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
Sean
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#

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