Q.Evaluate: $(\sin45° + \cos45°)^2$.

Both terms inside the bracket equal the same standard value, so the sum simplifies before squaring.

$\sin45°+\cos45° = \frac{1}{\sqrt2}+\frac{1}{\sqrt2} = \frac{2}{\sqrt2} = \sqrt2$, so $(\sin45°+\cos45°)^2 = (\sqrt2)^2 = 2$.

Step 1 — read the value. $\sin45°=\cos45°=\dfrac{1}{\sqrt2}$.

Step 2 — add.

Step 3 — square.

Alternatively, expand first: $(\sin45°+\cos45°)^2=\sin^245°+2\sin45°\cos45°+\cos^245° = 1 + 2\left(\frac1{\sqrt2}\right)\left(\frac1{\sqrt2}\right) = 1+1 = 2$, using $\sin^2\theta+\cos^2\theta=1$ from the identities section — the same answer by a route that leans on the Pythagorean identity instead of direct simplification.

A frequent error is squaring each term separately and adding ($\sin^245°+\cos^245°=1$) while forgetting the cross term $2\sin45°\cos45°$ from expanding the bracket — that shortcut only works when the bracket is a sum of squares, not a squared sum.