trigonometry problem plz?

Question

if 3sinA+5cosA=5
then prove, 5sinA-3cosA=+-3

Answer 1

3 ( 3sinA+5cos A) =3*5
5(5sinA - 3 cosA) =+- 5*3

9sinA +15cos A=15
25sinA-15 cos A= +-15
ADD THESES EQUATIONS TOGETHER
34sin A= 0 , 30
sinA= 0 , .88
A= 62 degree or 0 degree
1ST SOLUTION
now sin 62 =.88 , cos 62=.47
5 sin 62 - 3 cos 62 = +3
2ND SOLUTION
sin0= 0 , cos 0=1
5sin 0 - 3cos 0= -3
good luck

Answer 2

Given
3 sin A + 5 cos A = 5
Prove that
5 sin A - 3 cos A = ± 3

From the first equation, we can solve for sin A:
sin A = 5/3 (1 - cos A)

If we square both sides,
sin² A = 25/9 (1 - 2 cos A + cos² A)

Simplifying
sin² A = 25/9 (cos² A - 2 cos A + 10)

Since we also know that
sin² A = 1 - cos² A,

By the TPE,
25/9 (cos² A - 2 cos A + 1) = 1 - cos² A

We can now solve for cos A:
25 cos² A - 50 cos A + 25 = 9 - 9 cos² A
34 cos² A - 50 cos A + 16 = 0
17 cos² A - 25 cos A + 8 = 0
(17 cos A - 8)(cos A - 1) = 0

Therefore, the possible values for cos A is:
cos A = 8/17 or cos A = 1

Since
sin A = 5/3 (1 - cos A),

We can substitute these values to the expression:
5 sin A - 3 cos A
=5 (5/3)(1 - cos A) - 3 cos A
= 25/3 (1 - cos A) - 3 cos A

There are 2 possible values:
= 25/3 (1 - 8/17) - 3(8/17) or = 25/3 (1 - 1) - 3(1)
= 25/3 (9/17) - 24/17 or = 25/3 (0) - 3(1)
= (75 - 24)/17 or = 0 - 3(1)
= 51/17 or = -3(1)
= 3 or = -3

Thus,
5 sin A - 3 cos A = ±3

QED

^_^

Answer 3

if A=0 or A=360 degrees,
the first equation will be true.

sin A= sin 0= 0
cos A= cos 0=1

Hence
3. sin0-3 cos 0=0-3= - 3

Answer 4

Wish I understood math better, but I am not wired that way. Props to all of you. If this was a plant problem I could hook you up, though.

Sorry for the bother, but I am actually trying to get in touch with DHALLA W as I cannot reply to a question that was emailed to me. DHALLA W's settings block me out.

Answer 5

I tried... but its tough one...

sinA= 5*(1-cosA)/3

substituing thios value into the second expression does not give the result... I give up...

thanks.