How do I solve this system for a and b? (I was given the hint to use sin x and cos x as constants.) a cos x + b sin x = 0 -a sin x +b cos x = tan x

1 Answer
Nov 18, 2017

a=-tanxsinx

b=sinx

Explanation:

Yes, you need to use sinx and cosx as constants:

acosx+bsinx=0
-asinx+bcosx=tanx

One way to tackle this is to solve for something from the top equation and substitute it into the bottom one. Let's solve for a:

acosx=-bsinx

a=(-bsinx)/cosx

Now let's plug this into the bottom equation:

-((-bsinx)/cosx)sinx+bcosx=tanx

btanxsinx+bcosx=tanx

Now let's factor the b out:

b(tanxsinx+cosx)=tanx

Now let's divide both sides by (tanxsinx+cosx)

(b(tanxsinx+cosx))/(tanxsinx+cosx)=tanx/(tanxsinx+cosx)

b=tanx/(tanxsinx+cosx)

Now let's see if we can simplify this. Let's plug in sinx/cosx for tanx:

b=(sinx/cosx)/((sinx/cosx)sinx+cosx)

b=(sinx/cosx)/((sin^2x/cosx)+cosx)

Let's add the two terms together at the bottom by taking a common denominator:

b=(sinx/cosx)/((sin^x+cos^2x)/cosx)

But we know:

sin^2x+cos^2x=1

b=(sinx/cosx)/(1/cosx)or

b=(sinx/cosx)*(cosx/1)

b=(sinx/cancelcosx)(cancel cosx/1)

b=sinx

Now let's plug it into the top equation to solve for a:

acosx+sinxsinx=0

acosx+sin^2x=0

acosx=-sin^2x

Now we divide bith sides by cosx to solve for a:

acosx/cosx=-sin^2x/cosx

acancelcosx/cancelcosx=-sin^2x/cosx

a=-sin^2x/cosx=-(sinxsinx)/cosx=-tanxsinx