# If 3sinx + 4cosx = 5, then 4sinx - 3cosx =....??? 0 -5 1/5 5 plz explain me in detail... how to do questions like this...

KARAN ACHARYA
52 Points
12 years 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

BARATH P
18 Points
11 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

Chandrabhushan Reddy Chigarapalli
25 Points
8 years 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
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.

sumanth
32 Points
8 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

Lab Bhattacharjee
121 Points
8 years ago
HINT:  (3sinx +4cosx)^2+(3cosx-4sinx)^2=(3^2+4^2)(cos^2x+sin^2x)
ritesh
31 Points
7 years ago
easiest method....given 3sinx+4cosx=5..then find 4sinx-3cosx=?
a=3  b=4  c=5...let our ans to be k
a^2+b^2=c^2+k^2
after putting values nd solving we get...k=0...easy method...

3 years ago
Hello student
This question can be solved using two methods both are quite easy

Method-1
Given,
3sinx + 4cosx = 5
Squaring both sides,
9sin2x + 16cos2x + 24sinxcosx = 25
9sin2x + 16 cos2x + 24sinxcosx = 25(sin2x + cos2x)       [ 25 = 25 x1 = 25 x (sin2x + cos2x)
24sinxcosx = 16sin2x + 9 cos2x
or, (4sinx)2 + (3cosx)2 – 2(4sinx)(3cosx) = 0
or, [4sinx – 3cosx]2 = 0
hence, 4sinx – 3cosx =0

Method-2
3sinx + 4cosx = 5
as we can notice- 3, 4, 5 are pythagorean triplet therefore are a part of the same right angled triangle.
Dividing both sides by 5
3/5 sinx + 4/5 cosx = 1
If we take 3/5 = cosA ; then sinA = 4/5
or, sinx cosA + cosx sinA = 1
or, sin(x + A) = 1

therefore,
cos(x + A) = [1 – {sin(x + A)}2]1/2  = [ 1 – 1 ]½  = 0
cosx cosA – sinx sinA = 0
or, 3/5 cosx – 4/5 sinx = 0
multiplying both sides by -5, we get,
4sinx – 3cosx = 0

You can use either of these methods in questions like this. But the 2nd method might be easier in case of some problems.
Hope it helps.
Regards,
Kushagra
Vikas TU
14149 Points
3 years ago
Dear student
Question is not clear
Good luck
Cheers

Zaid
13 Points
2 years ago
3sinx + 4cosx = 5
as we can notice- 3, 4, 5 are pythagorean triplet therefore are a part of the same right angled triangle.
Dividing both sides by 5
3/5 sinx + 4/5 cosx = 1
If we take 3/5 = cosA ; then sinA = 4/5
or, sinx cosA + cosx sinA = 1
or, sin(x + A) = 1

therefore,
cos(x + A) = [1 – {sin(x + A)}2]1/2  = [ 1 – 1 ]½  = 0
cosx cosA – sinx sinA = 0
or, 3/5 cosx – 4/5 sinx = 0
multiplying both sides by -5, we get,
4sinx – 3cosx = 0