Question

Si se cumple x = y + 30°; calcula: P = sen(x + y) sen²x - sen²y
Calcula:
H = cos20° + cos 100° + cos220°
Las 3 por favor​

Answer 1

Verified answer

Respuesta:

1)[tex]\sqrt{3}[/tex]

2)0

3)[tex]2[/tex]

Explicación paso a paso:

1)x+y=30°

sen(x+3y)=sen(x+y+2y)=sen(30+2y)

sen(y+3x)=sen(x+y+2x)=sen(30+2x)

Piden:[tex]\frac{sen(x+3y)+sen(3x+y)}{sen2x+sen2y}[/tex]

Por transformación de suma de senos

numerador:

[tex]senA+senB=2sen(\frac{A+B}{2})cos(\frac{A-B}{2})\\\\sen(x+3y)+sen(3x+y)=sen(30+2y)+sen(30+2x)=\\\\2sen(\frac{30+2y+30+2x}{2})cos(\frac{30+2y-(30+2x)}{2})=2sen(\frac{60+2x+2y}{2})cos(\frac{2y-2x}{2})=\\\\2sen(30+x+y)cos(y-x)=2sen(30+30)cos(y-x)=2sen60cos(y-x)[/tex]

Denominador:

[tex]sen2x+sen2y=2sen(\frac{2x+2y}{2})cos(\frac{2x-2y}{2})=2sen(x+y)cos(x-y)\\\\\\=2sen(30)cos(x-y)\\[/tex]

Sabemos que: cos(-x)=cos(x)

entonces cos(y-x)=cos(-(y-x))=cos(x-y)

sen60:   [tex]\frac{\sqrt{3} }{2}[/tex], sen30:  [tex]\frac{1}{2}[/tex]

H=

[tex]\frac{2sen60cos(y-x)}{2sen30cos(x-y)} =\frac{sen60(x-y)}{sen30(x-y)}=\frac{sen60}{sen30}=\frac{\frac{\sqrt{3} }{2} }{\frac{1}{2} }=\sqrt{3}[/tex]

2)H=cos20+cos100+cos220=cos20+cos220+cos110

Se aplica transformación de la suma de cosenos:

[tex]cosA+cosB=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})\\\\cos20+cos220=2cos(\frac{20+220}{2})cos(\frac{20-220}{2})=2cos(\frac{240}{2})cos(\frac{-200}{2})\\\\2cos(120)(-100), cos(-100)=cos(100)\\\\2cos(120)(-100)=2cos(120)(100)\\[/tex]

cos120: [tex]-\frac{1}{2}[/tex]

En H:

[tex]2cos(120)cos(100)+cos100=2(-\frac{1}{2})cos100+cos100=-cos100+cos100=0[/tex]

3)x=y+30°

P: [tex]\frac{sen(x+y)}{sen^{2}x -sen^{2}y }[/tex]

Por Identidad auxiliar de suma de dos ángulos

[tex]sen^{2}x-sen^{2}y=sen(x+y)sen(x-y)[/tex]

Denominador:

[tex]sen^{2}x-sen^{2}y=sen(x+y)sen(x-y)[/tex]

En P:

x=y+30 , x-y=30

sen30= 1/2

[tex]\frac{sen(x+y)}{sen(x+y)sen(x-y)} =\frac{1}{sen(x-y)}=\frac{1}{sen30}=\frac{1}{\frac{1}{2} }=1*2=2[/tex]