Rockefeller
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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