Prove this identify :
2 cos^2 (45°-A ) = 1 +sin 2 A ​

I am guessing that you mean 2cos2(45° - A) = 1 + sin2A

According to a double angle identity, 2cos2θ - 1 = cos(2θ)

So, 2cos2(45° - A) - 1

= cos (90° - 2A) =cos90°cos2A + sin90°sin2A = (0)cos2A + (1)sin2A = sin2

Thus, 2cos2(45° -A ) = 1 + sin2A

given that triangle abc is transformed two times to get triangle

a"b"c"

first abc is transformed in to a'b'c' as follows:

the coordinates of a are (1,-1) and transformed into a'(-1,1)

similarly b (4,-2) became b'(-4,2) and

c(7,-2) became c'(-7,2)

i.e. (x,y) becomes (-x,-y)

this is nothing but reflection about a point here origin.

thus first transformation is reflection on the origin.

next is exactly shifting 3 units down

source:

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