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

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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

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Answer 1 · 2017-11-18T00:36:40

$a = - tan x sin x$ $a=-tanxsinx$

$b = sin x$ $b=sinx$

Explanation:

Yes, you need to use $sin x$ $sinx$ and $cos x$ $cosx$ as constants:

$a cos x + b sin x = 0$ $acosx+bsinx=0$
$- a sin x + b cos x = tan x$ $-asinx+bcosx=tanx$

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 $a$ $a$ :

$a cos x = - b sin x$ $acosx=-bsinx$

$a = \frac{- b sin x}{cos x}$ $a=(-bsinx)/cosx$

Now let's plug this into the bottom equation:

$- (\frac{- b sin x}{cos x}) sin x + b cos x = tan x$ $-((-bsinx)/cosx)sinx+bcosx=tanx$

$b tan x sin x + b cos x = tan x$ $btanxsinx+bcosx=tanx$

Now let's factor the $b$ $b$ out:

$b (tan x sin x + cos x) = tan x$ $b(tanxsinx+cosx)=tanx$

Now let's divide both sides by $(tan x sin x + cos x)$ $(tanxsinx+cosx)$

$\frac{b (tan x sin x + cos x)}{tan x sin x + cos x} = \frac{tan x}{tan x sin x + cos x}$ $(b(tanxsinx+cosx))/(tanxsinx+cosx)=tanx/(tanxsinx+cosx)$

$b = \frac{tan x}{tan x sin x + cos x}$ $b=tanx/(tanxsinx+cosx)$

Now let's see if we can simplify this. Let's plug in $\frac{sin x}{cos x}$ $sinx/cosx$ for $tan x$ $tanx$ :

$b = \frac{\frac{sin x}{cos x}}{(\frac{sin x}{cos x}) sin x + cos x}$ $b=(sinx/cosx)/((sinx/cosx)sinx+cosx)$

$b = \frac{\frac{sin x}{cos x}}{(\frac{{sin}^{2} x}{cos x}) + cos x}$ $b=(sinx/cosx)/((sin^2x/cosx)+cosx)$

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

$b = \frac{\frac{sin x}{cos x}}{\frac{{sin}^{x} + {cos}^{2} x}{cos x}}$ $b=(sinx/cosx)/((sin^x+cos^2x)/cosx)$

But we know:

${sin}^{2} x + {cos}^{2} x = 1$ $sin^2x+cos^2x=1$

$b = \frac{\frac{sin x}{cos x}}{\frac{1}{cos x}}$ $b=(sinx/cosx)/(1/cosx)$ or

$b = (\frac{sin x}{cos x}) \cdot (\frac{cos x}{1})$ $b=(sinx/cosx)*(cosx/1)$

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

$b=sinx$

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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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

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