How do you evaluate $cos (arctan (- 2) + arccos (\frac{5}{13}))$ $cos(arctan(-2) + arccos(5/13))$ ?

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Answer 1 · 2016-07-30T08:23:48

$= \frac{29}{13 \sqrt{5}}$ $=29/(13sqrt 5)$

Let $a = a r c tan (- 2) \in Q 3$ $a = arc tan (-2) in Q3$ ,

wherein sine and tangent are negative.

Then, $tan a = - 2, sin a = - \frac{2}{\sqrt{5}} and cos a = \frac{1}{\sqrt{5}}$ $tan a = -2, sin a =-2/sqrt 5 and cos a = 1/sqrt 5$

Let $b = a r c cos (\frac{5}{13}) \in Q 1$ $b = arc cos (5/13) in Q1$ , wherein sine is positive..

Then, $cos b = \frac{5}{13} and sin b = \frac{12}{13}$ $cos b = 5/13 and sin b =12/13$ .

Now, the given expression is

$cos (a + b)$ $cos(a+b)$

$= cos a cos b - sin a sin b$ $=cos a cos b - sin a sin b$

$= (\frac{1}{\sqrt{5}}) (\frac{5}{13}) - (- \frac{2}{\sqrt{5}}) (\frac{12}{13})$ $=(1/sqrt 5)(5/13)-(-2/sqrt 5)(12/13)$

$= \frac{29}{13 \sqrt{5}}$ $=29/(13sqrt 5)$

How do you evaluate cos(arctan(−2)+arccos(513))cos(arctan(-2) + arccos(5/13))?