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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... 

4 years ago

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Answers : (12)

										

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 


 


 


 PLEASE       APPROVE       MY       ANSWER       IF       YOU       LIKE       IT


 

4 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


 

3 years ago
										3sinx+4cosx=5Squaring on both sides.(3sinx+4cosx) 2 = 259sin2x +16 cos2x+24sinxcosx=259(1- cos2x)+16(1- sin2x)+ 24sinxcosx=259-9 cos2x+16-16 sin2x+24sinxcosx=2525-9 cos2x-16 sin2x+24sinxcosx=259 cos2x+16 sin2x-24sinxcosx=0(3cosx-4sinx) 2 = 04sinx-3cosx=0
										
11 months ago
										
3sinx+4cosx=5.
Do sqaring on both sides.
The equation will be as :- 9sin2x+16cos2x+24sinx.cosx=25.
Let the above equation be no.1
Think that the value of 4sinx-3cosx=p.
Do squaring on both sides.
The eqation will be as :- 16sin2x+9cos2x-24sinx.cosx=p2.
Let the above equation be no.2
Add equations no.1 and no.2
You will get as :- 25sin2x+25cos2x=25+p2
25(sin2x+cos2x)=25+p2
25=25+p2                                   (because sin2x+cos2x=1)
p2=0
p=0.
4sinx-3cosx=0.
Therefore answer is 0.
 
 
PLEASE  APPROVE  MY  ANSWER  IF IT IS RIGHT
11 months 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

11 months 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

11 months 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

 
10 months 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

 
10 months ago
										
 

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 

 

 

 
 
10 months ago
										
 

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 

 

 

 
 
10 months ago
										
 

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 

 

 

 
 
10 months ago
										
HINT:  (3sinx +4cosx)^2+(3cosx-4sinx)^2=(3^2+4^2)(cos^2x+sin^2x)
10 months ago

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