Question

ajutor va rog ......​

Answer 1

Răspuns:

Explicație pas cu pas:

1.

$sin^{2}x+cos^{2}x=1=> sin^{2}x+(\frac{-5}{13})^2=1=>sin^{2}x+\frac{25}{169}=1=>sin^{2}x=1-\frac{25}{169}=>sin^{2}x=\frac{169-25}{169} =>sin^{2}x=\frac{144}{169}=>sinx=\sqrt{\frac{144}{169}}$

sin x=±$\frac{12}{13}$

x∈(pi;$\frac{3pi}{2}$)=>sinx=-$\frac{12}{13}$

$tgx=\frac{sinx}{cosx}=\frac{-\frac{12}{13}}{-\frac{5}{13}}=(-\frac{12}{13})*(-\frac{13}{5})=\frac{12}{5}\\ctgx=\frac{1}{tgx}=\frac{1}{\frac{12}{5}}=\frac{5}{12} \\sin2x=2sinxcox=2*(-\frac{12}{13})*(-\frac{5}{13})=\frac{120}{169}\\cos2x=2cos^2x-1=2*(-\frac{5}{13})^2-1=2*\frac{25}{169}-1=\frac{50}{169}-\frac{169}{169}=-\frac{119}{169}$

2.

a)

$sin35-cos65=sin35-sin(90-65)=sin35-sin25=2sin(\frac{35-25}{2} )cos(\frac{35+25}{2})=2sin\frac{10}{2}cos\frac{60}{2}=2*\frac{\sqrt{3}}{2}sin5=\sqrt{3}sin5$

b)

$cos20+cos50+cos160+cos130=(cos160+cos20)+(cos130+cos50)=2cos\frac{160+20}{2}cos\frac{160-20}{2}+2cos\frac{150+30}{2}cos\frac{130-50}{2}=2cos\frac{180}{2}cos\frac{140}{2}+2cos\frac{180}{2}cos\frac{80}{2}=2cos90cos70+2cos90cos40=2*0*cos70+2*0*cos40=0$

c)

$sin^250+sin^240=sin^250+cos^2(90-40)=sin^250+cos^250=1$

d)

$sin^2100+cos^280=sin^2100+cos^2(180-100)=sin^2100+(-cos100)^2=sin^2100+cos^2100=1$

3.

$cos75+sin105=cos(45+30)+sin(60+45)=cos45cos30-sin45sin30+sin60cos45+cos60sin45=\frac{\sqrt{2}}{2}* \frac{\sqrt{3}}{2} -\frac{\sqrt{2}}{2}*\frac{1}{2}+\frac{\sqrt{3}}{2}*\frac{\sqrt{2}}{2}+\frac{1}{2}*\frac{\sqrt{2}}{2}=2*\frac{\sqrt{6} }{2}=\sqrt{6}$

4.

$(sinx+cosx)^2-2sinxcosx=sin^2x+cos^2x+2sinxcosx-2sinxcox=1$