Question

"Simple" systems of equations.

I saw this one in my Diff Equ book and cannot figure it out.Normally I just do things like this with my simple understanding of lin. alg. But no clue how to do it with trig functions. If you could help explain it to me I would be very greatful.

Acos(x)+Bsin(x)=0
A(-sin(x))+bcos(x)=tan(x)

The answer is a=-sin(x)tan(x)=cos(x)-sec(x) and b=cos(x)tan(x)=sin(x)

Answer 1

So you have the two equations:
(1) Asin(x) + Bcos(x) = 0
(2) -Asin(x) + Bcos(x) = tan(x)

Solve (1) for A:
(3) A = -Btan(x)

Substitute (3) into (2):
Btan(x) + Bcos(x) = tan(x)

Simplify:
B(1/cos(x)) = tan(x)

This gives B = cos(x)tan(x) = sin(x) as required...
Back-substitute into (3) and simplify

Answer 2

Wow, I guess I had a major brain fart there. Didn't see that tan(X)sin(x)+cos(x)=sec.. Thanks!

Answer 3

sometimes trig functions start dancing around on the page after you've looked at them for a while... :)

glad i could help.