De s e n θ ⋅ csc ⁡ θ = 1 \ sen\theta\cdot\csc\theta=1 s e n θ ⋅ csc θ = 1 se obtiene que:

s e n θ = 1 csc ⁡ θ sen\theta=\frac{1}{\csc\theta} s e n θ = csc θ 1

cos ⁡ θ ⋅ sec ⁡ θ = 1 \cos\theta\cdot\sec\theta=1 cos θ ⋅ sec θ = 1

tan ⁡ θ = 1 cot ⁡ θ \tan\theta=\frac{1}{\cot\theta} tan θ = cot θ 1

sec ⁡ θ = 1 cos ⁡ θ \sec\theta=\frac{1}{\cos\theta} sec θ = cos θ 1

De cos ⁡ θ ⋅ sec ⁡ θ = 1 \ \cos\theta\cdot\sec\theta=1 cos θ ⋅ sec θ = 1 se obtiene que:

sec ⁡ θ = 1 cos ⁡ θ \sec\theta=\frac{1}{\cos\theta} sec θ = cos θ 1

s e n θ = 1 csc ⁡ θ sen\theta=\frac{1}{\csc\theta} s e n θ = csc θ 1

tan ⁡ θ = 1 cot ⁡ θ \tan\theta=\frac{1}{\cot\theta} tan θ = cot θ 1

sec ⁡ θ = 1 cos ⁡ θ \sec\theta=\frac{1}{\cos\theta} sec θ = cos θ 1

De cos ⁡ θ ⋅ sec ⁡ θ = 1 \ \cos\theta\cdot\sec\theta=1 cos θ ⋅ sec θ = 1 se obtiene que:

sec ⁡ θ = 1 cos ⁡ θ \sec\theta=\frac{1}{\cos\theta} sec θ = cos θ 1

csc ⁡ θ = 1 s e n θ \csc\theta=\frac{1}{sen\theta} csc θ = s e n θ 1

s e n θ = 1 csc ⁡ θ sen\theta=\frac{1}{\csc\theta} s e n θ = csc θ 1

tan ⁡ θ = s e n θ cos ⁡ θ \tan\theta=\frac{sen\theta}{\cos\theta} tan θ = cos θ s e n θ

De s e n θ ⋅ csc ⁡ θ = 1 \ sen\theta\cdot\csc\theta=1 s e n θ ⋅ csc θ = 1 se obtiene que:

csc ⁡ θ = 1 s e n θ \csc\theta=\frac{1}{sen\theta} csc θ = s e n θ 1

cos ⁡ θ ⋅ sec ⁡ θ = 1 \cos\theta\cdot\sec\theta=1 cos θ ⋅ sec θ = 1

cot ⁡ θ = cos ⁡ θ s e n θ \cot\theta=\frac{\cos\theta}{sen\theta} cot θ = s e n θ cos θ

tan ⁡ θ = 1 cot ⁡ θ \tan\theta=\frac{1}{\cot\theta} tan θ = cot θ 1

De tan ⁡ θ ⋅ cot ⁡ θ = 1 \tan\theta\cdot\cot\theta=1 tan θ ⋅ cot θ = 1 se obtiene que:

tan ⁡ θ = 1 cot ⁡ θ \tan\theta=\frac{1}{\cot\theta} tan θ = cot θ 1

cos ⁡ θ ⋅ sec ⁡ θ = 1 \cos\theta\cdot\sec\theta=1 cos θ ⋅ sec θ = 1

cot ⁡ θ = cos ⁡ θ s e n θ \cot\theta=\frac{\cos\theta}{sen\theta} cot θ = s e n θ cos θ

csc ⁡ θ = 1 s e n θ \csc\theta=\frac{1}{sen\theta} csc θ = s e n θ 1

De tan ⁡ θ ⋅ cot ⁡ θ = 1 \tan\theta\cdot\cot\theta=1 tan θ ⋅ cot θ = 1 se obtiene que:

cot ⁡ θ = 1 tan ⁡ θ \cot\theta=\frac{1}{\tan\theta} cot θ = tan θ 1

cot ⁡ θ = cos ⁡ θ s e n θ \cot\theta=\frac{\cos\theta}{sen\theta} cot θ = s e n θ cos θ

tan ⁡ θ = 1 s e n θ \tan\theta=\frac{1}{sen\theta} tan θ = s e n θ 1

csc ⁡ θ = 1 s e n θ \csc\theta=\frac{1}{sen\theta} csc θ = s e n θ 1

tan ⁡ θ \tan\theta tan θ es igual a...

s e n θ cos ⁡ θ \frac{sen\theta}{\cos\theta} cos θ s e n θ

1 cos ⁡ θ \frac{1}{\cos\theta} cos θ 1

cos ⁡ θ s e n θ \frac{\cos\theta}{sen\theta} s e n θ cos θ

1 s e n θ \frac{1}{sen\theta} s e n θ 1

cot ⁡ θ \cot\theta cot θ es igual a...

cos ⁡ θ s e n θ \frac{\cos\theta}{sen\theta} s e n θ cos θ

s e n θ cos ⁡ θ \frac{sen\theta}{\cos\theta} cos θ s e n θ

1 cos ⁡ θ \frac{1}{\cos\theta} cos θ 1

1 s e n θ \frac{1}{sen\theta} s e n θ 1

El s e n ( α + β ) sen\left(\alpha+\beta\right) s e n ( α + β ) es igual a...

s e n α ⋅ cos ⁡ β ± s e n β ⋅ cos ⁡ α sen\alpha\cdot\cos\beta\pm sen\beta\cdot\cos\alpha s e n α ⋅ cos β ± s e n β ⋅ cos α

s e n α ⋅ cos ⁡ β − s e n α ⋅ cos ⁡ β sen\alpha\cdot\cos\beta-sen\alpha\cdot\cos\beta s e n α ⋅ cos β − s e n α ⋅ cos β

s e n α ⋅ cos ⁡ β ÷ s e n α ⋅ cos ⁡ β sen\alpha\cdot\cos\beta\div sen\alpha\cdot\cos\beta s e n α ⋅ cos β ÷ s e n α ⋅ cos β

El cos ⁡ ( α + β ) \cos\left(\alpha+\beta\right) cos ( α + β ) es igual a...

s e n α ⋅ cos ⁡ β ± s e n β ⋅ cos ⁡ α sen\alpha\cdot\cos\beta\pm sen\beta\cdot\cos\alpha s e n α ⋅ cos β ± s e n β ⋅ cos α

s e n α ⋅ cos ⁡ β − s e n α ⋅ cos ⁡ β sen\alpha\cdot\cos\beta-sen\alpha\cdot\cos\beta s e n α ⋅ cos β − s e n α ⋅ cos β

cos ⁡ α ⋅ cos ⁡ β ± s e n α ⋅ s e n β \cos\alpha\cdot\cos\beta\pm sen\alpha\cdot sen\beta cos α ⋅ cos β ± s e n α ⋅ s e n β

La tan ⁡ ( α ± β ) = \tan\left(\alpha\pm\beta\right)= tan ( α ± β ) = es igual a...

tan ⁡ α + tan ⁡ β 1 + tan ⁡ α ⋅ tan ⁡ β \frac{\tan\alpha+\tan\beta}{1+\tan\alpha\cdot\tan\beta} 1 + tan α ⋅ tan β tan α + tan β

tan ⁡ α ± tan ⁡ β 1 − tan ⁡ α ⋅ tan ⁡ β \frac{\tan\alpha\pm\tan\beta}{1-\tan\alpha\cdot\tan\beta} 1 − tan α ⋅ tan β tan α ± tan β

tan ⁡ α ± tan ⁡ β 1 − tan ⁡ α + tan ⁡ β \frac{\tan\alpha\pm\tan\beta}{1-\tan\alpha+\tan\beta} 1 − tan α + tan β tan α ± tan β

El ángulo doble s e n ( 2 θ ) = sen\left(2\theta\right)= s e n ( 2 θ ) = es igual a...

2 ⋅ s e n θ ⋅ cos ⁡ θ 2\cdot sen\theta\cdot\cos\theta 2 ⋅ s e n θ ⋅ cos θ

2 ⋅ s e n θ + cos ⁡ θ 2\cdot sen\theta+\cos\theta 2 ⋅ s e n θ + cos θ

2 − s e n θ ⋅ cos ⁡ θ 2-sen\theta\cdot\cos\theta 2 − s e n θ ⋅ cos θ

2 ⋅ s e n θ ⋅ s e n θ 2\cdot sen\theta\cdot sen\theta 2 ⋅ s e n θ ⋅ s e n θ

El angulo medio s e n ( θ 2 ) = sen\left(\frac{\theta}{2}\right)= s e n ( 2 θ ) = es igual a...

± 1 − cos ⁡ θ 2 \pm\sqrt{\frac{1-\cos\theta}{2}} ± 2 1 − cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

± 1 + cos ⁡ θ 2 \pm\sqrt{\frac{1+\cos\theta}{2}} ± 2 1 + cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

− 1 + cos ⁡ θ 2 -\sqrt{\frac{1+\cos\theta}{2}} − 2 1 + cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

N i n g u n a Ninguna N i n g u n a

El angulo medio cos ⁡ ( θ 2 ) = \cos\left(\frac{\theta}{2}\right)= cos ( 2 θ ) = es igual a...

± 1 + cos ⁡ θ 2 \pm\sqrt{\frac{1+\cos\theta}{2}} ± 2 1 + cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

± 1 − cos ⁡ θ 2 \pm\sqrt{\frac{1-\cos\theta}{2}} ± 2 1 − cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

± 1 ± cos ⁡ θ 2 \pm\sqrt{\frac{1\pm\cos\theta}{2}} ± 2 1 ± cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

N i n g u n a Ninguna N i n g u n a

El angulo medio tan ⁡ ( θ 2 ) = \tan\left(\frac{\theta}{2}\right)= tan ( 2 θ ) = es igual a...

± 1 − cos ⁡ θ 1 + cos ⁡ θ \pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} ± 1 + cos θ 1 − cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

1 − cos ⁡ θ s e n θ \frac{1-\cos\theta}{sen\theta} s e n θ 1 − cos θ

s e n θ 1 + cos ⁡ θ \frac{sen\theta}{1+\cos\theta} 1 + cos θ s e n θ

± 1 − cos ⁡ θ 2 \pm\sqrt{\frac{1-\cos\theta}{2}} ± 2 1 − cos θ <path d="M424,2478c-1.3,-0.7,-38.5,-172,-111.5,-514c-73, -342,-109.8,-513.3,-110.5,-514c0,-2,-10.7,14.3,-32,49c-4.7,7.3,-9.8,15.7,-15.5, 25c-5.7,9.3,-9.8,16,-12.5,20s-5,7,-5,7c-4,-3.3,-8.3,-7.7,-13,-13s-13,-13,-13, -13s76,-122,76,-122s77,-121,77,-121s209,968,209,968c0,-2,84.7,-361.7,254,-1079 c169.3,-717.3,254.7,-1077.7,256,-1081c4,-6.7,10,-10,18,-10H400000v40H1014.6 s-87.3,378.7,-272.6,1166c-185.3,787.3,-279.3,1182.3,-282,1185c-2,6,-10,9,-24,9 c-8,0,-12,-0.7,-12,-2z M1001 80H400000v40H1014z">

La identidad s e n α ⋅ cos ⁡ β = sen\alpha\cdot\cos\beta= s e n α ⋅ cos β = es igual a...

1 2 [ cos ⁡ ( α − β ) − cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha-\beta\right)-\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α − β ) − cos ( α + β ) ]

1 2 [ cos ⁡ ( α − β ) + cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha-\beta\right)+\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α − β ) + cos ( α + β ) ]

1 2 [ s e n ( α + β ) + s e n ( α − β ) ] \frac{1}{2}\left[sen\left(\alpha+\beta\right)+sen\left(\alpha-\beta\right)\right] 2 1 [ s e n ( α + β ) + s e n ( α − β ) ]

1 2 [ cos ⁡ ( α + β ) − cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha+\beta\right)-\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α + β ) − cos ( α + β ) ]

La identidad cos ⁡ α ⋅ cos ⁡ β = \cos\alpha\cdot\cos\beta= cos α ⋅ cos β = es igual a...

1 2 [ cos ⁡ ( α − β ) + cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha-\beta\right)+\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α − β ) + cos ( α + β ) ]

1 2 [ cos ⁡ ( α − β ) − cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha-\beta\right)-\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α − β ) − cos ( α + β ) ]

1 2 [ s e n ( α + β ) + s e n ( α − β ) ] \frac{1}{2}\left[sen\left(\alpha+\beta\right)+sen\left(\alpha-\beta\right)\right] 2 1 [ s e n ( α + β ) + s e n ( α − β ) ]

1 2 [ cos ⁡ ( α + β ) − cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha+\beta\right)-\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α + β ) − cos ( α + β ) ]

La identidad s e n α ⋅ cos ⁡ β = sen\alpha\cdot\cos\beta= s e n α ⋅ cos β = es igual a...

1 2 [ s e n ( α + β ) + s e n ( α − β ) ] \frac{1}{2}\left[sen\left(\alpha+\beta\right)+sen\left(\alpha-\beta\right)\right] 2 1 [ s e n ( α + β ) + s e n ( α − β ) ]

1 2 [ cos ⁡ ( α − β ) − cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha-\beta\right)-\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α − β ) − cos ( α + β ) ]

1 2 [ cos ⁡ ( α − β ) + cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha-\beta\right)+\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α − β ) + cos ( α + β ) ]

1 2 [ cos ⁡ ( α + β ) − cos ⁡ ( α + β ) ] \frac{1}{2}\left[\cos\left(\alpha+\beta\right)-\cos\left(\alpha+\beta\right)\right] 2 1 [ cos ( α + β ) − cos ( α + β ) ]

La identidad s e n α + s e n β = sen\alpha+sen\beta= s e n α + s e n β = es igual a...

2 ⋅ s e n ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ s e n ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ cos ⁡ ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot\cos\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ cos ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ s e n ( α − β 2 ) ⋅ cos ⁡ ( α + β 2 ) 2\cdot sen\left(\frac{\alpha-\beta}{2}\right)\cdot\cos\left(\frac{\alpha+\beta}{2}\right) 2 ⋅ s e n ( 2 α − β ) ⋅ cos ( 2 α + β )

− 2 ⋅ s e n ( α + β 2 ) ⋅ s e n ( α − β 2 ) -2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot sen\left(\frac{\alpha-\beta}{2}\right) − 2 ⋅ s e n ( 2 α + β ) ⋅ s e n ( 2 α − β )

La identidad cos ⁡ α + cos ⁡ β = \cos\alpha+\cos\beta= cos α + cos β = es igual a...

2 ⋅ cos ⁡ ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot\cos\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ cos ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ s e n ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ s e n ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ s e n ( α − β 2 ) ⋅ cos ⁡ ( α + β 2 ) 2\cdot sen\left(\frac{\alpha-\beta}{2}\right)\cdot\cos\left(\frac{\alpha+\beta}{2}\right) 2 ⋅ s e n ( 2 α − β ) ⋅ cos ( 2 α + β )

− 2 ⋅ s e n ( α + β 2 ) ⋅ s e n ( α − β 2 ) -2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot sen\left(\frac{\alpha-\beta}{2}\right) − 2 ⋅ s e n ( 2 α + β ) ⋅ s e n ( 2 α − β )

La identidad s e n α + s e n β = sen\alpha+sen\beta= s e n α + s e n β = es igual a...

2 ⋅ s e n ( α − β 2 ) ⋅ cos ⁡ ( α + β 2 ) 2\cdot sen\left(\frac{\alpha-\beta}{2}\right)\cdot\cos\left(\frac{\alpha+\beta}{2}\right) 2 ⋅ s e n ( 2 α − β ) ⋅ cos ( 2 α + β )

2 ⋅ cos ⁡ ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot\cos\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ cos ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ s e n ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ s e n ( 2 α + β ) ⋅ cos ( 2 α − β )

− 2 ⋅ s e n ( α + β 2 ) ⋅ s e n ( α − β 2 ) -2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot sen\left(\frac{\alpha-\beta}{2}\right) − 2 ⋅ s e n ( 2 α + β ) ⋅ s e n ( 2 α − β )

La identidad cos ⁡ α + cos ⁡ β = \cos\alpha+\cos\beta= cos α + cos β = es igual a...

− 2 ⋅ s e n ( α + β 2 ) ⋅ s e n ( α − β 2 ) -2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot sen\left(\frac{\alpha-\beta}{2}\right) − 2 ⋅ s e n ( 2 α + β ) ⋅ s e n ( 2 α − β )

2 ⋅ cos ⁡ ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot\cos\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ cos ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ s e n ( α + β 2 ) ⋅ cos ⁡ ( α − β 2 ) 2\cdot sen\left(\frac{\alpha+\beta}{2}\right)\cdot\cos\left(\frac{\alpha-\beta}{2}\right) 2 ⋅ s e n ( 2 α + β ) ⋅ cos ( 2 α − β )

2 ⋅ s e n ( α − β 2 ) ⋅ cos ⁡ ( α + β 2 ) 2\cdot sen\left(\frac{\alpha-\beta}{2}\right)\cdot\cos\left(\frac{\alpha+\beta}{2}\right) 2 ⋅ s e n ( 2 α − β ) ⋅ cos ( 2 α + β )

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IDENTIDADES TRIGONOMÉTRICAS

Authored by jannette quisbert

Mathematics

11th Grade

Used 10+ times

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35 questions

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MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De $\ sen\theta\cdot\csc\theta=1$ se obtiene que:

$\cos\theta\cdot\sec\theta=1$

$sen\theta=\frac{1}{\csc\theta}$

$\tan\theta=\frac{1}{\cot\theta}$

$\sec\theta=\frac{1}{\cos\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De $\ \cos\theta\cdot\sec\theta=1$ se obtiene que:

$\sec\theta=\frac{1}{\cos\theta}$

$sen\theta=\frac{1}{\csc\theta}$

$\tan\theta=\frac{1}{\cot\theta}$

$\sec\theta=\frac{1}{\cos\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De $\ \cos\theta\cdot\sec\theta=1$ se obtiene que:

$\csc\theta=\frac{1}{sen\theta}$

$sen\theta=\frac{1}{\csc\theta}$

$\sec\theta=\frac{1}{\cos\theta}$

$\tan\theta=\frac{sen\theta}{\cos\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De $\ sen\theta\cdot\csc\theta=1$ se obtiene que:

$\cos\theta\cdot\sec\theta=1$

$\cot\theta=\frac{\cos\theta}{sen\theta}$

$\tan\theta=\frac{1}{\cot\theta}$

$\csc\theta=\frac{1}{sen\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De $\tan\theta\cdot\cot\theta=1$ se obtiene que:

$\cos\theta\cdot\sec\theta=1$

$\cot\theta=\frac{\cos\theta}{sen\theta}$

$\tan\theta=\frac{1}{\cot\theta}$

$\csc\theta=\frac{1}{sen\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De $\tan\theta\cdot\cot\theta=1$ se obtiene que:

$\cot\theta=\frac{1}{\tan\theta}$

$\cot\theta=\frac{\cos\theta}{sen\theta}$

$\tan\theta=\frac{1}{sen\theta}$

$\csc\theta=\frac{1}{sen\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\tan\theta$ es igual a...

$\frac{sen\theta}{\cos\theta}$

$\frac{1}{\cos\theta}$

$\frac{\cos\theta}{sen\theta}$

$\frac{1}{sen\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

$\cot\theta$ es igual a...

$\frac{sen\theta}{\cos\theta}$

$\frac{1}{\cos\theta}$

$\frac{\cos\theta}{sen\theta}$

$\frac{1}{sen\theta}$

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

La identidad $sen^2\theta+\cos^2\theta$ es igual a...

ninguna

10.

MULTIPLE CHOICE QUESTION

30 sec • 1 pt

De la identidad $sen^2\theta+\cos^2\theta=1$ se obtiene...

$sen^2\theta=1-\cos^2\theta$

$sen\theta=1-\cos^2\theta$

$\tan\theta+1=\sec^2\theta$

ninguna

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IDENTIDADES TRIGONOMÉTRICAS

De senθ⋅csc⁡θ=1\ sen\theta\cdot\csc\theta=1 senθ⋅cscθ=1 se obtiene que:

De cos⁡θ⋅sec⁡θ=1\ \cos\theta\cdot\sec\theta=1 cosθ⋅secθ=1 se obtiene que:

De cos⁡θ⋅sec⁡θ=1\ \cos\theta\cdot\sec\theta=1 cosθ⋅secθ=1 se obtiene que:

De senθ⋅csc⁡θ=1\ sen\theta\cdot\csc\theta=1 senθ⋅cscθ=1 se obtiene que:

De tan⁡θ⋅cot⁡θ=1\tan\theta\cdot\cot\theta=1tanθ⋅cotθ=1 se obtiene que:

De tan⁡θ⋅cot⁡θ=1\tan\theta\cdot\cot\theta=1tanθ⋅cotθ=1 se obtiene que:

tan⁡θ\tan\thetatanθ es igual a...

cot⁡θ\cot\thetacotθ es igual a...

La identidad sen2θ+cos⁡2θsen^2\theta+\cos^2\thetasen2θ+cos2θ es igual a...

De la identidad sen2θ+cos⁡2θ=1sen^2\theta+\cos^2\theta=1sen2θ+cos2θ=1 se obtiene...

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Similar Resources on Wayground

Popular Resources on Wayground

Discover more resources for Mathematics

De $\ sen\theta\cdot\csc\theta=1$ se obtiene que:

De $\ \cos\theta\cdot\sec\theta=1$ se obtiene que:

De $\ \cos\theta\cdot\sec\theta=1$ se obtiene que:

De $\ sen\theta\cdot\csc\theta=1$ se obtiene que:

De $\tan\theta\cdot\cot\theta=1$ se obtiene que:

De $\tan\theta\cdot\cot\theta=1$ se obtiene que:

$\tan\theta$ es igual a...

$\cot\theta$ es igual a...

La identidad $sen^2\theta+\cos^2\theta$ es igual a...

De la identidad $sen^2\theta+\cos^2\theta=1$ se obtiene...