Question

What is `cos(arctan(-2) + arccos(5/13))`?

Answer 1

`cos(arctan(-2) + arccos(5/13)) =(29sqrt(5))/65`

Explanation:

First we expand that:
`cos(arctan(-2) + arccos(5/13)) =`
`cos(arctan(-2))cos(arccos(5/13)) - sin(arctan(-2))sin(arccos(5/13))`

We know that `cos(arccos(theta)) = theta` so we can rewrite that to
`(5cos(arctan(-2)))/13 - sin(arctan(-2))sin(arccos(5/13))`

We know that `sin^2(theta) = 1 - cos^2(theta)`, so
`sin^2(arccos(5/13)) = 1 - (5/13)^2 = 1 - 25/169 = 144/169`

Taking the root
`sin(arccos(5/13)) = 12/13`
It's positive because the sine is always positive in the arccosine range. It can let us rewrite the first expansion to:
`(5cos(arctan(-2)))/13 - (12sin(arctan(-2)))/13`

Now, from `sin^2(theta) + cos^2(theta) = 1`, if we divide both sides by `cos^2(theta)` we get `tan^2(theta) + 1 = 1/cos^2(theta)`. So:
`tan^2(arctan(-2)) + 1 = 1/cos^2(theta) rarr cos^2(theta) = 1/((-2)^2 + 1)`
`cos^2(theta) = 1/5 rarr cos(theta) = 1/sqrt(5) = sqrt(5)/5`

It's positive because on the range of the arctangent, the cosine is always postive. So we can rewrite it to:
`sqrt(5)/13 - (12sin(arctan(-2)))/13`

Since the tangent is negative, and the cosine positive, it means the sine is negative. Using `sin^2(theta) + cos^2(theta) =1` we have that
`sin^2(arctan(-2)) = 1 - 1/5 = 4/5`
`sin(arctan(-2)) = -2/sqrt(5) = -(2sqrt(5))/5`

Which means we can rewrite it all to:
`sqrt(5)/13 - (12(-2sqrt(5)/5))/13`

From there on it's algebra:
`(5sqrt(5))/65 + (24sqrt(5))/65 =`
`(29sqrt(5))/65`

Answer 2

Find: `cos (arctan (-2) + arccos (5/13))`

Ans: 1

Explanation:

Use calculator.
`tan x = (-2)`--> arc `x = -63.43` deg
`cos y = 5/13 = 0.38` --> arc `y = 67.38` deg
cos (-63.43 + 67.58) = cos 3.95 = 0.997 = 1