What is y=3sinx+4cosx in arccos?

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Answer 1 · 2018-07-13T08:46:09

The answer is #=arccos(y/5)+arctan(3/4)#

#y=3sinx+4cosx#

#y=4cosx+3sinx#

Write #y=rcos(x-phi)#

Then,

#y=rcosxcosphi+rsinxsinphi#

Comparing the #2# functions

#{(rcosphi=4),(rsinphi=3):}#

#=>#, #r^2(cos^2phi+sin^2phi)=4^2+3^2=16+9=25#

#=>#, #r=sqrt25=5#

Therefore,

#tanphi=3/4#

#phi=arctan(3/4)#

And

#y=5cos(x-phi)#

#cos(x-phi)=y/5#

#arccos(y/5)=x-phi#

#x=arccos(y/5)+phi#

#x=arccos(y/5)+arctan(3/4)#