Paritatea functiei cos si sin
cos(-x)=cos x
sin(-x)=-sin x
Stim ca:
cos (360+x)=cos x
sin(360+x)=sin x
cos(180+x)=-cos x
sin(180+x)=-sin x
cos(180-x)=-cos x
sin(180-x)=sin x
sin (360-x)=-sin x
cos(360-x)=cos x
sin(2kπ+x)=sin x
cos(2kπ+x)=cos x
c)
[tex]cos(-\frac{281\pi}{3} )=cos(\frac{281\pi}{3} )=cos( 300)\\\\[/tex]
cos 300°=cos (180+120)= -cos 120°
-cos 120°=-cos(180-60)=cos 60°=[tex]\frac{1}{2}[/tex]
[tex]cos(-\frac{281\pi}{3} )=\frac{1}{2}[/tex]
d)
[tex]sin(-\frac{407\pi}{4} )=-sin(\frac{407\pi}{4} )=-sin( 315)\\\\[/tex]
-sin 315°=-sin (360-45)=sin 45°=[tex]\frac{\sqrt{2} }{2}[/tex]
[tex]sin(-\frac{407\pi}{4} )=\frac{\sqrt{2} }{2}[/tex]
e)
[tex]sin(\frac{251\pi}{2} )=sin(270)[/tex]
sin 270°=sin (180+90)=-sin 90°=-1
[tex]sin(\frac{251\pi}{2} )=-1[/tex]
f)
[tex]cos(783\pi)=cos(180)=-1[/tex]
3)
[tex]cos\ x =-\frac{1}{5}[/tex]
d) Stim ca [tex]ctg\ x=\frac{cosx}{sinx}[/tex]
Aflam sin x din formula fundamentala a trigonometriei:
sin²x+cos²x=1
[tex]sin^2x+\frac{1}{25}=1\\\\ sin^2x=\frac{24}{25} \\\\sinx=-\frac{2\sqrt{6} }{5}[/tex] (este negativ, fiind in cadranul 3)
[tex]ctg\ x=\frac{cosx}{sinx}=\frac{-\frac{1}{5} }{-\frac{2\sqrt{6} }{5} }=\frac{1}{2\sqrt{6} }=\frac{\sqrt{6} }{12}[/tex]
4)
[tex]ctgx=\frac{4}{3}[/tex]
a)
[tex]ctg (a+b)=\frac{ctga\times ctgb-1}{ctga+ctgb}[/tex]
[tex]ctg (x+\frac{\pi}{4} )=ctg (x+45)=\frac{ctgx\times ctg45-1}{ctgx+ctg45}=\frac{ctgx-1}{ctgx+1}[/tex]
[tex]ctg (x+\frac{\pi}{4} )=\frac{\frac{4}{3}-1 }{\frac{4}{3}+1} =\frac{1}{7}[/tex]
b)
[tex]tgx=\frac{1}{ctgx} =\frac{3}{4}[/tex]
[tex]tg2x=\frac{2tgx}{1-tg^2x}[/tex]
[tex]tg2x=\frac{2\times \frac{3}{4} }{1-\frac{9}{16} } =\frac{\frac{3}{2} }{\frac{7}{16} }=\frac{24}{7}[/tex]
c)
[tex]\frac{cosx}{sinx} =\frac{4}{3} \\\\cosx=\sqrt{1-sin^2x}[/tex]
Notam sin x=y
[tex]\frac{4}{3} =\frac{\sqrt{1-y^2} }{y}[/tex]
[tex]4y=3\sqrt{1-y^2} \ \ \ \ |^2\\\\16y^2=9-9y^2\\\\25y^2=9\\\\y=sinx=-\frac{3}{5} \ (se\ afla\ in\ cadranul\ 3,\ deci\ este\ negativ)[/tex]
[tex]cosx=\sqrt{1-\frac{9}{25} } =-\frac{4}{5}[/tex]
d)
[tex]tgx=\frac{2tg\frac{x}{2} }{1-tg^\frac{x}{2} } \\\\Notam\ tg\frac{x}{2} =y[/tex]
[tex]\frac{3}{4} =\frac{2y}{1-y^2}[/tex]
3-3y²=8y
3y²+8y-3=0
Δ=b²-4ac
Δ=64+36=100
[tex]y=\frac{-b+\Delta}{2a}[/tex]
[tex]y=\frac{-8+10}{6} =\frac{1}{3} =tg\frac{x}{2}[/tex]
