# Trigonometric equation?

• Jul 18th 2010, 02:05 AM
D7236
Trigonometric equation?
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?

• Jul 18th 2010, 02:15 AM
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Quote:

Originally Posted by D7236
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?

You can use identity $\displaystyle \displaystyle \cos(\theta + \pi) = -\cos(\theta)$, to start out.

Edit: Well I suppose it should be $\displaystyle \displaystyle \cos(\theta - \pi) = -\cos(\theta)$, same idea.
• Jul 18th 2010, 02:23 AM
Prove It
You should know

$\displaystyle \sin{(\alpha \pm \beta)} = \sin{\alpha}\cos{\beta} \pm \cos{\alpha}\sin{\beta}$

and

$\displaystyle \cos{(\alpha \pm \beta)} = \cos{\alpha}\cos{\beta} \mp \sin{\alpha}\sin{\beta}$.

$\displaystyle 3\sin{\left(x + \frac{\pi}{3}\right)} = 2\cos{\left(x - \frac{2\pi}{3}\right)}$

$\displaystyle 3\left(\sin{x}\cos{\frac{\pi}{3}} + \cos{x}\sin{\frac{\pi}{3}}\right) = 2\left(\cos{x}\cos{\frac{2\pi}{3}} + \sin{x}\sin{\frac{2\pi}{3}}\right)$

$\displaystyle 3\left(\frac{1}{2}\sin{x} + \frac{\sqrt{3}}{2}\cos{x}\right) = 2\left(-\frac{1}{2}\cos{x} + \frac{\sqrt{3}}{2}\sin{x}\right)$

$\displaystyle \frac{3}{2}\sin{x} + \frac{3\sqrt{3}}{2}\cos{x} = -\cos{x} + \sqrt{3}\sin{x}$

$\displaystyle \left(\frac{3}{2} - \sqrt{3}\right)\sin{x} = \left(-1 - \frac{3\sqrt{3}}{2}\right)\cos{x}$

$\displaystyle \left(\frac{3 - 2\sqrt{3}}{2}\right)\sin{x} = -\left(\frac{2 + 3\sqrt{3}}{2}\right)\cos{x}$

$\displaystyle \frac{\sin{x}}{\cos{x}} = \frac{-\frac{2 + 3\sqrt{3}}{2}}{\phantom{-}\frac{3 - 2\sqrt{3}}{2}}$

$\displaystyle \tan{x} = -\frac{2 + 3\sqrt{3}}{3 - 2\sqrt{3}}$

$\displaystyle \tan{x} = -\frac{(2 + 3\sqrt{3})(3 + 2\sqrt{3})}{(3 - 2\sqrt{3})(3 + 2\sqrt{3})}$

$\displaystyle \tan{x} = -\frac{6 + 4\sqrt{3} + 9\sqrt{3} + 18}{9 - 12}$

$\displaystyle \tan{x} = \frac{-(24 + 13\sqrt{3})}{-3}$

$\displaystyle \tan{x} = \frac{24 + 13\sqrt{3}}{3}$.
• Jul 18th 2010, 02:34 AM
Prove It
Quote:

Originally Posted by undefined
You can use identity $\displaystyle \displaystyle \cos(\theta + \pi) = -\cos(\theta)$, to start out.

Edit: Well I suppose it should be $\displaystyle \displaystyle \cos(\theta - \pi) = -\cos(\theta)$, same idea.

Yes, we coud let $\displaystyle \theta = x + \frac{\pi}{3}$, but without the angle sum and difference identities we still can't reduce this to $\displaystyle \tan{x}$. So refer to my post above :D
• Jul 18th 2010, 02:46 AM
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Quote:

Originally Posted by D7236
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?

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

$\displaystyle \dfrac{-2}{3} = \dfrac{\tan x+\sqrt{3}}{1-\sqrt{3}\tan x}$

$\displaystyle -2(1-\sqrt{3}\tan x) = 3(\tan x+\sqrt{3})$

$\displaystyle -2+2\sqrt{3}\tan x = 3\tan x+3\sqrt{3}$

$\displaystyle (2\sqrt{3}-3)\tan x = 3\sqrt{3}+2$

$\displaystyle \tan x = \dfrac{2+3\sqrt{3}}{-3+2\sqrt{3}}$

$\displaystyle \tan x = \dfrac{(2+3\sqrt{3})(-3-2\sqrt{3})}{(-3+2\sqrt{3})(-3-2\sqrt{3})}$

$\displaystyle \tan x = \dfrac{24+13\sqrt{3}}{3}$

Quote:

Originally Posted by Prove It
Yes, we coud let $\displaystyle \theta = x + \frac{\pi}{3}$, but without the angle sum and difference identities we still can't reduce this to $\displaystyle \tan{x}$. So refer to my post above

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.