Trigonometric Identities: Numerical Problems Set (7)

Question

If $\sin\theta + \cos\theta = \sqrt{2}$, find the value of $\sin\theta\cos\theta$.

Solution — Step by Step

Why This Works

The “square the sum” technique exploits the Pythagorean identity. Whenever you see $\sin\theta + \cos\theta$ or $\sin\theta - \cos\theta$ given a specific value, squaring usually produces $\sin^2 + \cos^2 = 1$ as a free piece, leaving only the cross term $2\sin\theta\cos\theta$ to work with.

Many trig problems boil down to this trick. Memorise the pattern:

$(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta = 1 + \sin 2\theta$

$(\sin\theta - \cos\theta)^2 = 1 - 2\sin\theta\cos\theta = 1 - \sin 2\theta$

$(\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2$ (always)

Alternative Method

Use the substitution $\sin\theta + \cos\theta = \sqrt{2}\sin(\theta + \pi/4)$.

So $\sqrt{2}\sin(\theta + \pi/4) = \sqrt{2}$, meaning $\sin(\theta + \pi/4) = 1$, so $\theta + \pi/4 = \pi/2 \implies \theta = \pi/4$.

Then $\sin\theta\cos\theta = \sin(\pi/4)\cos(\pi/4) = (1/\sqrt{2})(1/\sqrt{2}) = 1/2$. Same answer.

Common Mistake

Students sometimes square one side and forget to square the other, writing $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = \sqrt{2}$. Wrong — both sides must be squared, so RHS becomes $2$, not $\sqrt{2}$.

Another trap: students multiply instead of squaring, or square but forget the cross term $2\sin\theta\cos\theta$. Always write $(a + b)^2 = a^2 + 2ab + b^2$ explicitly — don’t skip the middle term.