The Half That Everyone Forgets
Every student who meets $\sin A \cos A$ wants to jump straight to $\sin 2A$, and they are almost right. The double-angle identity says $\sin 2A = 2\sin A\cos A$, so the product of the sine and cosine of one angle is not the double-angle sine itself, but exactly half of it. That missing $\frac{1}{2}$ is the entire content of the sin A cos A formula, and it is the single most-dropped factor in the whole double-angle chapter.
The formula:
$$\sin A \cos A = \frac{1}{2}\sin 2A$$
Both trig functions share the same angle $A$, which is what separates this from the two-different-angles product $\sin A\cos B$. The result is compact and used constantly in integration and simplification.
What Is The Sin A Cos A Formula?
The sin A cos A formula expresses the product of the sine and cosine of one angle as half the sine of the doubled angle:
$$\sin A \cos A = \frac{1}{2}\sin 2A$$
The symbols are plain:
$A$ is any single angle, in degrees or radians.
$\sin A \cos A$ is the product of that angle's sine and cosine.
$\sin 2A$ is the sine of the doubled angle.
There is also a form written purely in terms of the tangent, useful when only $\tan A$ is known:
$$\sin A \cos A = \frac{\tan A}{1 + \tan^2 A}$$
Both belong to the trigonometric identities toolkit and are close relatives of the sin 2x double-angle identity.
How Is Sin A Cos A Different From Its Look-Alikes?
Three identities look almost the same, and mixing them up is the usual source of trouble. Here is the clean separation:
$\sin A \cos A = \frac{1}{2}\sin 2A$ - the product of one angle's sine and cosine, this article's identity.
$2\sin A \cos A = \sin 2A$ - the same product doubled, which lands exactly on the double-angle sine. This is the twin covered at 2 sin A cos A, and it is just this formula with both sides multiplied by $2$.
$\sin 2x = 2\sin x \cos x$ - the sin double angle formula read from left to right: given the single angle, build the doubled one.
And when the two angles are different, none of these apply; that is the job of the sin A cos B product-to-sum identity instead. The rule of thumb: same angle uses this formula, different angles use product-to-sum.
How Do You Derive The Sin A Cos A Formula?
The derivation is a single step from the double-angle identity. Start with:
$$\sin 2A = 2\sin A \cos A$$
Divide both sides by $2$:
$$\frac{1}{2}\sin 2A = \sin A \cos A$$
That is the whole proof. The tangent form takes one more step. Since $\sin 2A = \dfrac{2\tan A}{1 + \tan^2 A}$, substitute it in:
$$\sin A \cos A = \frac{1}{2} \cdot \frac{2\tan A}{1 + \tan^2 A} = \frac{\tan A}{1 + \tan^2 A}$$
The $\frac{1}{2}$ and the $2$ cancel, leaving the compact tangent form.
Examples Of Sin A Cos A
Example 1
Evaluate $\sin 30° \cos 30°$.
Apply the formula with $A = 30°$, so $2A = 60°$:
$$\sin 30° \cos 30° = \frac{1}{2}\sin 60° = \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{4}$$
Final answer: $\sin 30° \cos 30° = \dfrac{\sqrt{3}}{4}$.
Example 2
Evaluate $\sin 45° \cos 45°$. First the tempting route, then the correct one.
The reflex is to reach for the double angle and write $\sin 45° \cos 45° = \sin 90° = 1$, forgetting the $\frac{1}{2}$.
Check it directly: $\sin 45° = \cos 45° = \dfrac{1}{\sqrt{2}}$, so the product is $\dfrac{1}{\sqrt{2}} \cdot \dfrac{1}{\sqrt{2}} = \dfrac{1}{2}$, not $1$. The shortcut skipped the halving.
The correct route keeps the $\frac{1}{2}$:
$$\sin 45° \cos 45° = \frac{1}{2}\sin 90° = \frac{1}{2} \cdot 1 = \frac{1}{2}$$
Final answer: $\sin 45° \cos 45° = \dfrac{1}{2}$.
Example 3
Given $\sin 2A = \dfrac{2}{3}$, find $\sin A \cos A$.
Read the formula right to left: $\sin A \cos A = \dfrac{1}{2}\sin 2A$. Substitute $\sin 2A = \dfrac{2}{3}$:
$$\sin A \cos A = \frac{1}{2} \cdot \frac{2}{3} = \frac{1}{3}$$
Final answer: $\sin A \cos A = \dfrac{1}{3}$.
Example 4
If $\tan A = \dfrac{3}{4}$, find $\sin A \cos A$ using the tangent form.
Apply $\sin A \cos A = \dfrac{\tan A}{1 + \tan^2 A}$ with $\tan A = \dfrac{3}{4}$:
$$\sin A \cos A = \frac{\frac{3}{4}}{1 + \frac{9}{16}} = \frac{\frac{3}{4}}{\frac{25}{16}} = \frac{3}{4} \cdot \frac{16}{25} = \frac{12}{25}$$
Final answer: $\sin A \cos A = \dfrac{12}{25}$.
Example 5
Integrate $\displaystyle\int \sin x \cos x , dx$.
The product is awkward, but the formula turns it into a single sine. Replace $\sin x \cos x$ with $\dfrac{1}{2}\sin 2x$:
$$\int \sin x \cos x , dx = \frac{1}{2}\int \sin 2x , dx$$
Integrate the sine:
$$= \frac{1}{2}\left(-\frac{\cos 2x}{2}\right) + C = -\frac{\cos 2x}{4} + C$$
Final answer: $\displaystyle\int \sin x \cos x , dx = -\frac{\cos 2x}{4} + C$.
Example 6
Evaluate $\sin 15° \cos 15°$.
With $A = 15°$, the doubled angle is $30°$:
$$\sin 15° \cos 15° = \frac{1}{2}\sin 30° = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4}$$
Final answer: $\sin 15° \cos 15° = \dfrac{1}{4}$.
Where Is The Sin A Cos A Formula Used?
The identity earns its keep by collapsing a product into a single sine, which is exactly what most calculations need. Why not just multiply $\sin A$ by $\cos A$? Because a product of two functions resists integration and simplification, while a single sine gives way to both.
Integration. As Example 5 showed, $\int \sin x \cos x , dx$ has no clean antiderivative until $\sin x\cos x$ becomes $\frac{1}{2}\sin 2x$, after which it falls out at once. The same move helps across differentiation of trigonometric functions.
Simplifying identities. Longer proofs frequently produce a $\sin A\cos A$ pair, and folding it into $\frac{1}{2}\sin 2A$ keeps the algebra short.
Physics of oscillation. Power in an AC circuit involves a $\sin\theta\cos\theta$ term, and rewriting it as $\frac{1}{2}\sin 2\theta$ reveals that the power oscillates at twice the driving frequency.
Common Mistakes With Sin A Cos A
Mistake 1: Dropping The $\frac{1}{2}$
Where it slips in: Recognising the double angle and writing $\sin A \cos A = \sin 2A$.
Don't do this: $\sin A \cos A = \sin 2A$.
The correct way: $\sin A \cos A = \frac{1}{2}\sin 2A$. The double-angle identity is $\sin 2A = 2\sin A\cos A$, so solving for the product introduces the factor of $\frac{1}{2}$.
The habit that fixes this is deriving the formula from $\sin 2A = 2\sin A\cos A$ each time rather than memorising the result. A student who divides by $2$ on the spot never loses the half.
Mistake 2: Confusing It With Sin A Cos B
Where it slips in: Using this formula when the sine and cosine carry different angles.
Don't do this: Writing $\sin 5x \cos 3x = \frac{1}{2}\sin 8x$.
The correct way: Different angles need the product-to-sum identity: $\sin 5x\cos 3x = \frac{1}{2}[\sin 8x + \sin 2x]$. The single-angle formula only applies when both functions share the same angle.
Mistake 3: Mishandling The Sign In A Non-First Quadrant
Where it slips in: Evaluating $\sin A\cos A$ for an angle where the double angle lands in a quadrant with negative sine.
Don't do this: Assuming $\sin A\cos A$ is always positive.
The correct way: The sign follows $\sin 2A$. For $A = 120°$, $2A = 240°$ sits in the third quadrant, so $\sin A\cos A = \frac{1}{2}\sin 240° = -\frac{\sqrt{3}}{4}$, a negative value. The second-guesser who assumes a product of two magnitudes must be positive misses that $\cos 120°$ is itself negative.
This kind of sign oversight has echoes in engineering. AC power that oscillates at twice the line frequency, exactly the $\frac{1}{2}\sin 2\theta$ term, can drive mechanical resonance in transformers and rotating machinery, and mis-tracking that doubled frequency is a real design error, not a textbook footnote.
Key Takeaways
The sin A cos A formula is $\frac{1}{2}\sin 2A$, half the double-angle sine.
It is derived in one step by dividing $\sin 2A = 2\sin A\cos A$ by $2$.
The tangent form is $\dfrac{\tan A}{1 + \tan^2 A}$, handy when only $\tan A$ is known.
Keep the $\frac{1}{2}$, and use product-to-sum instead when the two angles differ.
To master double-angle and product identities with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for board and entrance prep, or live math tutoring with peers from 20+ countries.
A Practical Next Step
Practice these to solidify your understanding: evaluate $\sin 60° \cos 60°$, then find $\sin A\cos A$ given $\tan A = \dfrac{5}{12}$. If you get stuck, come back to the one-line derivation and rebuild the $\frac{1}{2}$ from the double-angle identity. Want a live Bhanzu trainer to work these through with you? Book a free demo class.
Read More
Pythagorean Identities — the $\sin^2\theta + \cos^2\theta = 1$ family behind the tan form.
Cos 2x — the cosine double-angle identity and its three forms.
Tan 2x Formula — the tangent double-angle identity.
Sin3x — the triple-angle identity for sine.
Double Angle Formula — sine, cosine, and tangent double angles in one place.
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