Question

Am cateva exercitii legate de vectori si trigonometrie la care ar fi de ajutor un raspuns, am incercuit cu rosu un subpunct deoarece doar acela m-a blocat. Multumesc in avans! Nu cred ca ar mai fi trebuit sa mentionez ca ofer coroana

Answer 1

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)

$cos(-\frac{281\pi}{3} )=cos(\frac{281\pi}{3} )=cos( 300)$

cos 300°=cos (180+120)= -cos 120°

-cos 120°=-cos(180-60)=cos 60°=$\frac{1}{2}$

$cos(-\frac{281\pi}{3} )=\frac{1}{2}$

d)

$sin(-\frac{407\pi}{4} )=-sin(\frac{407\pi}{4} )=-sin( 315)$

-sin 315°=-sin (360-45)=sin 45°=$\frac{\sqrt{2} }{2}$

$sin(-\frac{407\pi}{4} )=\frac{\sqrt{2} }{2}$

e)

$sin(\frac{251\pi}{2} )=sin(270)$

sin 270°=sin (180+90)=-sin 90°=-1

$sin(\frac{251\pi}{2} )=-1$

f)

$cos(783\pi)=cos(180)=-1$

3)

$cos\ x =-\frac{1}{5}$

d) Stim ca  $ctg\ x=\frac{cosx}{sinx}$

Aflam sin x din formula fundamentala a trigonometriei:

sin²x+cos²x=1

$sin^2x+\frac{1}{25}=1\\\\ sin^2x=\frac{24}{25} \\\\sinx=-\frac{2\sqrt{6} }{5}$ (este negativ, fiind in cadranul 3)

$ctg\ x=\frac{cosx}{sinx}=\frac{-\frac{1}{5} }{-\frac{2\sqrt{6} }{5} }=\frac{1}{2\sqrt{6} }=\frac{\sqrt{6} }{12}$

4)

$ctgx=\frac{4}{3}$

a)

$ctg (a+b)=\frac{ctga\times ctgb-1}{ctga+ctgb}$

$ctg (x+\frac{\pi}{4} )=ctg (x+45)=\frac{ctgx\times ctg45-1}{ctgx+ctg45}=\frac{ctgx-1}{ctgx+1}$

$ctg (x+\frac{\pi}{4} )=\frac{\frac{4}{3}-1 }{\frac{4}{3}+1} =\frac{1}{7}$

b)

$tgx=\frac{1}{ctgx} =\frac{3}{4}$

$tg2x=\frac{2tgx}{1-tg^2x}$

$tg2x=\frac{2\times \frac{3}{4} }{1-\frac{9}{16} } =\frac{\frac{3}{2} }{\frac{7}{16} }=\frac{24}{7}$

c)

$\frac{cosx}{sinx} =\frac{4}{3} \\\\cosx=\sqrt{1-sin^2x}$

Notam sin x=y

$\frac{4}{3} =\frac{\sqrt{1-y^2} }{y}$

$4y=3\sqrt{1-y^2} \ \ \ \ |^2\\\\16y^2=9-9y^2\\\\25y^2=9\\\\y=sinx=-\frac{3}{5} \ (se\ afla\ in\ cadranul\ 3,\ deci\ este\ negativ)$

$cosx=\sqrt{1-\frac{9}{25} } =-\frac{4}{5}$

d)

$tgx=\frac{2tg\frac{x}{2} }{1-tg^\frac{x}{2} } \\\\Notam\ tg\frac{x}{2} =y$

$\frac{3}{4} =\frac{2y}{1-y^2}$

3-3y²=8y

3y²+8y-3=0

Δ=b²-4ac

Δ=64+36=100

$y=\frac{-b+\Delta}{2a}$

$y=\frac{-8+10}{6} =\frac{1}{3} =tg\frac{x}{2}$