sin ⁡ ( α + β ) = \sin(\alpha+\beta)= sin ( α + β ) =

sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β \sin\alpha\cos\beta+\cos\alpha\sin\beta sin α cos β + cos α sin β

sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β \sin\alpha\cos\beta-\cos\alpha\sin\beta sin α cos β − cos α sin β

cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β \cos\alpha\cos\beta-\sin\alpha\sin\beta cos α cos β − sin α sin β

cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β \cos\alpha\cos\beta+\sin\alpha\sin\beta cos α cos β + sin α sin β

cos ⁡ ( α + β ) = \cos(\alpha+\beta)= cos ( α + β ) =

cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β \cos\alpha\cos\beta-\sin\alpha\sin\beta cos α cos β − sin α sin β

sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β \sin\alpha\cos\beta+\cos\alpha\sin\beta sin α cos β + cos α sin β

sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β \sin\alpha\cos\beta-\cos\alpha\sin\beta sin α cos β − cos α sin β

cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β \cos\alpha\cos\beta+\sin\alpha\sin\beta cos α cos β + sin α sin β

t g ( α + β ) = tg(\alpha+\beta)= t g ( α + β ) =

( t g α + t g β ) ( 1 − t g α t g β ) \frac{(tg\alpha+tg\beta)}{(1-tg\alpha tg\beta)} ( 1 − t g α t g β ) ( t g α + t g β )

( t g α − t g β ) ( 1 − t g α t g β ) \frac{(tg\alpha-tg\beta)}{(1-tg\alpha tg\beta)} ( 1 − t g α t g β ) ( t g α − t g β )

( 1 − t g α t g β ) ( t g α + t g β ) \frac{(1-tg\alpha tg\beta)}{(tg\alpha+tg\beta)} ( t g α + t g β ) ( 1 − t g α t g β )

( 1 + t g α t g β ) ( t g α + t g β ) \frac{(1+tg\alpha tg\beta)}{(tg\alpha+tg\beta)} ( t g α + t g β ) ( 1 + t g α t g β )

sin ⁡ α + sin ⁡ β = \sin\alpha+\sin\beta= sin α + sin β =

2 sin ⁡ α + β 2 cos ⁡ α − β 2 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 sin 2 α + β cos 2 α − β

2 sin ⁡ α − β 2 cos ⁡ α + β 2 2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2} 2 sin 2 α − β cos 2 α + β

2 cos ⁡ α + β 2 cos ⁡ α − β 2 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 cos 2 α + β cos 2 α − β

− 2 sin ⁡ α + β 2 sin ⁡ α − β 2 -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} − 2 sin 2 α + β sin 2 α − β

sin ⁡ ( α − β ) = \sin(\alpha-\beta)= sin ( α − β ) =

sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β \sin\alpha\cos\beta-\cos\alpha\sin\beta sin α cos β − cos α sin β

sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β \sin\alpha\cos\beta+\cos\alpha\sin\beta sin α cos β + cos α sin β

cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β \cos\alpha\cos\beta-\sin\alpha\sin\beta cos α cos β − sin α sin β

cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β \cos\alpha\cos\beta+\sin\alpha\sin\beta cos α cos β + sin α sin β

cos ⁡ ( α − β ) = \cos(\alpha-\beta)= cos ( α − β ) =

cos ⁡ α cos ⁡ β + sin ⁡ α sin ⁡ β \cos\alpha\cos\beta+\sin\alpha\sin\beta cos α cos β + sin α sin β

sin ⁡ α cos ⁡ β + cos ⁡ α sin ⁡ β \sin\alpha\cos\beta+\cos\alpha\sin\beta sin α cos β + cos α sin β

sin ⁡ α cos ⁡ β − cos ⁡ α sin ⁡ β \sin\alpha\cos\beta-\cos\alpha\sin\beta sin α cos β − cos α sin β

cos ⁡ α cos ⁡ β − sin ⁡ α sin ⁡ β \cos\alpha\cos\beta-\sin\alpha\sin\beta cos α cos β − sin α sin β

t g ( α − β ) = tg(\alpha-\beta)= t g ( α − β ) =

( t g α − t g β ) ( 1 + t g α t g β ) \frac{(tg\alpha-tg\beta)}{(1+tg\alpha tg\beta)} ( 1 + t g α t g β ) ( t g α − t g β )

( t g α + t g β ) ( 1 − t g α t g β ) \frac{(tg\alpha+tg\beta)}{(1-tg\alpha tg\beta)} ( 1 − t g α t g β ) ( t g α + t g β )

( 1 − t g α t g β ) ( t g α + t g β ) \frac{(1-tg\alpha tg\beta)}{(tg\alpha+tg\beta)} ( t g α + t g β ) ( 1 − t g α t g β )

( 1 + t g α t g β ) ( t g α + t g β ) \frac{(1+tg\alpha tg\beta)}{(tg\alpha+tg\beta)} ( t g α + t g β ) ( 1 + t g α t g β )

sin ⁡ α − sin ⁡ β = \sin\alpha-\sin\beta= sin α − sin β =

2 sin ⁡ α − β 2 cos ⁡ α + β 2 2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2} 2 sin 2 α − β cos 2 α + β

2 sin ⁡ α + β 2 cos ⁡ α − β 2 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 sin 2 α + β cos 2 α − β

2 cos ⁡ α + β 2 cos ⁡ α − β 2 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 cos 2 α + β cos 2 α − β

− 2 sin ⁡ α + β 2 sin ⁡ α − β 2 -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} − 2 sin 2 α + β sin 2 α − β

cos ⁡ α + cos ⁡ β = \cos\alpha+\cos\beta= cos α + cos β =

2 cos ⁡ α + β 2 cos ⁡ α − β 2 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 cos 2 α + β cos 2 α − β

2 sin ⁡ α + β 2 cos ⁡ α − β 2 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 sin 2 α + β cos 2 α − β

2 sin ⁡ α − β 2 cos ⁡ α + β 2 2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2} 2 sin 2 α − β cos 2 α + β

− 2 sin ⁡ α + β 2 sin ⁡ α − β 2 -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} − 2 sin 2 α + β sin 2 α − β

cos ⁡ α − cos ⁡ β = \cos\alpha-\cos\beta= cos α − cos β =

− 2 sin ⁡ α + β 2 sin ⁡ α − β 2 -2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2} − 2 sin 2 α + β sin 2 α − β

2 sin ⁡ α + β 2 cos ⁡ α − β 2 2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 sin 2 α + β cos 2 α − β

2 sin ⁡ α − β 2 cos ⁡ α + β 2 2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2} 2 sin 2 α − β cos 2 α + β

2 cos ⁡ α + β 2 cos ⁡ α − β 2 2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2} 2 cos 2 α + β cos 2 α − β

Selectați proprietățile funcției trigonometrice f ( x ) = sin ⁡ x f\left(x\right)=\sin x f ( x ) = sin x :

E ( sin ⁡ x ) = [ − 1 ; 1 ] E\left(\sin x\right)=\left[-1;1\right] E ( sin x ) = [ − 1 ; 1 ]

Perioada principală este T = 2 π T=2\pi T = 2 π

Punctele de maxim local x = π 2 + 2 k π , k ∈ Z x=\frac{\pi}{2}+2k\pi,\ k\in Z x = 2 π + 2 k π , k ∈ Z

Zerourile sunt numerele x = π 2 + k π , k ∈ Z x=\frac{\pi}{2}+k\pi,\ k\in Z x = 2 π + k π , k ∈ Z

Selectați proprietățile funcției trigonometrice f ( x ) = cos ⁡ x f\left(x\right)=\cos x f ( x ) = cos x :

Zerourile sunt numerele x = π 2 + k π , k ∈ Z x=\frac{\pi}{2}+k\pi,\ k\in Z x = 2 π + k π , k ∈ Z

Punctele de maxim local x = 2 k π , k ∈ Z x=2k\pi,\ k\in Z x = 2 k π , k ∈ Z

E ( cos ⁡ x ) = R E\left(\cos x\right)=R E ( cos x ) = R

Perioada principală este T = π T=\pi T = π

Selectați proprietățile funcției trigonometrice f ( x ) = t g x f\left(x\right)=tg\ x f ( x ) = t g x :

Zerourile sunt numerele x = k π , k ∈ Z x=k\pi,\ k\in Z x = k π , k ∈ Z

Ia valori pozitive în cadranele I și III

E ( t g x ) = R E\left(tg\ x\right)=R E ( t g x ) = R

Perioada principală este T = π T=\pi T = π

Selectați proprietățile funcției trigonometrice f ( x ) = ctg ⁡ x f\left(x\right)=\operatorname{ctg}\ x f ( x ) = c t g x :

Perioada principală este T = π T=\pi T = π

D ( ctg ⁡ x ) = R D\left(\operatorname{ctg}\ x\right)=R D ( c t g x ) = R

Zerourile sunt numerele x = k π , k ∈ Z x=k\pi,\ k\in Z x = k π , k ∈ Z

Care valori ale funcțiilor trigonometrice dau valoarea 2 2 \frac{\sqrt{2}}{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ?

cos ⁡ ( − π 4 ) \cos\left(-\frac{\pi}{4}\right) cos ( − 4 π )

sin ⁡ π 4 \sin\frac{\pi}{4} sin 4 π

t g π 4 tg\frac{\pi}{4} t g 4 π

ctg ⁡ ( − π 4 ) \operatorname{ctg}\left(-\frac{\pi}{4}\right) c t g ( − 4 π )

Care valori ale funcțiilor trigonometrice dau valoarea 3 \sqrt{3} 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z"> ?

t g π 3 tg\frac{\pi}{3} t g 3 π

ctg ⁡ π 6 \operatorname{ctg}\frac{\pi}{6} c t g 6 π

sin ⁡ π 3 \sin\frac{\pi}{3} sin 3 π

cos ⁡ π 6 \cos\frac{\pi}{6} cos 6 π

Valoarea unghiului α = 5 π 6 \alpha=\frac{5\pi}{6} α = 6 5 π în grade va fi:

a ° = 15 0 ° a^{\degree}=150^{\degree} a ° = 1 5 0 °

a ° = 14 0 ° a^{\degree}=140^{\degree} a ° = 1 4 0 °

a ° = 13 0 ° a^{\degree}=130^{\degree} a ° = 1 3 0 °

a ° = 12 0 ° a^{\degree}=120^{\degree} a ° = 1 2 0 °

Valoarea unghiului a ° = 11 0 ° a\degree=110^{\degree} a ° = 1 1 0 ° în radiani va fi:

α = 11 π 18 \alpha=\frac{11\pi}{18} α = 1 8 1 1 π

α = 13 π 18 \alpha=\frac{13\pi}{18} α = 1 8 1 3 π

α = 11 π 9 \alpha=\frac{11\pi}{9} α = 9 1 1 π

α = 13 π 9 \alpha=\frac{13\pi}{9} α = 9 1 3 π

Valoarea cos ⁡ 30 0 ° = \cos300^{\degree}= cos 3 0 0 ° =

2 2 \frac{\sqrt{2}}{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

− 2 2 -\frac{\sqrt{2}}{2} − 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Valorea sin ⁡ 5 π 6 = \sin\ \frac{5\pi}{6}= sin 6 5 π =

2 2 \frac{\sqrt{2}}{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 2 \frac{\sqrt{3}}{2} 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Valoarea t g 33 0 ° = tg\ 330^{\degree}= t g 3 3 0 ° =

− 3 3 -\frac{\sqrt{3}}{3} − 3 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 \sqrt{3} 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 3 \frac{\sqrt{3}}{3} 3 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

− 3 -\sqrt{3} − 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Valoarea ctg ⁡ 4 π 3 = \operatorname{ctg}\ \frac{4\pi}{3}= c t g 3 4 π =

3 3 \frac{\sqrt{3}}{3} 3 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 \sqrt{3} 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

sin ⁡ 73 ° cos ⁡ 1 3 ° − cos ⁡ 7 3 ° sin ⁡ 1 3 ° = \sin73\degree\cos13^{\degree}-\cos73^{\degree}\sin13^{\degree}= sin 7 3 ° cos 1 3 ° − cos 7 3 ° sin 1 3 ° =

3 2 \frac{\sqrt{3}}{2} 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 2 \frac{\sqrt{2}}{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

sin ⁡ π 8 ⋅ cos ⁡ π 8 = \sin\ \frac{\pi}{8}\cdot\cos\ \frac{\pi}{8}= sin 8 π ⋅ cos 8 π =

2 4 \frac{\sqrt{2}}{4} 4 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

2 2 \frac{\sqrt{2}}{2} 2 2 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

3 2 \frac{\sqrt{3}}{2} 2 3 <path d="M95,702c-2.7,0,-7.17,-2.7,-13.5,-8c-5.8,-5.3,-9.5, -10,-9.5,-14c0,-2,0.3,-3.3,1,-4c1.3,-2.7,23.83,-20.7,67.5,-54c44.2,-33.3,65.8, -50.3,66.5,-51c1.3,-1.3,3,-2,5,-2c4.7,0,8.7,3.3,12,10s173,378,173,378c0.7,0, 35.3,-71,104,-213c68.7,-142,137.5,-285,206.5,-429c69,-144,104.5,-217.7,106.5, -221c5.3,-9.3,12,-14,20,-14H400000v40H845.2724s-225.272,467,-225.272,467 s-235,486,-235,486c-2.7,4.7,-9,7,-19,7c-6,0,-10,-1,-12,-3s-194,-422,-194,-422 s-65,47,-65,47z M834 80H400000v40H845z">

Evaluare sumativă. Funcții și formule trigonometrice.

Authored by Natalia Pelevaniuc

Mathematics

10th Grade

Used 62+ times

Evaluare sumativă. Funcții și formule trigonometrice.

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30 questions

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MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\sin(\alpha+\beta)=$

$\sin\alpha\cos\beta+\cos\alpha\sin\beta$

$\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\cos\alpha\cos\beta-\sin\alpha\sin\beta$

$\cos\alpha\cos\beta+\sin\alpha\sin\beta$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\cos(\alpha+\beta)=$

$\sin\alpha\cos\beta+\cos\alpha\sin\beta$

$\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\cos\alpha\cos\beta-\sin\alpha\sin\beta$

$\cos\alpha\cos\beta+\sin\alpha\sin\beta$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$tg(\alpha+\beta)=$

$\frac{(tg\alpha+tg\beta)}{(1-tg\alpha tg\beta)}$

$\frac{(tg\alpha-tg\beta)}{(1-tg\alpha tg\beta)}$

$\frac{(1-tg\alpha tg\beta)}{(tg\alpha+tg\beta)}$

$\frac{(1+tg\alpha tg\beta)}{(tg\alpha+tg\beta)}$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\sin\alpha+\sin\beta=$

$2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$

$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\sin(\alpha-\beta)=$

$\sin\alpha\cos\beta+\cos\alpha\sin\beta$

$\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\cos\alpha\cos\beta-\sin\alpha\sin\beta$

$\cos\alpha\cos\beta+\sin\alpha\sin\beta$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\cos(\alpha-\beta)=$

$\sin\alpha\cos\beta+\cos\alpha\sin\beta$

$\sin\alpha\cos\beta-\cos\alpha\sin\beta$

$\cos\alpha\cos\beta-\sin\alpha\sin\beta$

$\cos\alpha\cos\beta+\sin\alpha\sin\beta$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$tg(\alpha-\beta)=$

$\frac{(tg\alpha+tg\beta)}{(1-tg\alpha tg\beta)}$

$\frac{(tg\alpha-tg\beta)}{(1+tg\alpha tg\beta)}$

$\frac{(1-tg\alpha tg\beta)}{(tg\alpha+tg\beta)}$

$\frac{(1+tg\alpha tg\beta)}{(tg\alpha+tg\beta)}$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\sin\alpha-\sin\beta=$

$2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$

$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\cos\alpha+\cos\beta=$

$2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$

$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

10.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

$\cos\alpha-\cos\beta=$

$2\sin\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$2\sin\frac{\alpha-\beta}{2}\cos\frac{\alpha+\beta}{2}$

$2\cos\frac{\alpha+\beta}{2}\cos\frac{\alpha-\beta}{2}$

$-2\sin\frac{\alpha+\beta}{2}\sin\frac{\alpha-\beta}{2}$

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Evaluare sumativă. Funcții și formule trigonometrice.

sin⁡(α+β)=\sin(\alpha+\beta)=sin(α+β)=

cos⁡(α+β)=\cos(\alpha+\beta)=cos(α+β)=

tg(α+β)=tg(\alpha+\beta)=tg(α+β)=

sin⁡α+sin⁡β=\sin\alpha+\sin\beta=sinα+sinβ=

sin⁡(α−β)=\sin(\alpha-\beta)=sin(α−β)=

cos⁡(α−β)=\cos(\alpha-\beta)=cos(α−β)=

tg(α−β)=tg(\alpha-\beta)=tg(α−β)=

sin⁡α−sin⁡β=\sin\alpha-\sin\beta=sinα−sinβ=

cos⁡α+cos⁡β=\cos\alpha+\cos\beta=cosα+cosβ=

cos⁡α−cos⁡β=\cos\alpha-\cos\beta=cosα−cosβ=

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$\sin(\alpha+\beta)=$

$\cos(\alpha+\beta)=$

$tg(\alpha+\beta)=$

$\sin\alpha+\sin\beta=$

$\sin(\alpha-\beta)=$

$\cos(\alpha-\beta)=$

$tg(\alpha-\beta)=$

$\sin\alpha-\sin\beta=$

$\cos\alpha+\cos\beta=$

$\cos\alpha-\cos\beta=$