Q.The maximum value of $\sin x \cdot \cos x$ is:
(A) $\dfrac{1}{4}$
(B) $\dfrac{1}{2}$
(C) $\sqrt{2}$
(D) $2\sqrt{2}$

Concept: Maximum value of $\sin x \cos x$ — use the double-angle identity to rewrite as a single sine function.

Step 1: Recall the identity $\sin 2x = 2 \sin x \cos x$.

Thus $\sin x \cos x = \frac{1}{2} \sin 2x$. …

The problem asks for the maximum value of $\sin x \cdot \cos x$. At first glance, you might think of trying values like $x = 45^\circ$ (where both sine and cosine are $\frac{1}{\sqrt{2}}$) and get $\frac{1}{2}$. That’s a good guess — but let’s confirm it rigorously and understand why it’s the absolute maximum.

The key insight is that a product of two different trig functions can often be simplified into a single sine or cosine function using an identity. Here, the double-angle formula for sine is your best friend:

This means $\sin x \cos x = \frac{1}{2} \sin 2x$.

Now the problem becomes much simpler. Instead of juggling two functions, we just need to find the maximum of $\frac{1}{2} \sin 2x$.

1. Recall the range of sine. For any real angle $\theta$, $\sin \theta$ lies between $-1$ and $1$. So $\sin 2x$ also lies between $-1$ and $1$.

2. Scale by the constant. Multiplying by $\frac{1}{2}$ scales the entire range: $\frac{1}{2} \sin 2x$ lies between $-\frac{1}{2}$ and $\frac{1}{2}$.

3. When does the maximum occur? The maximum of $\sin 2x$ is $1$, which happens when $2x = \frac{\pi}{2} + 2\pi k$ (i.e., $x = \frac{\pi}{4} + \pi k$). At those points, $\sin x \cos x = \frac{1}{2} \cdot 1 = \frac{1}{2}$. …

Method: Maximizing a Trigonometric Product Using a Double-Angle Identity

Use this whenever you need the maximum or minimum of a product like $\sin x \cos x$ (or similar), rather than a sum.

Steps

Step 1: Recognise the product form and recall the relevant double-angle identity.

The identity $\sin 2x = 2\sin x\cos x$ converts a product of $\sin x$ and $\cos x$ into a single sine function of a doubled angle.

Step 2: Rewrite the expression as a constant multiple of a single trig function.

Once expressed as $\frac{1}{2}\sin 2x$, the problem reduces from a two-variable-looking product to the range of one familiar function.

Step 3: Apply the known range $-1 \le \sin(\text{anything}) \le 1$. …

Common Mistakes

Mistake 1: Assuming the maximum of a product equals the product of each factor's individual maximum.

Why it's wrong: $\sin x$ and $\cos x$ each reach their own maximum of $1$, but not at the same value of $x$ — at $x=\tfrac{\pi}{2}$, $\sin x=1$ but $\cos x=0$, giving a product of $0$, not $1$. Correct approach: never multiply the separate maxima of two different trig functions; instead combine them algebraically (here, via the double-angle identity) before finding the true joint maximum.

Mistake 2: Forgetting the $\tfrac{1}{2}$ scaling factor after applying the double-angle identity. …