Xueqin!

Originally Posted by delicious french cuisine

rate 5! for math

Lol.

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?