# If cos(50)=a, then how do you express tan(130) in terms of a?

Feb 8, 2015

One method

Assume $a = \cos {50}^{0}$
$\tan \left({130}^{0}\right) = \sin \frac{{130}^{0}}{\cos} \left({130}^{0}\right)$
$\sin \left({130}^{0}\right) = \sin \left({90}^{0} + {40}^{0}\right)$
Expand using $\sin \left(A + B\right)$ formula
This will give you
$\sin \left({130}^{0}\right) = \cos \left({40}^{0}\right)$
$\cos \left({40}^{0}\right) = \sin \left({50}^{0}\right)$ therefore
$\sin \left({130}^{0}\right) = \sin \left({50}^{0}\right)$
$\sin \left({130}^{0}\right) = + \sqrt{1 - {\cos}^{2} \left({50}^{0}\right)}$
$\sin \left({130}^{0}\right) = \sqrt{1 - {a}^{2}}$
Similarly for $\cos \left({130}^{0}\right)$ we get
$\cos \left({130}^{0}\right) = - \sin \left({40}^{0}\right)$
$\cos \left({50}^{0}\right) = \sin \left({40}^{0}\right)$ hence
$\cos \left({130}^{0}\right) = - \cos \left({50}^{0}\right)$
Combining them gives

$\tan \left({130}^{0}\right) = \frac{- \sqrt{1 - {a}^{2}}}{a}$

Aug 31, 2015

You can use some identities to make this easier.

$\cos \left({50}^{o}\right) = \cos \left(- {50}^{o}\right) = - \cos \left({130}^{o}\right)$

due to

$\cos \left(x\right) = \cos \left(- x\right)$
$\cos \left(x\right) = - \cos \left(x + \pi\right)$

Thus, you have:

$\tan \left({130}^{o}\right) = \frac{\sin \left({130}^{o}\right)}{\cos \left({130}^{o}\right)} = \frac{- \sin \left({130}^{o}\right)}{- \cos \left({130}^{o}\right)}$

Now, as for determining $- \sin \left({130}^{o}\right)$... Note that ${\sin}^{2} \left({130}^{o}\right) + {\cos}^{2} \left({130}^{o}\right) = 1$. Therefore:

$\sin \left({130}^{o}\right) = \sqrt{1 - {\cos}^{2} \left({130}^{o}\right)}$

$\implies - \sin \left({130}^{o}\right) = - \sqrt{1 - {\cos}^{2} \left({130}^{o}\right)}$

Finally, you get:

$\textcolor{b l u e}{\tan \left({130}^{o}\right)} = \frac{- \sin \left({130}^{o}\right)}{- \cos \left({130}^{o}\right)} = \frac{- \sqrt{1 - {\left(\cos \left({130}^{o}\right)\right)}^{2}}}{- \cos \left({130}^{o}\right)}$

$= \frac{- \sqrt{1 - {\left(- \cos \left({50}^{o}\right)\right)}^{2}}}{\cos \left({50}^{o}\right)}$

$= \frac{- \sqrt{1 - {\cos}^{2} \left({50}^{o}\right)}}{\cos \left({50}^{o}\right)}$

$= \textcolor{b l u e}{\frac{- \sqrt{1 - {a}^{2}}}{a}}$