If 3sinx + 4cosx = 5, then 4sinx - 3cosx =....???



  1. 0

  2. -5

  3. 1/5

  4. 5


plz explain me in detail... how to do questions like this... 

3 years ago

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Answers : (2)

                                        

ANS    3sinx+4cosx=5


     => 3/5 sinx +4/5 cosx = 1


     let cosA=3/5   => sinA=4/5


     =>  cosAsinx + sinAcosx = 1


     =>  sin(x+A) = 1


Now,


       4sinx - 3cosx


  =   5(4/5sinx - 3/5 cosx)     [multipying numerator and denominator by 5]


  =   5(sinAsinx - cosAcosx)


  =   -5{cos(x+A)} = -5[root{1-(sin(x+A)^2)}] = -5 x 0 = 0          Ans 0


 


TRICK: if asinx+bcosx is given then multiply numerator and denominator by root(a^2 +b^2)


            this method is useful  in many questions 


 


 


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3 years ago
                                        

3sinx+4cosx=5


Squaring on both sides.


(3sinx+4cosx) 2 = 25


9sin2x +16 cos2x+24sinxcosx=25


9(1- cos2x)+16(1- sin2x)+ 24sinxcosx=25


9-9 cos2x+16-16 sin2x+24sinxcosx=25


25-9 cos2x-16 sin2x+24sinxcosx=25


9 cos2x+16 sin2x-24sinxcosx=0


(3cosx-4sinx) 2 = 0


4sinx-3cosx=0


 

one year ago

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