cos A + cos 2A + cos 3A = cos 2A(1-cos A)

Printable View

- October 20th 2009, 08:32 PMreiwardProve
cos A + cos 2A + cos 3A = cos 2A(1-cos A)

- October 20th 2009, 08:45 PMmr fantastic
- October 21st 2009, 12:58 AMreiward
OH! Then how come it appeared on our final exams in Trigo? I've answered every item there except that one.

:O Professor tricked me!! - October 21st 2009, 01:13 AMpacman
yeah, not all identities questions are meant to be. Before you start solving, test it first before jumping. The graph showed it is a trigonometric equation with multiple roots. When you found the counter example . . . . don't proceed solving.

- October 21st 2009, 01:16 AMreiward
nah, it is stated there to prove the identities/equations.

I have proved it actually but then I've realized that I was wrong and then I couldn't get the process anymore.

cos A + cos 2A + cos 3A = cos 2A(1-cos A)

What I did was to factor cos A, thus it makes cos A (1 + cos A + cos 2A)

then I've realized that it is wrong. Though I aririved at cos 2A(1-cos A) - October 21st 2009, 03:20 AMmr fantastic
- October 21st 2009, 04:38 AMpacman
**reiward**,**Mr Fantastic**is correct. What you have is a trigonometric equation, not an identity. His counter-example was when A = 0, that will give you 3 = 0, which is wrong. So it merits to be an equation, not an identity as what you claimed it to be. Se my graph . . . . - October 21st 2009, 04:43 AMmr fantastic
Are you meant to solve this for A (which is OK) or prove it, that is, show it's true for all A (which is not OK because it's not an identity). Chipce of words is important. Prove (the word you used) does not mean the same thing as solve (the word you might have meant ....)

- October 21st 2009, 06:51 AMreiward
I'm sure it is meant for proving. But anyway, thank you !