Trigonometric equation?

**D7236** · July 18th 2010, 02:05 AM

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.

**undefined** · July 18th 2010, 02:15 AM

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?

Thanks in advance.

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

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

**Prove It** · July 18th 2010, 02:23 AM

You should know

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

and

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

So in your case

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

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

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

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

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

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

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

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

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

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

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

$\tan{x} = \frac{24 + 13\sqrt{3}}{3}$ .

**Prove It** · July 18th 2010, 02:34 AM

Originally Posted by undefined

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

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

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

**undefined** · July 18th 2010, 02:46 AM

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?

Thanks in advance.

Since Prove It posted a full solution, here's another way.

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

$\tan\left(x+\dfrac{\pi}{3} \right) = \dfrac{-2}{3}$

Use

$\displaystyle \tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}$

if you know it, otherwise use sin and cos angle sum like Prove It.

Here

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

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

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

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

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

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

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

Originally Posted by Prove It

Yes, we coud let $\theta = x + \frac{\pi}{3}$ , but without the angle sum and difference identities we still can't reduce this to $\tan{x}$ . 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.

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