If cos x = 12/13 and x is an acute angle, how do you find the value of cos (180 - x)?

Sep 24, 2016

$\cos \left({180}^{\circ} - x\right) = - \frac{12}{13}$

Explanation:

We will use the cosine angle subtraction formula:

$\cos \left(a - b\right) = \cos \left(a\right) \cos \left(b\right) + \sin \left(a\right) \sin \left(b\right)$

Thus:

$\cos \left({180}^{\circ} - x\right) = \cos \left({180}^{\circ}\right) \cos \left(x\right) + \sin \left({180}^{\circ}\right) \sin \left(x\right)$

Note that $\cos \left({180}^{\circ}\right) = - 1$ and $\sin \left({180}^{\circ}\right) = 0$:

$\cos \left({180}^{\circ} - x\right) = \left(- 1\right) \cos \left(x\right) + \left(0\right) \sin \left(x\right)$

$\cos \left({180}^{\circ} - x\right) = - \cos \left(x\right)$

And since in this scenario $\cos \left(x\right) = \frac{12}{13}$:

$\cos \left({180}^{\circ} - x\right) = - \frac{12}{13}$