If 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?
I started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?
Thanks in advance.
If 3sin(x + pi/3) = 2cos(x - 2pi/3), find tanx?
I started doing it using the sum and difference formulas but didn't end up getting the correct answer. Could someone show me how to do it using the sum and difference formulas?
Thanks in advance.
You should know
$\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$
and
$\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$.
So in your case
$\displaystyle 3\sin{\left(x + \frac{\pi}{3}\right)} = 2\cos{\left(x - \frac{2\pi}{3}\right)}$
$\displaystyle 3\left(\sin{x}\cos{\frac{\pi}{3}} + \cos{x}\sin{\frac{\pi}{3}}\right) = 2\left(\cos{x}\cos{\frac{2\pi}{3}} + \sin{x}\sin{\frac{2\pi}{3}}\right)$
$\displaystyle 3\left(\frac{1}{2}\sin{x} + \frac{\sqrt{3}}{2}\cos{x}\right) = 2\left(-\frac{1}{2}\cos{x} + \frac{\sqrt{3}}{2}\sin{x}\right)$
$\displaystyle \frac{3}{2}\sin{x} + \frac{3\sqrt{3}}{2}\cos{x} = -\cos{x} + \sqrt{3}\sin{x}$
$\displaystyle \left(\frac{3}{2} - \sqrt{3}\right)\sin{x} = \left(-1 - \frac{3\sqrt{3}}{2}\right)\cos{x}$
$\displaystyle \left(\frac{3 - 2\sqrt{3}}{2}\right)\sin{x} = -\left(\frac{2 + 3\sqrt{3}}{2}\right)\cos{x}$
$\displaystyle \frac{\sin{x}}{\cos{x}} = \frac{-\frac{2 + 3\sqrt{3}}{2}}{\phantom{-}\frac{3 - 2\sqrt{3}}{2}}$
$\displaystyle \tan{x} = -\frac{2 + 3\sqrt{3}}{3 - 2\sqrt{3}}$
$\displaystyle \tan{x} = -\frac{(2 + 3\sqrt{3})(3 + 2\sqrt{3})}{(3 - 2\sqrt{3})(3 + 2\sqrt{3})}$
$\displaystyle \tan{x} = -\frac{6 + 4\sqrt{3} + 9\sqrt{3} + 18}{9 - 12}$
$\displaystyle \tan{x} = \frac{-(24 + 13\sqrt{3})}{-3}$
$\displaystyle \tan{x} = \frac{24 + 13\sqrt{3}}{3}$.
Since Prove It posted a full solution, here's another way.
$\displaystyle 3\sin\left(x + \dfrac{\pi}{3}\right) = 2\cos\left(x - \dfrac{2\pi}{3}\right) = -2 \cos\left(x + \dfrac{\pi}{3}\right)$
$\displaystyle \tan\left(x+\dfrac{\pi}{3} \right) = \dfrac{-2}{3}$
Use
$\displaystyle \displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$
if you know it, otherwise use sin and cos angle sum like Prove It.
Here
$\displaystyle \dfrac{-2}{3} = \dfrac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x} $
$\displaystyle -2(1-\sqrt{3}\tan x) = 3(\tan x+\sqrt{3})$
$\displaystyle -2+2\sqrt{3}\tan x = 3\tan x+3\sqrt{3}$
$\displaystyle (2\sqrt{3}-3)\tan x = 3\sqrt{3}+2$
$\displaystyle \tan x = \dfrac{2+3\sqrt{3}}{-3+2\sqrt{3}}$
$\displaystyle \tan x = \dfrac{(2+3\sqrt{3})(-3-2\sqrt{3})}{(-3+2\sqrt{3})(-3-2\sqrt{3})}$
$\displaystyle \tan x = \dfrac{24+13\sqrt{3}}{3}$
I was busy typing and didn't see your recent post until just now. I think your way is a bit easier than mine, anyway.