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

*print*Print*list*Cite

### 3 Answers

cos(pi/4+x) = cos(pi/4)cos(x) - sin(pi/4)sin(x)

sin(pi/4) = cos(pi/4) = sqrt(2)/2

so

cos(pi/4 + x) = sqrt(2)/2 (cos(x) - sin(x))

so

2/sqrt(2) cos(pi/4 + x) = cos(x) - sin(x) and 2/sqrt(2) = (sqrt(2))^2/sqrt(2) = sqrt(2)

so

sqrt(2) cos(pi/4 + x) = cos(x) - sin(x)

So we have proven the identity.

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.**

thank you both so much for your help!