SOLUTION: Express −5.9sin(x)+6.6cos(x) in the form ksin(x+ϕ)
K=
ϕ=
Note: ϕ should be given in radians in the interval −π<ϕ<π
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Trigonometry-basics
-> SOLUTION: Express −5.9sin(x)+6.6cos(x) in the form ksin(x+ϕ)
K=
ϕ=
Note: ϕ should be given in radians in the interval −π<ϕ<π
Log On
(1) -5.9sin(x)+6.6cos(x) in the form ksin(x+ϕ)
That makes us think of the right side of the identity:
cos(x+ϕ) = cos(x)cos(ϕ) - sin(x)sin(ϕ)
multiplied by some constant k:
(2) k*cos(x+ϕ) = k[cos(x)cos(ϕ) - sin(x)sin(ϕ)]
Let's get
(1) -5.9sin(x)+6.6cos(x)
in the form of the right side of (2)
k[cos(x)cos(ϕ) - sin(x)sin(ϕ)]
(1) -5.9sin(x)+6.6cos(x) =
cos(x)(6.6) - sin(x)(5.9)
Let A be some positive non-zero constant. Then by multiplying
then dividing by A
We find A and ϕ such that
and
We see that ϕ must be in QIV because that's where the
cosine is positive and the sine is positive. We draw
an angle in QI to represent ϕ.
We can find A by the Pythagorean theorem:
We can find ϕ from
using the inverse tangent feature on a calculator
in radian mode:
ϕ = 0.7294565922
So
now becomes
which is equivalent to
or
Edwin