# Math Help - Trig Identities Problem

1. ## Trig Identities Problem

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?

2. First - recognize that what you have are two angles - call them alpha and beta, where:

$
\alpha = tan ^{-1} (\frac 5 {12}), \ \ \beta = cos ^{-1} (\frac 2 3 )
$

You are aked to find $cos(\alpha - \beta)$. So use the trig identity: $cos(\alpha - \beta) = cos \alpha \cdot \cos \beta + sin \alpha \cdot sin \beta$

You can figure the values of these sines and cosines from the information you've been given. For example, $cos \alpha = \frac {12} {13}$. BTW - the side legth of the second triangle is $\sqrt { 3^2 - 2^2} = \sqrt 5$

3. Hello, hellojellojw!

Smplify: . $\cos\left(\arctan\tfrac{5}{12} - \arccos\tfrac{2}{3}\right)$ .[1]

Let: . $\alpha \:=\: \arctan\tfrac{5}{12} \quad\Rightarrow\quad \tan\alpha \:=\:\tfrac{5}{12} \:=\:\tfrac{opp}{hyp}$

We have: . $opp = 5,\;adj = 12 \quad\Rightarrow\quad hyp = 13$

. . Hence: . $\sin\alpha \,=\,\tfrac{5}{13},\;\cos\alpha \,=\,\tfrac{12}{13}$ .[2]

Let: . $\beta \,=\,\arccos\tfrac{2}{3} \quad\Rightarrow\quad \cos\beta \:=\:\tfrac{2}{3} \:=\:\tfrac{adj}{hyp}$

We have: . $adj = 2,\;hyp = 3 \quad\Rightarrow\quad opp = \sqrt{5}$

. . Hence: . $\sin\beta \,=\,\tfrac{\sqrt{5}}{3},\;\cos\beta \,=\,\tfrac{2}{3}$ .[3]

From [1], we have: . $\cos(\alpha - \beta) \;=\;\cos\alpha\cos\beta + \sin\alpha\sin\beta$

Substitute [2] and [3]: . $\cos(\alpha + \beta) \;=\;\left(\tfrac{12}{13}\right)\left(\tfrac{2}{3} \right) + \left(\tfrac{5}{13}\right)\left(\tfrac{\sqrt{5}}{3 }\right) \;=\;\frac{24+5\sqrt{5}}{39}$

4. why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values?

5. Originally Posted by BugzLooney
why does arctan being 5/12 and arccos being 2/3 imply that the tangent and cosine are also those values?
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 $\pi/4$. Now find the tangent of $\pi/4$ - it's 1. So you see that tan(arctan(x)) = x.