If cos theta + sin theta = root 2 cos thetha prove that cos theta - sin theta = root 2 sin theta ​

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.

logv3(x+2)=1

rewrite in exponential form:

logb(x) = y is equal to b^y =x

b = 3, y = 1

now solve for x:

3^1 = x+2)

x = 3-2

x = 1

see the picture for the graph.