if cosx-sinx=root 2 sinx, then what is the value of cosx+sinx. How

√2sinx + sinx = cosx

sinx(√2+1)= cosx
sinx/cosx = 1/(√2+1)
tanx = 1/(√2+1)

now construct a right-angled triange with sides 1 and √2+1
let the hypotenuse be r
r^2 = 1^2 + (√2+1)^2 = 1 + 2 + 2√2 + 1 = 4 + 2√2
r = √(4+2√2))

cosx + sinx = (√2+1)√(4+2√2) + 1/(√(4+2√2)

= (√2 + 2)/(√(4 + 2√2)

You might want to rationalize this to get a different looking answer.

To find the value of cosx+sinx when given cosx-sinx=root 2 sinx, we'll have to first solve the given equation and then substitute the value of cosx into the expression.

Let's solve the equation cosx-sinx=root 2 sinx:

Rearrange the equation by moving all terms involving sinx to one side:
cosx - root 2 sinx - sinx = 0

Now, combine like terms:
cosx - (1 + root 2)sinx = 0

Factor out sinx:
cosx - sinx(1 + root 2) = 0

Divide through by (1 + root 2):
(cosx - sinx(1 + root 2)) / (1 + root 2) = 0 / (1 + root 2)
cosx / (1 + root 2) - sinx = 0

Now, let's substitute the value of cosx from the equation cosx-sinx=root 2 sinx:
(root 2 sinx + sinx) / (1 + root 2) - sinx = 0

Combine like terms again:
((root 2 + 1)sinx - sinx(1 + root 2)) / (1 + root 2) = 0

Further simplify:
(sinx(root 2 + 1) - sinx(1 + root 2)) / (1 + root 2) = 0

We can see that both terms are equal to sinx, so we can cancel them out:
(0) / (1 + root 2) = 0

Now, we have 0 = 0, which is true. This means that the equation cosx-sinx=root 2 sinx holds for any value of x.

Since the given equation holds for any value of x, there is no specific value for cosx or sinx. Therefore, we cannot determine the value of cosx+sinx based solely on the given equation.