if cosx sinx =root2cosx, then prove that cosx-sinx =root2sinx
To prove that cos(x) - sin(x) = √2sin(x), we need to start with the given equation cos(x)sin(x) = √2cos(x) and manipulate it algebraically to obtain the desired expression.
1. Start with the given equation:
cos(x)sin(x) = √2cos(x)
2. Divide both sides of the equation by cos(x):
sin(x) = √2
3. Square both sides of the equation to eliminate the square root:
sin^2(x) = 2
4. Rewrite sin^2(x) as 1 - cos^2(x) using the Pythagorean identity sin^2(x) + cos^2(x) = 1:
1 - cos^2(x) = 2
5. Rearrange the equation by subtracting 1 from both sides:
-cos^2(x) = 1
6. Multiply both sides of the equation by -1:
cos^2(x) = -1
7. Take the square root of both sides of the equation:
cos(x) = ±√(-1)
8. Since the square root of a negative number is not a real number, we can conclude that there are no real solutions for cos(x) = ±√(-1).
Hence, it is not possible to prove that cos(x) - sin(x) = √2sin(x) based on the given equation cos(x)sin(x) = √2cos(x).
cosx sinx = √2 cosx
sinx = √2
sorry - cannot happen
However,
cosx-sinx
= √2 (1/√2 cosx - 1/√2 sinx)
= √2 sin(π/4 + x)