Write 3cosx-4sinx in the form k(sinx-β)?

2 Answers
Narad T.
May 21, 2018

Please see the explanation below.

Explanation:

We need

sin(a-b)=sinacosb-sinbcosasin(a−b)=sinacosb−sinbcosa

cos^2x+sin^2x=1cos2x+sin2x=1

Therefore,

3cosx-4sinx=ksin(x-beta)3cosx−4sinx=ksin(x−β)

3cosx-4sinx=k(sinxcosbeta-sinbetacosx)3cosx−4sinx=k(sinxcosβ−sinβcosx)

So,

{(-4=kcosbeta),(3=-ksinbeta):}

<=>, {(cosbeta=-4/k),(sinbeta=-3/k):}

cos^2beta+sin^2beta=16/k^2+9/k^2=25/k^2=1

=>, k^2=25

k=+-5

k=5,=>, {(cosbeta=-4/k=-4/5),(sinbeta=-3/k=-3/5):}

=>, tanbeta=3/4

=>, beta=216.9^@

=5sin(x-216.9^@)

k=-5,=>, {(cosbeta=-4/k=4/5),(sinbeta=-3/k=3/5):}

=>, tanbeta=3/4

=>, beta=36.9^@

=-5sin(x-36.9^@)

Nghi N
May 21, 2018

f(x) = (1.8)sin (x - 36^@87)

Explanation:

f(x) = 3cos x - 4sin x = 3(cos x - (4/3)sin x) (1)
Call cot b = cos b/(sin b) = 4/3 --> tan b = 3/4
Calculator gives -->
b = 36^@87 --> sin b = 0.6
f(x) becomes:
f(x) = 3(cos x - (cos b/(sin b))sin x)
f(x) = 3sin b(cos x.sin b - cos b.sin x)
f(x) = 3(0.6)sin (x - b) = (1.8)sin (x - 36^@87)

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