how wud u solve the following?
\[\cos (\tan^{-1} (-5/12)+\tan^{-1} (15/8))\]
Plz help this is in 15 min!

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- anonymous

- chestercat

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- anonymous

ok do the numbers (5,12, ) and (8,15, ) remind you of something?

- anonymous

these numbers appear in the problem for good reason

- anonymous

umm im not sure way u mean

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## More answers

- anonymous

ok let me fill in the blanks: (5,12,13), (8,15,17). do these numbers look familiar?

- anonymous

these r (x y r) values?

- anonymous

pythagorean triples. i.e. integer side lengths for right triangles

- anonymous

ok then wat do i do w/this #?

- anonymous

don't mean to rush but this is due online in 10 min exactly

- anonymous

so you know [\tan^{-1} (-5/12)] refers to the 5,12,13 triangle,
and [\tan^{-1} (15/8)] refers to the 8,15,17 triangle

- anonymous

ugh sry not familiar with the mathtype syntax here... i apologize for poorly formatted equations

- anonymous

its ok don't worry bout its understandable

- anonymous

now the 2 inverse tangents give you to angles. let's call them a and b. the angle a comes from the 5,12,13 triangle, and the angle b comes from the 8,15,17 triangle

- anonymous

draw your triangles, and identify which angle is a, and which angles is b

- anonymous

now back at the original equation, since i'm lazy and don't want to write out all the inverse tangents, i'll just write a and b:
cos(a+b) = cos(a) cos(b) - sin(a) sin(b)

- anonymous

from your triangles, you should be able to identify what cos(a), cos(b), sin(a), and sin(b) are

- anonymous

so y for the 1st one is 5 n x is 12 while the 2nd y is 15 n 2nd x is 8?

- anonymous

|dw:1353483877024:dw|

- anonymous

sry, diagram not drawn to scale

- anonymous

thats fine but I'm still stuck :(

- anonymous

ok back to original equation
remember a = tan-1(5/12) and b = tan-1(15/8)
so the original equation: cos(a+b), can be expanded as cos(a)cos(b) - sin(a)sin(b)

- anonymous

now look at the pictures, what's cos(a)? what's cos(b)? what's sin(a)? what's sin(b)?
(remember cos is adjacent/hypotenuse, sin is opposite/hypotenuse)

- anonymous

you should make a change of variables

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