# sin(-Ø)-6cos(-Ø)+3tan(2Ø)

• Feb 26th 2013, 09:17 AM
hermeticcharm
sin(-Ø)-6cos(-Ø)+3tan(2Ø)
Edit: Added some tex tags
This is a question on a review I am doing, my answer doesn't match up with the answer key and I am looking for some input on what I am doing wrong.

If $cos\theta = \frac{1}{6}$ with $\theta$ in quadrant IV find:

$sin(-\theta)-6cos(-\theta)+3tan(2\theta)$

I solved for $sin\theta = \frac{-\sqrt{35}}{6}$ $cos\theta = \frac{1}{6}$ and $tan\theta = -\sqrt{35}$

Using negative angle identities and the double angle identity I end up with:

$-sin\theta-6cos\theta+3(\frac{2tan\theta}{1-tan^2\theta})$

Plugging in my values gives me:

$\frac{\sqrt{35}}{6} - \frac{6}{6} + 3[\frac{2(-\sqrt{35})}{1-35}]$

Simplified to:

$\frac{35(\sqrt{35}) -102}{102}$

The answer key gives: $\frac{42+25\sqrt{35}}{42}$

Not sure where I am messing up =\ any help is much appreciated.
• Feb 26th 2013, 10:06 AM
ebaines
Re: sin(-Ø)-6cos(-Ø)+3tan(2Ø)

Given $\cos \phi = \frac 1 6$ in the fourth quadrant then $\phi$ = -1.4 radians = -80.45 degrees.

$\sin(- \phi) - 6 \cos( - \phi) + 3 \tan (2 \phi) = 0.986 - 1+ 1.044 = 1.030$

$\frac { 35 \sqrt {35} - 102}{102} = 1.030$
$\frac {42+ 25 \sqrt{35}}{42} = 4.521$
I think I may have found where whoever came up with the answer key made mistakes: if (a) they used $+ 6\cos(\- \phi)$ instead of $-6 \cos (- \phi)$, and also (b) instead of simplifiying the $\tan (2 \phi)$ term to $\frac { 3 \sqrt{35}}{17}$ as you have it they made it $\frac { 3 \sqrt {35}}{7}$, that would explain it.