if
[tex]if \: \tan( \beta + \alpha ) - (3 + 2 \sqrt{2} ) \tan( \beta ) = \: 0 \: \\: show \: that \: \sin(2 \beta + \alpha ) = \sqrt{2} \sin( \alpha ) [/tex]
plzz helppppp​

Respuesta :

Step-by-step explanation:

tan(β+α) − (3+2√2) tan β = 0

Convert to sine and cosine:

sin(β+α) / cos(β+α) − (3+2√2) sin β / cos β = 0

sin(β+α) / cos(β+α) = (3+2√2) sin β / cos β

Cross multiply:

sin(β+α) cos β = (3+2√2) sin β cos(β+α)

sin(2β+α)

Rearrange:

sin(β+(β+α))

Angle sum formula:

sin β cos(β+α) + sin(β+α) cos β

Substitute:

sin β cos(β+α) + (3+2√2) sin β cos(β+α)

(4+2√2) sin β cos(β+α)

Rearrange:

(2+√2) (2 sin β cos(β+α))

Product to sum:

(2+√2) (sin(2β+α) + sin(-α))

Reflection:

(2+√2) (sin(2β+α) − sin α)

Since this equals sin(2β+α) from the beginning:

(2+√2) (sin(2β+α) − sin α) = sin(2β+α)

(2+√2) sin(2β+α) − (2+√2) sin α = sin(2β+α)

(1+√2) sin(2β+α) − (2+√2) sin α = 0

(1+√2) sin(2β+α) = (2+√2) sin α

sin(2β+α) = (2+√2) / (1+√2) sin α

Multiply by the conjugate:

sin(2β+α) = (2+√2)(1−√2) / ((1+√2)(1−√2)) sin α

sin(2β+α) = (2−2√2+√2−2) / (1−√2+√2−2) sin α

sin(2β+α) = (-√2) / (-1) sin α

sin(2β+α) = √2 sin α

ACCESS MORE