Squaring a Sum

A common mistake: ${\left(\sin\left(x\right)+\cos\left(x\right)\right)}^{2}$ is not ${\sin\left(x\right)}^{2}+{\cos\left(x\right)}^{2}$, so it is not 1. At 45°, $\sin\left(45^{\circ}\right)+\cos\left(45^{\circ}\right) = \sqrt{2}$, and √2 squared is 2.

Use ${\left(a+b\right)}^{2} = {a}^{2}+2ab+{b}^{2}$. Keep the middle term. So ${\left(\sin\left(x\right)+\cos\left(x\right)\right)}^{2} = {\sin\left(x\right)}^{2}+2\sin\left(x\right)\cos\left(x\right)+{\cos\left(x\right)}^{2}$. The two squares add up to 1. That leaves $1+2\sin\left(x\right)\cos\left(x\right)$. At 45° that is 1 + 1.