# How do you calculate Cos^-1(5/13) - Cos^-1(8/17)?

Jun 15, 2015

With a calculator. A TI-8# can do this with 2nd+COS. 13 and 17 are prime numbers, so 5/13 and 8/17 will give you irrational values. This isn't easily doable enough by hand to be something you need to do by hand.

$\arccos \left(\frac{5}{13}\right) - \arccos \left(\frac{8}{17}\right) = {67.38}^{o} - {61.93}^{o} = {5.45}^{o} \approx 0.095 \text{rad}$

Jul 3, 2015

$= {\cos}^{- 1} \left(\frac{220}{221}\right)$

#### Explanation:

Another way would be to let $A = {\cos}^{- 1} \left(\frac{5}{13}\right)$ and $B = {\cos}^{- 1} \left(\frac{8}{17}\right)$

$\implies \cos A = \frac{5}{13}$

And , $\cos B = \frac{8}{17}$

From the identity ${\cos}^{2} x + {\sin}^{2} x = 1$
We can obtain that,

$\sin x = \sqrt{1 - {\cos}^{2} x}$

$\implies \sin A = \sqrt{1 - {\left(\frac{5}{13}\right)}^{2}} = \sqrt{\frac{144}{169}} = \frac{12}{13}$

Similarly , $\implies \sin B = \sqrt{1 - {\left(\frac{8}{17}\right)}^{2}} = \sqrt{\frac{225}{289}} = \frac{15}{17}$

Now, say $\text{ } C = A - B$

$\implies \cos C = \cos \left(A - B\right) = \cos A \cos B + \sin A \sin B = \frac{5}{13} \cdot \frac{8}{17} + \frac{12}{13} \cdot \frac{15}{17} = \frac{40}{221} + \frac{180}{221} = \frac{220}{221}$

$\implies \cos C = \frac{220}{221}$

$\implies C = {\cos}^{- 1} \left(\frac{220}{221}\right)$

$\implies A - B = {\cos}^{- 1} \left(\frac{220}{221}\right)$

$= {\cos}^{- 1} \left(\frac{5}{13}\right) - {\cos}^{- 1} \left(\frac{8}{17}\right) = {\cos}^{- 1} \left(\frac{220}{221}\right)$