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Ok so I was absent from school yesterday because long story short I had no way to get to class b/c something happened last minute. I'm pretty sure this is easy but I keep getting the wrong answer for $\tan2x$.

Find $\sin 2x$, $\cos 2x$, and $\tan 2x$ from the given information.

$\tan x = -\frac{4}{3}$, $x$ in Quadrant II

So I used the double angle formulas and got

$\sin2x = -24/25$

$\cos2x = -7/25$

But I keep getting wrong answers for tangent (I got both $-200/27$ and $-24/7$ somehow). Can someone do a step by step guide on how to get $\tan2x$? Thank you.

Michael Hardy
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TheNewGuy
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2 Answers2

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\begin{align} \tan(2x) & = \frac{2\tan x}{1-\tan^2 x} = \frac{2(-4/3)}{1-(-4/3)^2} = \frac{-8/3}{1-16/9} = \frac{-24}{-7} = \frac{24} 7. \\[30pt] \tan(2x) & = \frac{\sin(2x)}{\cos(2x)} = \frac{-24/25}{-7/25} = \frac{24} 7. \end{align}

Michael Hardy
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If one is having difficulty with a problem like this, then perhaps going back to the original principles would be prudent.

Consider the $\sin 2\theta$ problem

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Using the sin law we have $$\frac{\sin 2\theta}{2\sin\theta} = \frac{\sin \theta^C}{1}$$ Which reduces to $$\sin 2\theta = 2\sin\theta\cos\theta$$ Using the law of cosines we have $$(2\sin\theta)^2 =1^2+1^2-2\cdot1\cdot1\cdot\cos 2\theta$$ Which reduces to $$\cos 2\theta = \cos^2\theta-\sin^2\theta$$

To find the $\tan$ identity, just divide $$\tan 2\theta =\frac{\sin 2\theta}{\cos 2\theta}= \frac{2\sin\theta\cos\theta}{\cos^2\theta-\sin^2\theta}=\frac{\frac{2\sin\theta\cos\theta}{\cos^2\theta}}{\frac{\cos^2\theta-\sin^2\theta}{\cos^2\theta}}=\frac{2\tan\theta}{1-\tan^2\theta}$$ There is no need to memorize double angle formulas; just draw a few triangles. Math suddenly becomes a lot of fun. :)

Community
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John Joy
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