Question

for the first derivative of f(theta)=2cos(theta)+cos^2(theta),
I got
f'(theta)=-2sin(theta)+2cos(theta)-sin(theta)=0
After simplifying it, the book had
(-sin(theta))(1+cos(theta))=0 ... how did they get (1+cos(theta))??? the answers is prob really simple but i am stuck on it if anyone could kindly help. @Mathematics

Answer 1

I think you derived incorrectly.

\[f'(\theta)=2(-\sin(\theta))\cos(\theta)-2\sin(\theta)\]
Is what I got

Answer 2

i double checked in the solutions book and it had the same f'

Answer 3

f'\[\Theta = -2\sin \Theta + 2\cos \Theta \left( -\sin \theta \right)\]

Answer 4

I'm pretty horrible at trig identities, but wolfram takes my f'(theta) and puts it to something with cos(theta)+1 also.

Answer 5

http://www.wolframalpha.com/input/?i=d%2Fdtheta+2cos%28theta%29%2Bcos%5E2%28theta%29

Answer 6

it has something to do with trig identities huh?

Answer 7

anytime you end up with something with (trig function) +1 or -1... it's from an identity (most times at least)... I never took any trig classes so I always struggle with them

Answer 8

let me keep on playing with it, thanks!

Answer 9

probably has to do with double or half angle formulae
http://www.sosmath.com/trig/douangl/douangl.html

Answer 10

\[f(\theta)=2\cos(\theta)+\cos ^{2}(\theta)\]\[f\prime(\theta)=-2\sin(\theta)+(2\cos(\theta)*(-\sin(\theta))=-2\sin(\theta)-2\cos(\theta)\sin(\theta)\]\[f\prime(\theta)=-2\sin(\theta)(1+\cos(\theta))\]now if derivative is zero, then you get\[-2\sin(\theta)(1+\cos(\theta))=0\]divide both sides by 2\[-\sin(\theta)(1+\cos(\theta))=0\]