Trigonometry question solving help

[Rockefeller]

Homework Statement

Use the algebraic method to write √2 sin x - √2 cos x in the form
kcos(x + ϕ) where k>0, 0≤ ϕ ≤ 2π

Homework Equations

auxiliary form??

The Attempt at a Solution

my working out:

1st:

A= √2 B= -√2

K = √a^2 + b^2

= √(√2)^2 + (-√2)^2

= √2+2 = √4 = 2 therefore k = 2

2nd: find ϕ

let √2 sin - √2 cos x = k cos (x+ϕ )

= k [ cosx cosϕ - sinx sinϕ ]

therefore √2 sinx - √2 cosx = kcosx cosϕ -ksinx sinϕ

equate: sinx => √2 = - ksinϕ

sinϕ = - √2/k

therfore sinϕ = - √2/2 (k = 2)

cosx => -√2 = kcosϕ

cos ϕ = -√2/K

cos ϕ= -√2/√2

sinϕ = -√2/√2 < 0 therefore ϕ in 3rd quadrant

cosϕ = -√2/√2 < 0 therefore ϕ in 3rd quadrant

tan x = -1

x tan ^-1 (1) = 45

= 225. π/180

ϕ = 5π/4

therefore √2sinx--√2cosx = k cos(x+ϕ)

= √2sinx - √2cosx = 2cos (x + 5π/4)

therefore 2cos (x+ 5π/4)

is this correct??
and is there any easier way to show this or solve this ?? if so can u show me the full method

Thanks

 

[mjsd]

welcome!

looks good to me