Tan (a+b+c) =????

D

Damoo

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I know that
SIN (A+B+C) = SIN ((A+B)+C) = SIN (A+B)COS C + COS (A+B) SIN C

What is TAN (A+B+C) =?

I'm so confused any help would be appreciated.



EDIT:
Sorry I didn't think it mattered since it seemed like the prof was rearranging the equation in the form of ((A+B)+C)


The actual question was find TAN (arctan 3 + arctan 5 + arctan 7).




I know the answer to a similar question but with a SIN (arctan2 + arctan3 +arctan4)

and the formula the prof used was "SIN (A+B+C) = SIN ((A+B)+C) = SIN (A+B)COS C + COS (A+B) SIN C"

WHICH became : (sinAcosB+cosAsinB)cosC +(cosAcosB-sinAsinB)sinC.

Then he'd plug in all the numbers for the arctan triangles and there you have the solution.

I was wondering if the formula differed at all when it was TAN. I know that TAN (A+B) = SIN (A+B)/Cos (A+B) afterwards but the C does it become a cos C for tan as well?
 
Last edited:
Damoo said:
I know that
SIN (A+B+C) = SIN ((A+B)+C) = SIN (A+B)COS C + COS (A+B) SIN C

What is TAN (A+B+C) =?

I'm so confused any help would be appreciated.
Click to expand...

You have not told us what the directions are for your problem! If you are supposed to express tan(a + b + c) in terms of sine and cos, consider that tanx = sinx/cosx.
 
Hello, Damoo!

What is \(\displaystyle \tan(A+B+C)\) ?

The actual question was find: .\(\displaystyle \tan(\arctan 3 + \arctan 5 + \arctan 7)\)
Click to expand...

You can derive that formula if you are very careful.

We know that: .\(\displaystyle \tan(\alpha + \beta) \:=\:\dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)

We have:

.. \(\displaystyle \tan(A + B + C) \;=\;\tan\big[(A+B) + C\big]\)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\tan(A+B) + \tan C}{1 - \tan(A+B)\tan C}\)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\frac{\tan A+\tan B}{1-\tan A\tan B} + \tan C}{1 - \frac{\tan A+\tan B}{1-\tan A\tan B}\tan C} \)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\tan A + \tan B + \tan C(1 - \tan A\tan B)}{1-\tan A\tan B - \tan C(\tan A+\tan B)}\)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B - \tan B\tan C - \tan A\tan C}\)
 
Wow...

You are a math god.

Formula checks out, finished one part of the assignment all thanks to you.

Just gotta understand it before the exam. :)

A million thanks for the formula!
 
soroban said:
Hello, Damoo!


You can derive that formula if you are very careful.

We know that: .\(\displaystyle \tan(\alpha + \beta) \:=\:\dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)

We have:

.. \(\displaystyle \tan(A + B + C) \;=\;\tan\big[(A+B) + C\big]\)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\tan(A+B) + \tan C}{1 - \tan(A+B)\tan C}\)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\frac{\tan A+\tan B}{1-\tan A\tan B} + \tan C}{1 - \frac{\tan A+\tan B}{1-\tan A\tan B}\tan C} \)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\tan A + \tan B + \tan C(1 - \tan A\tan B)}{1-\tan A\tan B - \tan C(\tan A+\tan B)}\)

. . . . . . . . . . . . . . .\(\displaystyle =\;\dfrac{\tan A + \tan B + \tan C - \tan A\tan B\tan C}{1 - \tan A\tan B - \tan B\tan C - \tan A\tan C}\)
Click to expand...


I have a question if you don't mind. Basically Since \(\displaystyle \tan(\alpha + \beta) \:=\:\dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}\)



EDIT: NVM I GOT IT, understand now, not the same thing at all!!!!




So its wrong of me to say:


TAN ((A+B)+C) = \(\displaystyle \:\dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta} +\tan C\)
 
Last edited:
Damoo said:
Wow...

You are a math god.

Formula checks out, finished one part of the assignment all thanks to you.

Just gotta understand it before the exam. :)

A million thanks for the formula!
Click to expand...
Do not memorize formulas. Learn methods.
 
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