Question

How do you prove `(sinx+cosx)^2 + (sinx-cosx)^2 = 2`?

Answer 1

Let's see:
opening the brackets by evaluating the squares:
`sin^2(x)+2sin(x)cos(x)+cos^2(x)+sin^2(x)-2sin(x)cos(x)+cos^2(x)=2`
simplifying the `2sin(x)cos(x)` and considering that `sin^2(x)+cos^2(x)=1` you get:
`sin^2(x)+cos^2(x)+sin^2(x)+cos^2(x)=2`
`1+1=2`