Find sin(4x) if tan(x) = 4

**dapore** · January 10th 2010, 11:58 AM

if $tanx=4$ find $sin4x$

**bandedkrait** · January 10th 2010, 12:06 PM

I can solve it in two steps,

First find $\tan(2x)$ using the identity,

$\tan(2x) = 2\tan(x)/(1-\tan^2(x))$

so $\tan(2x)$ = 2(4)/(1-16) = -8/15

Now find $\sin(4x)$ using,

$\sin(2x) =2\tan(x)/[1+\tan^2(x)]$

Hence, $\sin(4x) =2\tan(2x)/[1+\tan^2(2x)]$

$=(-16/15)/([225+64]/225)$
= $=(-16)(15)/([225+64]$
= $-240/289$ ????
= -

**dapore** · January 10th 2010, 12:28 PM

Thank you very much

**Soroban** · January 10th 2010, 02:07 PM

Hello, dapore!

bandedkrait is absolutely correct!
Here's my approach . . .

If $\tan x = 4$ , find $\sin4x$

We have: . $\tan x \:=\:\frac{1}{4} \:=\:\frac{opp}{adj}$

Then: . $opp = 1,\;adj = 4\quad\Rightarrow\quad hyp = \sqrt{17}$

. . Hence: . $\sin x = \frac{4}{\sqrt{17}},\;\cos x \,=\,\frac{1}{\sqrt{17}}$

And we have: . $\begin{array}{ccccccc}\sin2x &=& 2\sin x\cos x &=&2\left(\frac{4}{\sqrt{17}}\right)\left(\frac{1} {\sqrt{17}}\right) &=& \dfrac{8}{17} \\ \\[-3mm]<br /> \cos2x &=& 1-2\sin^2\!x &=& 1 - 2\left(\frac{4}{\sqrt{17}}\right)^2 &=& \text{-}\dfrac{15}{17} \end{array}$

Therefore: . $\sin4x\;=\;2\sin2x\cos2x \;=\;2\left(\frac{8}{17}\right)\left(-\frac{15}{17}\right) \;=\;-\frac{240}{289}$

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