# Thread: [SOLVED] Trigonometric identity

1. ## [SOLVED] Trigonometric identity

cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) +...+cos{A +(n-1)B} =

sin nB/2
---------- . cos{A + (n-1)B/2}
sin B/2

Can any of you guys prove this for me?
Yours truly,
fanofandrew

2. Originally Posted by fanofandrew
cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) +...+cos{A +(n-1)B} =

sin nB/2
---------- . cos{A + (n-1)B/2}
sin B/2

Can any of you guys prove this for me?
Yours truly,
fanofandrew
$\cos A + \cos (A + B) + \cos (A + 2B) + \cos (A + 3B) + ... + \cos (A +[n-1]B)$

$= \cos A + \{\cos A \cos B - \sin A \sin B\} + \{ \cos A \cos (2B) - \sin A \sin (2B)\}$

$+ \{\cos A \cos (3B) - \sin A \sin (3B)\} + ... + \{\cos A \cos ([n-1]B) - \sin A \sin ([n-1]B)\}$

$= \cos A + \cos A \{\cos B + \cos (2B) + \cos (3B) ... + \cos ([n-1]B) \}$

$- \sin A \{ \sin B + \sin (2B) + \sin (3B) + ... + \sin ([n-1]B) \}$.

Now use the two useful results given and proved by topsquark in this thread ( aha! I knew they'd be useful in the future )

3. ## Thank you!

Thank you for the solution.
Since then I have seen another solution.
Let S = cosA + cos(A + B) + cos(A + 2B) + cos(A + 3B) + ... + cos(A + (n - 1)B).

2S sin(B/2) = sin(A + B/2) - sin(A - B/2) + sin(A + 3B/2) - sin(A + B/2) + sin(A + 5B/2) - sin(A + 3B/2) + ... + sin(A + (n - 1/2)B) - sin(A + (n - 3/2)B)
= -sin(A - B/2) + sin(A + (n - 1/2)B)
= 2cos(A + (n - 1)B/2)sin(nB/2).

sin(nB/2)
So S = ----------- . cos(A + (n - 1)B/2).
sin(B/2)