cos(arcTan (5/12) - arcCos(2/3))
so basically I set up the two triangles and get the hypotenuse 13 and the side length 2sqrt5. I'm supposed to evaluate without using a calculator, what do I do next?
First - recognize that what you have are two angles - call them alpha and beta, where:
$\displaystyle
\alpha = tan ^{-1} (\frac 5 {12}), \ \ \beta = cos ^{-1} (\frac 2 3 )
$
You are aked to find $\displaystyle cos(\alpha - \beta) $. So use the trig identity: $\displaystyle cos(\alpha - \beta) = cos \alpha \cdot \cos \beta + sin \alpha \cdot sin \beta $
You can figure the values of these sines and cosines from the information you've been given. For example, $\displaystyle cos \alpha = \frac {12} {13} $. BTW - the side legth of the second triangle is $\displaystyle \sqrt { 3^2 - 2^2} = \sqrt 5 $
Hello, hellojellojw!
Smplify: .$\displaystyle \cos\left(\arctan\tfrac{5}{12} - \arccos\tfrac{2}{3}\right)$ .[1]
Let: .$\displaystyle \alpha \:=\: \arctan\tfrac{5}{12} \quad\Rightarrow\quad \tan\alpha \:=\:\tfrac{5}{12} \:=\:\tfrac{opp}{hyp}$
We have: .$\displaystyle opp = 5,\;adj = 12 \quad\Rightarrow\quad hyp = 13$
. . Hence: .$\displaystyle \sin\alpha \,=\,\tfrac{5}{13},\;\cos\alpha \,=\,\tfrac{12}{13}$ .[2]
Let: .$\displaystyle \beta \,=\,\arccos\tfrac{2}{3} \quad\Rightarrow\quad \cos\beta \:=\:\tfrac{2}{3} \:=\:\tfrac{adj}{hyp} $
We have: .$\displaystyle adj = 2,\;hyp = 3 \quad\Rightarrow\quad opp = \sqrt{5}$
. . Hence: .$\displaystyle \sin\beta \,=\,\tfrac{\sqrt{5}}{3},\;\cos\beta \,=\,\tfrac{2}{3}$ .[3]
From [1], we have: .$\displaystyle \cos(\alpha - \beta) \;=\;\cos\alpha\cos\beta + \sin\alpha\sin\beta$
Substitute [2] and [3]: .$\displaystyle \cos(\alpha + \beta) \;=\;\left(\tfrac{12}{13}\right)\left(\tfrac{2}{3} \right) + \left(\tfrac{5}{13}\right)\left(\tfrac{\sqrt{5}}{3 }\right) \;=\;\frac{24+5\sqrt{5}}{39} $
Definition of the arctan: it's the angle whose tangent is the given value. Think of arctan as the inverse function of tan. Hence the tangent of the arctangent of x is simply x. You can see this with a simple example: consider the angle whose tangent = 1. You can probably see right away that the angle is $\displaystyle \pi/4 $. Now find the tangent of $\displaystyle \pi/4 $ - it's 1. So you see that tan(arctan(x)) = x.