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-sinbcosa#

#cos^2x+sin^2x=1#

Therefore,

#3cosx-4sinx=ksin(x-beta)#

#3cosx-4sinx=k(sinxcosbeta-sinbetacosx)#

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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