If #3sinx+5cosx=4, then #(3cosx-5sinx)^(2)=#?

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Answer 1 · 2017-12-02T08:46:38

The answer is #=18#

We need

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

As,

#3sinx+5cosx=4#

Then

#(3sinx+5cosx)^2=4^2=16#

#9sin^2x+25cos^2x+30sinxcosx=16#.........#(1)#

Let ,

#(3cosx-5sinx)^2=p#

Then

#9cos^2x+25sin^2x-30sinxcosx=p#.............#(2)#

Adding equations #(1)# and #(2)#

#9(sin^2x+cos^2x)+25(sin^2x+cos^2x)=p+16#

As the #30sinxcosx# cancels

Therefore,

#9+25=p+16#

#p=34-16=18#