# Xueqin!

#### Maths help

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by , 31st July 2012 at 07:16 AM (266 Views)
2cos^2((π/4)-A) = 1 + sin(2A)
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Originally Posted by delicious french cuisine
rate 5! for math
Lol.
Updated 31st July 2012 at 09:23 AM by Xueqin!
Xueqin! long time!
Also the answer is potato. My logic: Everything +1-8/6Triangle+random figures I don't under stand = potato
2cos^2((π/4)-A) = 1 + sin(2A)
2cos^2((π/4)-A) - 1 = sin(2A)
cos(π/2 - 2A) = sin(2A)

that's it I guess, as cos(π/2 - x) = sin(x) for any x.
Sry, im in algebra not trigonometry. i think
Well, skipping a few less steps (Firebird's answer is correct, but he skipped a million steps):

2 cos2(pi/4 - A) =

= 2 (cos(pi/4 - A))2 - 1 + 1 = (by adding and subtracting 1, the result doesn't change)

= 2 (cos(pi/4 - A))2 - ((sin(pi/4 - A))2 + (cos(pi/4 - A))2) + 1 = (because sin2 x + cos2 x = 1 for any x)

= (cos(pi/4 - A))2 - (sin(pi/4 - A))2 + 1. (just subtracted the cos(...) from 2cos(...))

Let's hold that there for a sec.

Now, take any value, x; let's calculate cos(2x):

cos(2x) =
= cos(x + x) =
= cos x cos x - sin x sin x = (because cos(a + b) = cos a cos b - sin a sin b)
= cos2 x - sin2 x.

So, for any x, cos2 x - sin2 x = cos(2x).

Now, if x = pi/4 - A, we have that our former equation, (cos(pi/4 - A))2 - (sin(pi/4 - A))2 + 1, can be rewritten as cos(2(pi/4 - A)) + 1.

So, we have cos(2(pi/4 - A)) + 1 = cos(pi/2 - 2A) + 1. Since, for any x, cos(pi/2 - x) = sin x, then cos(pi/2 - 2A) + 1 = sin(2A) + 1, which is what we wanted to prove.

That way, it's proven that 2cos2(pi/4 - A) = 1 + sin(2A).
Originally Posted by aaaaaa123456789
Well, skipping a few less steps (Firebird's answer is correct, but he skipped a million steps):

2 cos2(pi/4 - A) =

= 2 (cos(pi/4 - A))2 - 1 + 1 = (by adding and subtracting 1, the result doesn't change)

= 2 (cos(pi/4 - A))2 - ((sin(pi/4 - A))2 + (cos(pi/4 - A))2) + 1 = (because sin2 x + cos2 x = 1 for any x)

= (cos(pi/4 - A))2 - (sin(pi/4 - A))2 + 1. (just subtracted the cos(...) from 2cos(...))

Let's hold that there for a sec.

Now, take any value, x; let's calculate cos(2x):

cos(2x) =
= cos(x + x) =
= cos x cos x - sin x sin x = (because cos(a + b) = cos a cos b - sin a sin b)
= cos2 x - sin2 x.

So, for any x, cos2 x - sin2 x = cos(2x).

Now, if x = pi/4 - A, we have that our former equation, (cos(pi/4 - A))2 - (sin(pi/4 - A))2 + 1, can be rewritten as cos(2(pi/4 - A)) + 1.

So, we have cos(2(pi/4 - A)) + 1 = cos(pi/2 - 2A) + 1. Since, for any x, cos(pi/2 - x) = sin x, then cos(pi/2 - 2A) + 1 = sin(2A) + 1, which is what we wanted to prove.

That way, it's proven that 2cos2(pi/4 - A) = 1 + sin(2A).
@Bold How is that true?