Al aplicar la identidad para s e n 135 ° sen\ 135°\ s e n 1 3 5 ° sería:

s e n 180 ° cos ⁡ 45 ° − cos ⁡ 180 ° s e n 45 ° sen\ 180°\ \cos\ 45°-\cos\ 180°\ sen\ 45° s e n 1 8 0 ° cos 4 5 ° − cos 1 8 0 ° s e n 4 5 °

s e n 120 ° cos ⁡ 15 ° + cos ⁡ 120 ° s e n 15 ° sen\ 120°\ \cos\ 15°\ +\ \cos\ 120°\ sen\ 15° s e n 1 2 0 ° cos 1 5 ° + cos 1 2 0 ° s e n 1 5 °

El resultado de s e n 180 ° cos ⁡ 45 ° − cos ⁡ 180 ° s e n 45 ° sen\ 180°\ \cos\ 45°-\cos\ 180°\ sen\ 45° s e n 1 8 0 ° cos 4 5 ° − cos 1 8 0 ° s e n 4 5 ° es:

= ( 0 ) ( 2 2 ) − ( − 1 ) ( 2 2 ) = 2 2 =\left(0\right)\left(\frac{\sqrt{2}}{2}\right)-\left(-1\right)\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2} = ( 0 ) ( 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"> ) − ( − 1 ) ( 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 <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">

= ( 0 ) ( 2 2 ) − ( − 1 ) ( 2 2 ) = 2 2 2 = 2 =\left(0\right)\left(\frac{\sqrt{2}}{2}\right)-\left(-1\right)\left(\frac{\sqrt{2}}{2}\right)=\frac{2\sqrt{2}}{2}=\sqrt{2} = ( 0 ) ( 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"> ) − ( − 1 ) ( 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 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 <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">

Al aplicar la identidad para cos ⁡ 120 ° \cos\ 120° cos 1 2 0 ° se tiene:

cos ⁡ 90 ° cos ⁡ 30 ° − s e n 90 ° s e n 30 ° \cos\ 90°\ \cos\ 30°-\ sen\ 90°\ sen\ 30° cos 9 0 ° cos 3 0 ° − s e n 9 0 ° s e n 3 0 °

cos ⁡ 90 ° cos ⁡ 30 ° + s e n 90 ° s e n 30 ° \cos\ 90°\ \cos\ 30°\ +\ sen\ 90°\ sen\ 30° cos 9 0 ° cos 3 0 ° + s e n 9 0 ° s e n 3 0 °

cos ⁡ 150 ° cos ⁡ 30 ° − s e n 150 ° s e n 30 ° \cos\ 150°\ \cos\ 30°-\ sen\ 150°\ sen\ 30° cos 1 5 0 ° cos 3 0 ° − s e n 1 5 0 ° s e n 3 0 °

El resultado de cos ⁡ 90 ° cos ⁡ 30 ° − s e n 90 ° s e n 30 ° \cos\ 90°\ \cos\ 30°-\ sen\ 90°\ sen\ 30° cos 9 0 ° cos 3 0 ° − s e n 9 0 ° s e n 3 0 ° es:

= ( 0 ) ( 3 2 ) − ( 1 ) ( 1 2 ) = 1 2 =\left(0\right)\left(\frac{\sqrt{3}}{2}\right)-\left(1\right)\left(\frac{1}{2}\right)\ =\frac{1}{2} = ( 0 ) ( 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"> ) − ( 1 ) ( 2 1 ) = 2 1

= ( 0 ) ( 3 2 ) − ( 2 2 ) ( 1 2 ) = 2 4 =\left(0\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\ =\frac{\sqrt{2}}{4} = ( 0 ) ( 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 <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 1 ) = 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">

IDENTIDADES TRIGO + Y - ANGULOS

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Mathematics

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10th Grade

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Practice Problem

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Medium

Vanessa Sánchez

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6 Slides • 8 Questions

IDENTIDADES TRIGO + Y - ANGULOS

IDENTIDADES DE SUMA Y RESTA ÁNGULOS

Siempre expresar los ángulos en términos de los ángulos notables

Multiple Choice

La identidad de suma de sen es:

$sen\ \left(\alpha+\beta\right)=sen\alpha\ \cos\beta+sen\alpha\ \cos\beta$

$sen\ \left(\alpha+\beta\right)=\ sen\ \alpha\ \cos\beta+\cos\alpha\ sen\beta$

Multiple Choice

La identidad de suma de cos es:

$\cos\ \left(\alpha+\beta\right)=\ \cos\alpha\cos\beta-sen\ \alpha\ sen\ \beta$

$\cos\ \left(\alpha+\beta\right)=\ \cos\alpha\ \cos\beta+sen\ \alpha\ sen\ \beta$

Multiple Choice

La identidad de resta sen es:

$sen\ \left(\alpha-\beta\right)=sen\ \alpha\ \cos\beta+\ \cos\ \alpha\ sen\ \beta$

$sen\ \left(\alpha-\beta\right)=sen\ \alpha\ \cos\beta-\ \cos\ \alpha\ sen\ \beta$

Multiple Choice

La identidad de resta de cos es:

$\cos\ \left(\alpha-\beta\right)=\cos\alpha\ \cos\beta\ +\ sen\ \alpha\ sen\ \beta$

$\cos\ \left(\alpha-\beta\right)=\ sen\ \alpha\ \cos\beta\ +\cos\ \alpha\ sen\ \beta$

Recorderis de ángulos notables

y el truco es....?

Ejemplo aplicación identidad de suma en sen

$\text{sen (150°)= sen 90° cos 60° + cos 90° sen 60°}$
$sen\ 150°=\ \left(1\right)\left(\frac{1}{2}\right)\ +\left(0\right)\left(\frac{\sqrt{3}}{2}\right)$
$sen\ 150°=\ \frac{1}{2}$

Ejemplo de aplicación identidad resta en sen

$sen\ 15°=sen\ 45°\ \cos\ 30°\ -\ \cos\ 45°\ sen\ 30°$
$sen\ 15°=\ \left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{2}}{2\ }\right)\left(\frac{1}{2}\right)\$
$sen\ 15°=\frac{\sqrt{6}}{4\ }-\frac{\sqrt{2}}{4}=\frac{\left(\sqrt{6}-\sqrt{2}\right)}{4}$

Multiple Choice

Al aplicar la identidad para $sen\ 135°\$ sería:

$sen\ 180°\ \cos\ 45°-\cos\ 180°\ sen\ 45°$

$sen\ 120°\ \cos\ 15°\ +\ \cos\ 120°\ sen\ 15°$

Multiple Choice

El resultado de $sen\ 180°\ \cos\ 45°-\cos\ 180°\ sen\ 45°$ es:

$=\left(0\right)\left(\frac{\sqrt{2}}{2}\right)-\left(-1\right)\left(\frac{\sqrt{2}}{2}\right)=\frac{2\sqrt{2}}{2}=\sqrt{2}$

$=\left(0\right)\left(\frac{\sqrt{2}}{2}\right)-\left(-1\right)\left(\frac{\sqrt{2}}{2}\right)=\frac{\sqrt{2}}{2}$

Ejemplo de identidad de suma de cos:

$\cos\ \left(105°\right)=\ \cos\ 45°\ \cos\ 60°-sen\ 45°\ sen\ 60°$
$\cos\ 105°=\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\ -\left(\frac{\sqrt{2}}{2}\right)\left(\frac{\sqrt{3}}{2}\right)$
$\cos\ 105°=\frac{\sqrt{2}}{4}-\frac{\sqrt{6}}{4}=\frac{\left(\sqrt{2}-\sqrt{6}\right)}{4}$

Multiple Choice

Al aplicar la identidad para $\cos\ 120°$ se tiene:

$\cos\ 90°\ \cos\ 30°-\ sen\ 90°\ sen\ 30°$

$\cos\ 90°\ \cos\ 30°\ +\ sen\ 90°\ sen\ 30°$

$\cos\ 150°\ \cos\ 30°-\ sen\ 150°\ sen\ 30°$

Multiple Choice

El resultado de $\cos\ 90°\ \cos\ 30°-\ sen\ 90°\ sen\ 30°$ es:

$=\left(0\right)\left(\frac{\sqrt{3}}{2}\right)-\left(\frac{\sqrt{2}}{2}\right)\left(\frac{1}{2}\right)\ =\frac{\sqrt{2}}{4}$

$=\left(0\right)\left(\frac{\sqrt{3}}{2}\right)-\left(1\right)\left(\frac{1}{2}\right)\ =\frac{1}{2}$

IDENTIDADES TRIGO + Y - ANGULOS

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