Proving cos⁡(x)-sin⁡(x)=√2 cos⁡(pi/4+x)

We know that sin x = cos (pi/2 - x), therefore we can
create a difference of two matching trigonometric functions, to the left
side:

cos x - cos (pi/2 - x)=√2
cos⁡(pi/4+x)

We'll transform the formula into a
product:

cos a - cos b =
2sin[(a+b)/2]*sin[(b-a)/2]

cos x - cos (pi/2 - x)=2sin[(x +
pi/2 - x)/2]*sin(pi/2 - 2x)/2

cos x - cos (pi/2 -
x)=2sin[(pi/2)/2]*sin(pi/2 - 2x)/2

cos x - cos (pi/2 -
x)=2sin(pi/4)*sin(pi/4-2x/2)

But sin pi/4 =
√2/2

cos x - cos (pi/2 -
x)=2√2sin(pi/4-x)/2

We'll
simplify:

cos x - cos (pi/2 -
x)=√2sin(pi/4-x)

But sin(pi/4-x) = cos (pi/4 + x),
therefore cos x - cos (pi/2 - x)=√2cos (pi/4 +
x)

Therefore, the given identity cos x -sin x
= √2cos (pi/4 + x) is verified.