Expert: Paul Klarreich - 2/27/2009

Question
Sin (a + b + c) + sin (a - b - c)    tan b tan c - 1
--------------------------------- =  ----------------
cos (a + b + c) - cos (a - b - c)    tan b + tan c

a classmate told me the  left side is broken down to:

sin 2a
---------
cos 2b + cos 2c


I have no idea if thats right but Im totally lost.

Answer
Questioner:   Michael
Category:  Calculus
Private:  Yes
 
Subject:  Verify the Identity
Question:  
Sin (a + b + c) + sin (a - b - c)    tan b tan c - 1
--------------------------------- =  ----------------
cos (a + b + c) - cos (a - b - c)    tan b + tan c

a classmate told me the  left side is broken down to:

sin 2a
---------
cos 2b + cos 2c


I have no idea if thats right but Im totally lost.
......................................  
 
Check this site for your standard list of identities and look for the
Sum-to-product section.

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

OR just expand using  sin(x +- y), and  cos(x +- y)
...........................
Interesting: the right side has no 'a' in it.
Remember:  a - b - c = a - (b + c), as we expand.

Lots of writing here -- I should charge by the letter.
BUT make sure to use Courier font.

..................

Expanding: Top of left side:

Sin (a + b + c) +
sin (a - b - c) =

sin(a) cos(b + c) + cos (a) sin(b + c) +
sin(a) cos(b + c) - cos (a) sin(b + c) =

2 sin a cos(b + c) =

2 sin a [cos b cos c - sin b sin c]

......................
Expanding: Bottom of left side:

cos (a + b + c) -
cos (a - b - c)  =

cos(a) cos(b + c) - sin (a) sin (b + c) -
[cos(a) cos(b + c) + sin(a) sin(b + c)] =

cos(a) cos(b + c) - sin (a) sin (b + c)
- cos(a) cos(b + c) - sin(a) sin(b + c) =

-2 sin(a) sin(b + c) =

-2 sin a [sin b cos c + cos b sin c]
.......................................
OK, let's write the fraction again:


2 sin a [cos b cos c - sin b sin c]
-------------------------------------
-2 sin a [sin b cos c + cos b sin c]

yes, the 'a' disappears. [How about that?]

 cos b cos c - sin b sin c
-----------------------------
- sin b cos c - cos b sin c
.......................
Now the right side has tan's and tan = sin/cos, so try for that; divide each term by cos b cos c:

 cos b cos c/cos b cos c - sin b sin c/cos b cos c
---------------------------------------------------- =
- sin b cos c/cos b cos c - cos b sin c/cos b cos c


 1 - tan b tan c
------------------ =
- tan b + tan c

You can fix it from here.