Zad. 1
[tex]a = 1cm\\b = 4cm\\a^2+b^2=c^2\\1^2+4^2=c^2\\1+16=c^2\\17=c^2\\c=\sqrt{17}\\\\sin\alpha = \frac{1}{\sqrt{17}}=\frac{\sqrt{17}}{17}\\cos\alpha=\frac{4}{\sqrt{17}}=\frac{4\sqrt{17}}{17}\\tg\alpha=\frac{1}4\\ctg\alpha=\frac{4}1=4\\\\sin\beta=\frac{4}{\sqrt{17}}=\frac{4\sqrt{17}}{17}\\cos\beta=\frac{1}{\sqrt{17}}=\frac{\sqrt{17}}{17}\\tg\beta=\frac{4}1=4\\ctg\beta=\frac{1}4[/tex]
Zad. 2
[tex]a = 1cm\\b = 3cm\\a^2+b^2=c^2\\1^2+3^2=c^2\\1+9=c^2\\c=\sqrt{10}\\\\sin\alpha=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}\\cos\alpha=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}\\tg\alpha=\frac{1}{3}\\ctg\alpha=\frac{3}1=3\\\\sin\beta=\frac{3}{\sqrt{10}}=\frac{3\sqrt{10}}{10}\\cos\beta=\frac{1}{\sqrt{10}}=\frac{\sqrt{10}}{10}\\tg\beta=\frac{3}1=3\\ctg\beta=\frac{1}3[/tex]
Zad. 3
[tex]a=1cm\\c = 6cm\\a^2+b^2=c^2\\1^2+b^2=6^2\\1+b^2=36 /-1\\b^2=35\\b=\sqrt{35}\\\\sin\alpha=\frac{1}6\\cos\alpha=\frac{\sqrt{35}}6\\tg\alpha=\frac{1}{\sqrt{35}}=\frac{\sqrt{35}}{35}\\ctg\alpha=\frac{\sqrt{35}}1=\sqrt{35}\\\\sin\beta=\frac{\sqrt{35}}6\\cos\beta=\frac{1}{6}\\tg\beta=\frac{\sqrt{35}}1=\sqrt{35}\\ctg\beta=\frac{1}{\sqrt{35}}=\frac{\sqrt{35}}{35}[/tex]
Zad. 4
[tex]a = 5cm\\c = 10cm\\a^2+b^2=c^2\\5^2+b^2=10^2\\25+b^2=100 /-25\\b^2=75\\b=\sqrt{75}=5\sqrt{3}\\\\sin\alpha=\frac{5}{10}=\frac12\\cos\alpha=\frac{5\sqrt3}{10}=\frac12\sqrt3\\tg\alpha = \frac{5}{5\sqrt3}=\frac{1}{\sqrt3}=\frac{\sqrt{3}}3\\ctg\alpha=\frac{5\sqrt3}5=\sqrt3\\\\sin\beta=\frac{5\sqrt3}{10}=\frac12\sqrt3\\cos\beta=\frac5{10}=\frac12\\tg\beta=\frac{5\sqrt3}5=\sqrt3\\ctg\beta=\frac{5}{5\sqrt3}=\frac{1}{\sqrt3}=\frac{\sqrt3}3[/tex]
Zad. 5
a)
[tex]sin\alpha=\frac25\\a = 2\\c=5\\2^2+b^2=5^2\\4+b^2=25 /-4\\b^2=21\\b=\sqrt{21}\\cos\alpha=\frac{\sqrt{21}}5\\tg\alpha=\frac{2}{\sqrt{21}}=\frac{2\sqrt{21}}{21}\\ctg\alpha=\frac{\sqrt{21}}2[/tex]
b)
[tex]sin\alpha=\frac35\\a=3\\c=5\\3^2+b^2=5^2\\9+b^2=25\\b^2=16\\b=4\\cos\alpha=\frac{4}5\\tg\alpha=\frac{3}4\\ctg\alpha=\frac{4}3[/tex]
c)
[tex]cos\alpha=\frac23\\b=2\\c=3\\a^2+2^2=3^2\\a^2+4=9\\a^2=9-4\\a^2=5\\a=\sqrt5\\sin\alpha=\frac{\sqrt5}3\\tg\alpha=\frac{\sqrt5}2\\ctg\alpha=\frac{2}{\sqrt5}=\frac{2\sqrt5}5[/tex]
d)
[tex]tg\alpha=\frac14\\a=1\\b=4\\1^2+4^2=c^2\\1+16=c^2\\17=c^2\\c=\sqrt{17}\\sin\alpha=\frac{1}{\sqrt{17}}=\frac{\sqrt{17}}{17}\\cos\alpha=\frac{4}\sqrt{17}}=\frac{4\sqrt{17}}{17}\\ctg\alpha=\frac{4}1=4[/tex]
e)
[tex]tg\alpha=\frac34\\a=3\\b=4\\3^2+4^2=c^2\\9+16=c^2\\25=c^2\\c=5\\sin\alpha=\frac{3}5\\cos\alpha=\frac45\\ctg\alpha=\frac43[/tex]
f)
[tex]cos\alpha=\frac{\sqrt2}2\\b=\sqrt2\\c=2\\a^2+(\sqrt2)^2=2^2\\a^2+2=4 \\a^2=2\\a=\sqrt2\\sin\alpha=\frac{\sqrt2}2\\tg\alpha=\frac{\sqrt2}{\sqrt2}=1\\ctg\alpha=\frac{\sqrt2}{\sqrt2}=1[/tex]
Zad. 6
[tex]a)\\\frac{sin150-2sin30}{cos120}=\frac{sin(180-30)-2sin30}{cos(90+30)}=\frac{sin30-2sin30}{-sin30}=\frac{-sin30}{-sin30}=1[/tex]
b)
[tex]4cos150-2tg135=4cos(180-30)-2tg(180-45)=4*(-cos30)-2*(-tg45)=4*(-\frac{\sqrt3}2)-2*(-1)=-2\sqrt3+2[/tex]
c)
[tex]sin20*cos70+cos^220=sin20*cos(90-20)+cos^220=sin20*sin20+cos^220=sin^220+cos^220=1 \\\text{Jedynka trygonometryczna! : } sin^2\alpha+cos^2\alpha=1[/tex]
d)
[tex]\frac{sin120-2sin60}{cos150}=\frac{sin(180-60)-2sin60}{cos(180-30)}=\frac{sin60-2sin60}{-cos30}=\frac{-sin60}{-cos30}=\frac{-(\frac{\sqrt3}2)}{-(\frac{\sqrt3}2)}=1[/tex]
e)
[tex]4cos120-2tg135=4cos(180-60)-2tg(180-45)=4*(-cos60)-2*(-tg45)=4*(-\frac12)-2*(-1)=-2+2=0[/tex]
f)
[tex]sin25*cos65+cos^225=sin25*cos(90-25)+cos^225=sin25*sin25+cos^225=sin^225+cos^225=1\\\text{Jedynka trygonometryczna!}[/tex]
g)
[tex]sin120+cos150=sin(180-60)+cos(180-30)=sin60+(-cos30)=\frac{\sqrt3}2+(-\frac{\sqrt3}2)=0[/tex]
h)
[tex]sin30*cos120+2tg120=sin30*cos(180-60)+2tg(180-60)=sin30*(-cos60)+2*(-tg60)=\frac12*(-\frac12)+2*(-\sqrt3)=-\frac14-2\sqrt3[/tex]
i)
[tex]sin60*cos150+6tg150=sin60*cos(180-30)+6tg(180-30)=sin60*(-cos30)+6*(-tg30)=\frac{\sqrt3}2*(-\frac{\sqrt3}2)+6*(-\frac{\sqrt3}3)=-\frac{3}4-2\sqrt3[/tex]
j)
[tex]sin60*cos120+3tg120=sin60*cos(180-60)+3tg(180-60)=sin60*(-cos60)+3*(-tg60)=\frac{\sqrt3}2*(-\frac12)+3*(-\sqrt3)=-\frac{\sqrt3}4-3\sqrt3=-\frac{\sqrt3}4-\frac{12\sqrt3}{4}=\frac{-\sqrt3-12\sqrt3}4=-\frac{13\sqrt3}4[/tex]
Zad. 7
[tex]sin\alpha=\frac25\\cos\alpha=\frac14\\sin^2\alpha+cos^2\alpha=1\\(\frac25)^2+(\frac14)^2=1\\\frac{4}{25}+\frac{1}{16}=1\\\frac{64}{400}+\frac{25}{400}=1\\\frac{89}{400} \neq 1\\\text{Taki kat ostry nie istnieje}[/tex]