If $\sin\theta + \cos\theta = 1$, then $\sin\theta \cdot \cos\theta$ equals:
- $\dfrac{1}{2}$
- $1$
- $0$
- $-1$
Show answer and solution
Correct answer: C. $0$
Square both sides of $\sin\theta + \cos\theta = 1$:
$(\sin\theta + \cos\theta)^2 = 1^2$
$\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1$
Using $\sin^2 + \cos^2 = 1$:
$1 + 2\sin\theta\cos\theta = 1$
$\sin\theta \cos\theta = 0$.
Verify: the original equation holds only when $\theta = 0°$ (so $\sin = 0$, $\cos = 1$) or $\theta = 90°$ (so $\sin = 1$, $\cos = 0$). In both cases the product is 0. ✓