How do we Find the maximum and minimum vlaues of #1/(3sinx-4cosx+7)#?

Question

11495 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2018-03-20T08:30:46

Let

#y=1/(3sinx-4cosx+7)#

#=>y=1/((5(3/5sinx-4/5cosx)+7)#

Considering
#3/5=cosalpha#

So #4/5=sinalpha#

And #alpha=tan^-1(4/3)#

#=>y=1/((5(cosalphasinx-sinalphacosx)+7)#

#=>y=1/((5sin(x-alpha)+7))#

We know #-1<=sin(x-alpha)<=+1#

Hence #y# will be maximum when #sin(x+alpha)=-1#

So #=y_"max"=1/(5(-1)+7)=1/2#

And #y# will be minimum when #sin(x+alpha)=1#

So #=y_"min"=1/(5*1+7)=1/12#