Answer:
We have:
If:
Cos(θ) + Sin(θ) = √2*cos(θ)
We want to prove that:
Cos(θ) - Sin(θ) = √2*Sin(θ)
Well, let's start with the first relation:
Cos(θ) + Sin(θ) = √2*cos(θ)
Now we can subtract 2*Sin(θ) and we will get:
Cos(θ) + Sin(θ) - 2*Sin(θ) = √2*cos(θ) - 2*sin(θ)
Cos(θ) - Sin(θ) = √2*cos(θ) - 2*sin(θ)
Now, we also can rewrite the first equation as:
Cos(θ) + Sin(θ) = √2*cos(θ)
Cos(θ) - √2*cos(θ) = - Sin(θ)
Cos(θ)*( 1 - √2) = -Sin(θ)
Cos(θ) = -Sin(θ)/( 1 - √2) = Sin(θ)/(- 1 + √2)
We can replace this in the right side of theequation:
Cos(θ) - Sin(θ) = √2*cos(θ) - 2*sin(θ)
Cos(θ) - Sin(θ) = √2* Sin(θ)/(- 1 + √2) - 2*sin(θ)
Cos(θ) - Sin(θ) = (√2/(- 1 + √2) - 2)*Sin(θ)
Now we have this:
√2/(- 1 + √2)) - 2 = a
if we multiply all by (-1 + √2) we get:
√2 - 2*(-1 + √2) = a*(-1 + √2)
√2 + 2 - 2*√2 = a*(-1 + √2)
2 - √2 = a*(-1 + √2)
√2*(√2 - 1) = a*(-1 + √2)
√2*(√2 - 1)/(-1 + √2) = a
√2 = a
Then:
√2/(- 1 + √2)) - 2 = √2
If we replace this in the equation:
Cos(θ) - Sin(θ) = (√2/(- 1 + √2) - 2)*Sin(θ)
We get:
Cos(θ) - Sin(θ) = √2*Sin(θ)
Which is the thing we wanted to get.
