# Trig questions help

• Sep 9th 2005, 09:49 AM
curtains
Trig questions help
Hiya, i'm kind of stuck on this problem:

Using cos2A = cos(squared)A - sin(squared) A or otherwise

prove cos2a = 1/2(1+cos2A)

any help appreciated, thanks!
• Sep 9th 2005, 10:26 AM
ticbol
prove cos2a = 1/2(1+cos2A) ?

Is cos2a the same as cos2A ?
if yes,
cos2a = 1/2(1+cos2A)
Multiply both sides by 2,
2cos2a = 1 +cos2A
2cos2a -cos2A = 1
cos2a = 1 ----------------No, that is not true.

That means cos2a is not the same as cos2A.

I think your cos2a should have been cos^2(A), or (cosA)^2.
That is the same as your "cos(squared)A"

Let us see,
cos^2(A) = (1/2)[1 +cos(2A)]
cos^2(A) = (1/2)[{sin^2(A) +cos^2(A)} +{cos^2(A) -sin^2(A)}]
cos^2(A) = (1/2)[sin^2(A) +cos^2(A) +cos^2(A) -sin^2(A)]
cos^2(A) = (1/2)[2cos^2(A)]
cos^2(A) = cos^2(A) -----------***
Proven.

---------
We used also the trig identity
sin^2(A) +cos^2(A) = 1
• Sep 9th 2005, 11:00 AM
curtains
Sorry! Big oops, I pasted the wrong part of the question, god I feel like such an eejit!

Yes that was a cos(squared)A on the left. I proved that fine, it was the next part that got me:

hence or otherwise solve the equation -
1+cos2x=cosx where 0 degrees </= x </= 360
• Sep 9th 2005, 11:19 AM
ticbol
hence or otherwise solve the equation -
1+cos2x=cosx where 0 degrees </= x </= 360

Don't feel too bad. Everyone commits mistakes.

1 +cos(2X) = cosX
1 +cos^2(X) -sin^2(X) = cosX
We make them all in cosX,
1 +cos^2(X) -[1 -cos^2(X)] = cosX
1 +cos^2(X) -1 +cos^2(X) = cosX
2cos^2(X) = cosX
2cos^2(X) -cosX = 0
(cosX)(2cosX -1) = 0

cosX = 0
X = arccos(0)
x = 90deg or 270deg -------***

2cosX -1 = 0
2cosX = 1
cosX = 1/2 --------positive. Cosine is positive in the 1st and 4th quadrants.
X = arccos(1/2)
X = 60deg ---in the 1st quadrant.
X = 360 -60 = 300deg ---in the 4th quadrant.

Therefore, X = 60, 90, 270, or 300 degrees. --------answer.