0
$\begingroup$

Eliminate B from each pair of equations:

x=sinB -3cosB and y=sinB+2cosB

  • I've tried solving this simultaneously just as the textbook has guided me through, but it still doesn't work. My initial working out was moving sinB to the left hand side and everything else to the right and solve simultaneously. But that didn't work out, I must have done something wrong, can someone please help me solve this. Thanks!
$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

HINT:

Solve for $\sin B,\cos B$

Then use $\sin^2B+\cos^2B=1$

$\endgroup$
14
  • $\begingroup$ Can you please elaborate as too how to solve for sin B? $\endgroup$
    – Red Queen10101
    Jul 8, 2013 at 15:21
  • 1
    $\begingroup$ @RedQueen10101,$$\sin B+2\cos B-(\sin B-3\cos B)=y-x\implies 5\cos B=y-x\implies \cos B=\frac{y-x}5$$ and from the first equation $$\sin B=x+3\cos B=x+3\cdot \frac{y-x}5=\frac{2x+3y}5$$ Now, use the last identity in the answer $\endgroup$
    – lab bhattacharjee
    Jul 8, 2013 at 15:46
  • 1
    $\begingroup$ $\sin^2B+\cos^2B=1$ $\endgroup$
    – lab bhattacharjee
    Jul 8, 2013 at 16:09
  • 1
    $\begingroup$ @RedQueen10101, yes. $\endgroup$
    – lab bhattacharjee
    Jul 8, 2013 at 16:18
  • 1
    $\begingroup$ @RedQueen10101, $$\left(\frac{y-x}5\right)^2+\left(\frac{2x+3y}5\right)^2=1\implies (y-x)^2+(2x+3y)^2=25\implies 5(x^2+2y^2+2xy)=25\implies x^2+2y^2+2xy=5$$ $\endgroup$
    – lab bhattacharjee
    Jul 8, 2013 at 16:27

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .