If cos(alpha) + sin(alpha) = (2)^1/2cos(alpha) , then find the value of cos(alpha) - sin(alpha).?

-sinA=(1-√2)cosA -> tanA =√2 - 1 -> A= 22.%º -> cosA - sinA = 0.541

If you mean it as on identity (true for all α), then your premise is wrong.
cosα + sinα is not identical to √2 cosα.

If it's an equation that you want to solve for α, then you can still use the following trick.

There's a neat graphical trick to find the identity, that uses vectors in the plane:
Replace α with θ, just for familiarity.
Then on an x-y Cartesian plot, draw the vector v₁=î=<0,1> from (0,0) to (0,1).
Now draw the vector v₂=jˆ=<1,0> from (0,1) to (1,1).
Finally, draw the vector sum
v₃ = v₁ + v₂ = î + jˆ = <1,1>
from (0,0) to (1,1).

Now think of that triangle as a rigid figure that rotates about the origin.
When it has rotated through an angle θ,
• v₁ has x-coordinate cosθ;
• v₂ adds to that x-coordinate, +sinθ;
• and the vector sum, v₃, has x-coordinate cosθ + sinθ.

But because of the 45º angle, v₃ has length √2, and makes an angle, θ+45º, with the +x-axis. So from this you can see that
cosθ + sinθ ≡ √2 cos(θ+45º)

Similarly,
cosθ - sinθ ≡ √2 cos(θ-45º) ≡ √2 cos(θ+45º-90º)

So for values of θ that satisfy:
cosα + sinα = √2 cosα

. . you can see that:
cosα - sinα = √2 cos(α-90º) = √2 sinα

cos a - sin a = sqrt(1+ cos 2a - 2 sin 2a)

cosa + sina = 2^1/2 cosa
squaring on both sides
cos^2a+ 2cosa sina + sin^2a = 2 cos^2a
or cos^2a- sin^2a= 2 cosa sina
or ( cosa -sina) ( cosa+ sina) =2sina cosa
or cosa- sina = 2 sina cosa/(cosa + sina)
= 2 sina cosa/2^1/2 cosa
=2^1/2sina
so cosa- sina = 2^1/2 sina ANSWER

Hi Akash,

Adding to the four answers you have, let me add mine also as below.

i) Given: cos(α) + sin(α) = √2*cos(α)
Squaring this, cos²α + sin²α + 2*cos(α)*sin(α) = 2*cos²α ------------ (1)

ii) Let required, cos(α) - sin(α) = x
Squaring this also, cos²α + sin²α - 2*cos(α)*sin(α) = x² ----------- (2)

iii) Adding (1) & (2):
2 = 2*cos²α + x²

==> x² = 2 - 2*cos²α = 2(1 - cos²α) = 2*sin²α

So, x = ±√2*sin(α)

Assuming α is in (0, π/4), cos(α) - sin(α) = √2*sin(α)

Square it
(cos a +sin a)^2 = cos^2 a + sin^2 a+ 2 sin a cos a= 2 cos^2 a
(cos a-sin a)^2 = cos^2 a + sin ^2 a - 2 sin a cos a
from the above equation, replace cos^2 a + sin^2 a
= 2 cos^2 a - 2 cos a sin a - 2 sin a cos a
= 2 cos^2 a - 4 cos a sin a= 1+ cos 2a - 2 sin 2 a
This gives
cos a - sin a = sqrt(1+ cos 2a - 2 sin 2a)

I assumed you mean sqrt(2)*cos(alpha)
so cos(alpha) - sin(alpha) = sqrt(2)*cos(alpha) - 2sin(alpha)