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!

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- Sep 9th 2005, 09:49 AMcurtainsTrig 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 AMticbol
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 AMcurtains
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 AMticbol
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.