrt{2} }{3}[/tex]
tg x = [tex]\frac{sinx}{cosx} = \frac{\frac{2\sqrt{2} }{3}}{\frac{1}{3} }[/tex] = 2[tex]\sqrt{2}[/tex]
tg(x-y) = [tex]\frac{tgx-tgy}{1+tgx*tgy}[/tex] = [tex]\frac{2\sqrt{2}+\sqrt{5} }{1 + 2\sqrt{2}*\sqrt{5} }[/tex] = [tex]\frac{2\sqrt{2}+\sqrt{5} }{1+2\sqrt{10} }[/tex]
4. ctg(x+y) = ?
tg x = -3
ctg x = [tex]tgx^{-1}[/tex] = [tex]-\frac{1}{3}[/tex]
ctg y = [tex]-\frac{2}{3}[/tex]
ctg(x+y) = [tex]\frac{ctg x*ctgy -1 }{ctgx + ctgy}[/tex]
ctg(x+y) = [tex]\frac{-\frac{1}{3}*-\frac{2}{3}-1 }{-\frac{1}{3}- \frac{2}{3} }[/tex] = [tex]\frac{\frac{2}{9}-1}{-\frac{3}{3} } = \frac{\frac{2}{9} - \frac{9}{9} }{-1} = \frac{7}{9}[/tex]
