2 Sin A Cos A Formula — Sin 2A Proof, Examples

The Identity That Folds Two Factors Into One Angle

A radar dish sweeping a beam, a pendulum at the bottom of its swing, an AC generator at peak output — each is governed by a quantity that doubles an angle, and the 2 sin A cos A formula is what collapses that doubling into a single clean term.

The 2 sin A cos A formula says:

$$\boxed{;2\sin A \cos A = \sin 2A;}$$

In words: twice the product of the sine and cosine of an angle equals the sine of double that angle. It is one of the three double-angle identities (the sine one), and it falls directly out of the sum and difference formulas when the two angles are equal.

What Does 2 Sin A Cos A Equal?

The expression $2\sin A \cos A$ equals $\sin 2A$ — the sine of twice the angle. This is the bottom line, and it holds for every real value of $A$, in degrees or radians, with no domain restriction. The factor of $2$ is not decoration; it is what makes the two sides match exactly rather than off by half.

Read it in either direction. Left to right, it simplifies a product into a single sine — useful in integration. Right to left, it expands a double angle into factors — useful when you know $A$ but not $2A$.

How Is the 2 Sin A Cos A Formula Proved?

The proof is one substitution into a formula you already have.

From the sine sum formula. The sine of a sum is:

$$\sin(A + B) = \sin A \cos B + \cos A \sin B.$$

Set $B = A$. The left side becomes $\sin(A + A) = \sin 2A$, and the right side becomes:

$$\sin A \cos A + \cos A \sin A = 2\sin A \cos A.$$

Equating the two:

$$\sin 2A = 2\sin A \cos A.$$

That is the whole derivation — the double-angle identity is the angle-sum identity looking at itself.

Double-Anchoring — Right Triangle and Unit Circle

The identity reads two ways, and seeing both keeps it from feeling like a memorized string.

From the right triangle. Take an acute angle $A$ in a right triangle with hypotenuse $1$. Then $\sin A$ is the opposite side and $\cos A$ is the adjacent side. The product $2\sin A \cos A$ is twice the area of that triangle (base $\cos A$, height $\sin A$, doubled) — and that doubled area equals the height $\sin 2A$ of the angle $2A$ on the unit circle. The geometry and the algebra agree.

From the unit circle. Place angle $A$ so its terminal point is $(\cos A, \sin A)$. The angle $2A$ has terminal point $(\cos 2A, \sin 2A)$. The $y$-coordinate of that doubled angle — its sine — is exactly $2\sin A \cos A$. For example, at $A = 30°$ ($\pi/6$): $2\sin 30° \cos 30° = 2 \cdot \tfrac{1}{2} \cdot \tfrac{\sqrt{3}}{2} = \tfrac{\sqrt{3}}{2}$, which is $\sin 60° = \sin 2A$. The unit circle confirms what the triangle suggested.

The Tangent Form

When only the tangent of $A$ is known, the identity rewrites without sine or cosine. Divide numerator and denominator of $2\sin A \cos A$ by $\cos^2 A$ and use $1 + \tan^2 A = \sec^2 A$:

$$\sin 2A = 2\sin A \cos A = \frac{2\tan A}{1 + \tan^2 A}.$$

This form is the one that shows up in calculus substitutions and in the trigonometric identities used to integrate rational functions of sine and cosine.

Examples of the 2 Sin A Cos A Formula

Example 1

Simplify $2\sin 30° \cos 30°$.

By the formula, this is $\sin(2 \times 30°) = \sin 60° = \dfrac{\sqrt{3}}{2}$.

In radians, $2\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{6} = \sin\dfrac{\pi}{3} = \dfrac{\sqrt{3}}{2}$.

Final answer: $\dfrac{\sqrt{3}}{2}$.

Example 2

Evaluate $2\sin 15° \cos 15°$.

Wrong attempt. A student recalls "$\sin$ of double is two $\sin$" and writes $2\sin 15° \cos 15° = 2\sin 15° = \sin 30°$, then reports $\dfrac{1}{2}$ — getting the right number by a broken route.

The break. The shortcut "$\sin 2A = 2\sin A$" drops the cosine factor entirely. Test it on $A = 45°$: it would give $\sin 90° = 2\sin 45° = \sqrt{2} \approx 1.41$, but sine can never exceed $1$. The shortcut is false; it only looked right at $15°$ by coincidence, because $\cos 15°$ happened not to be needed once the formula was applied correctly.

Correct. Apply the identity in full: $2\sin 15° \cos 15° = \sin(2 \times 15°) = \sin 30° = \dfrac{1}{2}$. The cosine factor is what completes the double angle; it is not optional.

In radians, $2\sin\dfrac{\pi}{12}\cos\dfrac{\pi}{12} = \sin\dfrac{\pi}{6} = \dfrac{1}{2}$.

Final answer: $\dfrac{1}{2}$.

Example 3

Find $\sin 120°$ using the formula.

Write $120° = 2 \times 60°$, so $\sin 120° = 2\sin 60° \cos 60°$. Substitute $\sin 60° = \dfrac{\sqrt{3}}{2}$, $\cos 60° = \dfrac{1}{2}$:

$$\sin 120° = 2 \cdot \frac{\sqrt{3}}{2} \cdot \frac{1}{2} = \frac{\sqrt{3}}{2}.$$

In radians, $\sin\dfrac{2\pi}{3} = \dfrac{\sqrt{3}}{2}$.

Final answer: $\dfrac{\sqrt{3}}{2}$.

Example 4

Given $\sin A = \dfrac{3}{5}$ and $\cos A = \dfrac{4}{5}$, find $\sin 2A$.

Apply the formula directly:

$$\sin 2A = 2\sin A \cos A = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}.$$

Final answer: $\dfrac{24}{25} = 0.96$.

Example 5

Use the tangent form to find $\sin 2A$ when $\tan A = \dfrac{1}{2}$.

$$\sin 2A = \frac{2\tan A}{1 + \tan^2 A} = \frac{2 \cdot \frac{1}{2}}{1 + \frac{1}{4}} = \frac{1}{\frac{5}{4}} = \frac{4}{5}.$$

Final answer: $\dfrac{4}{5} = 0.8$.

Example 6

Integrate $\displaystyle\int 2\sin x \cos x , dx$.

Recognise the integrand as $\sin 2x$:

$$\int 2\sin x \cos x , dx = \int \sin 2x , dx = -\frac{1}{2}\cos 2x + C.$$

The identity turns a product into a single sine that integrates in one step.

Final answer: $-\dfrac{1}{2}\cos 2x + C$.

Where the Sine Double-Angle Identity Earns Its Keep

This identity is not a textbook curiosity — it is the algebra behind doubling in the physical world.

• Projectile range. The horizontal range of a projectile is $R = \dfrac{v^2 \sin 2\theta}{g}$, and the $\sin 2\theta$ term — rewritten as $2\sin\theta\cos\theta$ — is why a $45°$ launch maximizes distance: that is where $\sin 2\theta$ peaks at $1$.

• AC power. Instantaneous power in an alternating-current circuit involves a product of a sine voltage and a sine current; the $2\sin A \cos A$ rewrite is how engineers split it into average power plus an oscillating $\sin 2\omega t$ term.

• Signal mixing. Multiplying two waves of the same frequency produces a component at twice that frequency — the doubling that the identity describes, used in frequency doublers and laser optics.

• Pendulum and oscillation analysis. The energy of a swinging system carries $\sin 2\theta$ terms whenever displacement and velocity multiply.

The destination is clear: any time a system's output doubles an angle, this identity is the tool that exposes the structure.

The Mathematicians Behind the 2 Sin A Cos A Formula

Claudius Ptolemy (c. 100–170 CE, Greco-Egyptian) built the first usable trigonometric tables in the Almagest using a chord theorem that contains the angle-sum identity — and therefore the double-angle case — in geometric disguise. His tables drove astronomical prediction for over a thousand years.

Bhaskara II (1114–1185, India) stated the double-angle results for sine and cosine explicitly in the Siddhanta-Shiromani (1150), deriving them as special cases of the addition formulas centuries before the modern notation existed.

Why Students Get the Sine Double Angle Wrong

Mistake 1: Writing $\sin 2A = 2\sin A$

Where it slips in: A reader drops the cosine factor and treats doubling the angle as doubling the sine.

Don't do this: Pull the $2$ out front and discard $\cos A$.

The correct way: $\sin 2A = 2\sin A \cos A$ — both factors stay. The quickest disproof: at $A = 45°$, "$2\sin 45°$" gives $\sqrt{2} > 1$, impossible for a sine.

Mistake 2: Confusing $\sin^2 A$ with $\sin 2A$

Where it slips in: A reader reads $2\sin A \cos A$ and writes $\sin^2 A$ somewhere in the chain, conflating "sin squared" with "sin of two A."

Don't do this: Treat $\sin 2A$ and $\sin^2 A$ as interchangeable — they are different objects.

The correct way: $\sin 2A = 2\sin A \cos A$ is the double angle; $\sin^2 A = (\sin A)^2$ is the square. The memorizer who learned both notations on the same day is the one who blends them under exam pressure.

Mistake 3: Mixing degree and radian values mid-problem

Where it slips in: A reader computes $2\sin(\pi/6)\cos(\pi/6)$ but plugs in calculator values taken in degree mode.

Don't do this: Switch units between the substitution and the evaluation.

The correct way: Pick one unit and hold it. $2\sin 30° \cos 30° = 2\sin\dfrac{\pi}{6}\cos\dfrac{\pi}{6}$ — the same number written two ways — so state the unit beside the answer and keep the calculator mode matched to it.

Key Takeaways

• The 2 sin A cos A formula is $2\sin A \cos A = \sin 2A$, the sine double-angle identity.

• It comes from the sine sum formula with both angles equal — no separate proof to memorize.

• The tangent form is $\sin 2A = \dfrac{2\tan A}{1 + \tan^2 A}$.

• The most common mistake is dropping the cosine and writing $\sin 2A = 2\sin A$, which can give impossible values above $1$.

• The identity drives projectile range, AC power, and signal doubling — anywhere an angle gets doubled.

Practice These Three

1. Simplify $2\sin 45° \cos 45°$ in degrees and radians.

2. Given $\sin A = \dfrac{5}{13}$ and $\cos A = \dfrac{12}{13}$, find $\sin 2A$.

3. Show that $\dfrac{2\sin A \cos A}{1 - 2\sin^2 A} = \tan 2A$.

If Problem 1 returns $\sin 90° = 1$, the doubling landed exactly at the peak of sine. Want a live Bhanzu trainer to walk your child through the double-angle identities and the Class 11 trigonometry chapter? Book a free demo class — online globally.