Question

Proszę niech ktoś mi powie czy 1 zad a) dobrze jest zrobione bo nie wiem co z tym Minusem ma być jeżeli x leży w drugiej ćwiartce to jest dodatni tak?
Ja nie wiem czy cos²a = - 7/9 czy ten minus tez w pierwiastek czy przed ma zostać proszę pomocy!!!!!
Niech ktoś zrobi dwa pierwsze podpunkty żebym wiedziała jak mam dalej robić...a), b)

Answer 1

Verified answer

Korzystamy ze wzorów:

[tex]sin\alpha = \frac{y}{r}\\\\cos\alpha = \frac{x}{r}\\\\tg\alpha = \frac{y}{x}\\\\ctg\Lph = \frac{x}{y}\\\\r = \sqrt{x^{2}+y^{2}}[/tex]

a)

[tex]P = (-3,4) \ \ \rightarrow \ \ x = -3, \ y = 4\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(-3)^{2}+4^{2}}=\sqrt{9+16} = \sqrt{25} = 5\\\\\underline{r = 5}\\\\sin\alpha = \frac{y}{r} =\boxed{\frac{4}{5}}\\\\cos\alpha=\frac{x}{r} = \frac{-3}{5} =\boxed{ -\frac{3}{5}}\\\\tg\alpha=\frac{y}{x} = \frac{4}{-3} =\boxed{ -\frac{4}{3}}\\\\ctg\alpha =\frac{x}{y}=\frac{-3}{4} = \boxed{-\frac{3}{4}}[/tex]

b)

[tex]P = (4,-3) \ \ \rightarrow \ \ x = 4, \ y = -3\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{4^{2}+(-3)^{2}} = \sqrt{16+9} = \sqrt{25} = 5\\\\\underline{r = 5}\\\\sin\alpha = \frac{-3}{5}=\boxed{-\frac{3}{5}}\\\\cos\alpha= \boxed{\frac{4}{5}}\\\\tg\alpha = \frac{-3}{4} =\boxed{ -\frac{3}{4}}\\\\ctg\alpha =\boxed{ -\frac{4}{3}}[/tex]

c)

[tex]P = (-2,-1) \ \ \rightarrow \ \ x = -2, \ y = -1\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(-2)^{2}+(-1)^{2}} = \sqrt{4+1} - \sqrt{5}\\\\sin\alpha = \frac{-1}{\sqrt{5}} = -\frac{1}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} =\boxed{ -\frac{\sqrt{5}}{5}}\\\\cos\alpha = \frac{-2}{\sqrt{5}} = -\frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} =\boxed{-\frac{2\sqrt{5}}{5}}\\\\tg\alpha = \frac{-1}{-2} = \boxed{\frac{1}{2}}\\\\ctg\alpha = \frac{-2}{-1} = \boxed{2}[/tex]

d)

[tex]P = (-4,3) \ \ \rightarrow \ \ x = -4, \ y = 3\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(-4)^{2}+3^{2}} = \sqrt{16+9} = \sqrt{25} =5\\\\sin\alpha = \boxed{\frac{3}{5}}\\\\cos\alpha = \frac{-4}{5} =\boxed{ -\frac{4}{5}}\\\\tg\alpha = \frac{3}{-4} =\boxed{ -\frac{3}{4}}\\\\ctg\alpha = \frac{x}{y}=\frac{-4}{3} =\boxed{ -\frac{4}{3}}[/tex]

e)

[tex]P = (5,-11) \ \ \rightarrow \ \ x = 5, \ y = -11\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{5^{2}+(-11)^{2}} = \sqrt{25+121} = \sqrt{146}\\\\sin\alpha = \frac{-11}{\sqrt{146}} = -\frac{11}{\sqrt{146}}\cdot\frac{\sqrt{146}}{\sqrt{146}} =\boxed{-\frac{11\sqrt{146}}{146}}\\\\cos\alpha = \frac{5}{\sqrt{146}}\cdot\frac{\sqrt{146}}{\sqrt{146}} = \boxed{\frac{5\sqrt{146}}{146}}\\\\tg\alpha = \frac{-11}{5} =\boxed{-\frac{11}{5}}\\\\ctg\alpha = \frac{5}{-11} = \boxed{-\frac{5}{11}}[/tex]

f)

[tex]P = (-2,4) \ \ \rightarrow \ \ x = -2, \ y = 4\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(-2)^{2}+4^{2}} = \sqrt{4+16} = \sqrt{20} = \sqrt{4\cdot5} = 2\sqrt{5}\\\\sin\alpha = \frac{4}{2\sqrt{5}} = \frac{2}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} =\boxed{ \frac{2\sqrt{5}}{5}}\\\\cos\alpha = \frac{-2}{2\sqrt{5}} = -\frac{1}{\sqrt{5}}\cdot\frac{\sqrt{5}}{\sqrt{5}} = \boxed{-\frac{\sqrt{5}}{5}}\\\\tg\alpha = \frac{4}{-2} =\boxed{-2}\\\\ctg\alpha = \frac{-2}{4} = \boxed{-\frac{1}{2}}[/tex]

g)

[tex]P = (-\sqrt{3},1) \ \ \rightarrow \ \ x = -\sqrt{3}, \ y = 1\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(-\sqrt{3})^{2}+1^{2}} = \sqrt{3+1} = \sqrt{4} = 2\\\\sin\alpha =\boxed{ \frac{1}{2}}\\\\cos\alpha = \frac{-\sqrt{3}}{2} =\boxed{ -\frac{\sqrt{3}}{2}}\\\\tg\alpha = \frac{1}{-\sqrt{3}} =-\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} =\boxed{ -\frac{\sqrt{3}}{3}}\\\\ctg\alpha = \frac{-\sqrt{3}}{1} = \boxed{-\sqrt{3}}[/tex]

h)

[tex]P = (\sqrt{3},-1) \ \ \rightarrow \ \ x = \sqrt{3}, \ y = -1\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(\sqrt{3})^{2}+(-1)^{2}} = \sqrt{3+1}} = \sqrt{4} = 2\\\\sin\alpha = \frac{-1}{2} =\boxed{ -\frac{1}{2}}\\\\cos\alpha =\boxed{ \frac{\sqrt{3}}{2}}\\\\tg\alpha = \frac{-1}{\sqrt{3}} = -\frac{1}{\sqrt{3}}\cdot\frac{\sqrt{3}}{\sqrt{3}} = \boxed{-\frac{\sqrt{3}}{3}}\\\\ctg\alpha = \frac{\sqrt{3}}{-1} = \boxed{-\sqrt{3}}[/tex]

i)

[tex]P = (-\sqrt{2},2) \ \ \rightarrow \ \ x = -\sqrt{2}, \ y = 2\\\\r = \sqrt{x^{2}+y^{2}} = \sqrt{(-\sqrt{2})^{2}+2^{2}} = \sqrt{2+4} = \sqrt{6}\\\\sin\alpha = \frac{2}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt6}}=\frac{2\sqrt{6}}{6} = \boxed{\frac{\sqrt{6}}{3}}\\\\cos\alpha = \frac{-\sqrt{2}}{\sqrt{6}} = -\frac{\sqrt{2}}{\sqrt{6}}\cdot\frac{\sqrt{6}}{\sqrt{6}} = -\frac{\sqrt{2\cdot6}}{6} =-\frac{\sqrt{12}}{6}=-\frac{\sqrt{4\cdot3}}{6} = -\frac{2\sqrt{3}}{6} =\boxed{ -\frac{\sqrt{3}}{3}}[/tex]

[tex]tg\alpha = \frac{2}{-\sqrt{2}} = -\frac{2}{\sqrt{2}}\cdot\frac{\sqrt{2}}{\sqrt{2}} = -\frac{2\sqrt{2}}{2} = \boxed{-\sqrt{2}}\\\\ctg\alpha = \frac{-\sqrt{2}}{2}=\boxed{-\frac{\sqrt{2}}{2}}[/tex]