Sin A Cos B Formula : Derivation & Examples

The Identity That Turns Multiplication Into Addition

Radio engineers had a problem long before computers: multiplying two wave signals by hand was slow and error-prone, but adding them was quick. The product-to-sum identities were the workaround, and sin A cos B is the one that rewrites a sine-times-cosine product as a clean sum of two sines. That single move, trading a multiplication for an addition, is why the identity still earns its place in signal processing and calculus today.

The formula reads:

$$\sin A \cos B = \frac{1}{2}\left[\sin(A+B) + \sin(A-B)\right]$$

Here $A$ and $B$ are any two angles, in degrees or radians. The product on the left becomes the average of two sines, one at the sum of the angles and one at the difference.

What Is The Sin A Cos B Formula?

The sin A cos B formula is a trigonometric identity that expresses the product $\sin A \cos B$ as a sum:

$$\sin A \cos B = \frac{1}{2}\left[\sin(A+B) + \sin(A-B)\right]$$

It belongs to the family of product-to-sum formulas, which convert a product of two trigonometric ratios into a sum or difference. Each symbol carries a plain meaning:

• $A$ is the angle inside the sine.

• $B$ is the angle inside the cosine.

• $A+B$ and $A-B$ are the sum and difference angles the product resolves into.

The identity holds for every value of $A$ and $B$, so it works for numeric angles like $75°$ and $15°$ as well as variable angles like $3x$ and $x$. It sits alongside its three companions in the trigonometric identities toolkit: $\cos A \cos B$, $\sin A \sin B$, and $\cos A \sin B$.

One quick warning about order: $\sin A \cos B$ is not the same as $\cos A \sin B$. Swapping which angle sits in the sine and which sits in the cosine changes the result, because $\cos A \sin B = \frac{1}{2}\left[\sin(A+B) - \sin(A-B)\right]$ carries a minus sign where this one carries a plus.

How Do You Derive The Sin A Cos B Formula?

The derivation starts from the sum and difference identities for sine. Write both of them out:

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

$$\sin(A-B) = \sin A \cos B - \cos A \sin B$$

Now add the two equations. The $\cos A \sin B$ terms have opposite signs, so they cancel:

$$\sin(A+B) + \sin(A-B) = 2\sin A \cos B$$

Divide both sides by $2$:

$$\sin A \cos B = \frac{1}{2}\left[\sin(A+B) + \sin(A-B)\right]$$

That is the whole proof. Why does adding the two identities work so cleanly? Because sine of a difference flips the sign of only the $\cos A \sin B$ piece, so stacking the sum on the difference deletes exactly the term you do not want and doubles the term you do.

Examples Of Sin A Cos B

Example 1

Express $\sin 3x \cos x$ as a sum.

Identify $A = 3x$ and $B = x$, then apply the formula:

$$\sin 3x \cos x = \frac{1}{2}\left[\sin(3x+x) + \sin(3x-x)\right]$$

$$= \frac{1}{2}\left[\sin 4x + \sin 2x\right]$$

Final answer: $\sin 3x \cos x = \dfrac{1}{2}\left[\sin 4x + \sin 2x\right]$.

Example 2

Evaluate $\sin 75° \cos 15°$ exactly. First the tempting route, then the correct one.

The first instinct many students reach for is to collapse the product into a single sine, writing $\sin 75° \cos 15° = \sin(75° - 15°) = \sin 60° = \dfrac{\sqrt{3}}{2} \approx 0.866$.

Check that against a calculator: $\sin 75° \approx 0.966$ and $\cos 15° \approx 0.966$, so the true product is about $0.933$, not $0.866$. The shortcut is wrong because a product of two ratios is not the sine of anything by itself.

The rescue is the identity. With $A = 75°$ and $B = 15°$:

$$\sin 75° \cos 15° = \frac{1}{2}\left[\sin 90° + \sin 60°\right]$$

$$= \frac{1}{2}\left[1 + \frac{\sqrt{3}}{2}\right] = \frac{2 + \sqrt{3}}{4} \approx 0.933$$

Final answer: $\sin 75° \cos 15° = \dfrac{2 + \sqrt{3}}{4}$.

Example 3

Rewrite $2\sin 5\theta \cos 3\theta$ as a sum.

The leading $2$ cancels the $\frac{1}{2}$, so $2\sin A \cos B = \sin(A+B) + \sin(A-B)$. With $A = 5\theta$ and $B = 3\theta$:

$$2\sin 5\theta \cos 3\theta = \sin 8\theta + \sin 2\theta$$

Final answer: $2\sin 5\theta \cos 3\theta = \sin 8\theta + \sin 2\theta$.

Example 4

Integrate $\displaystyle\int \sin 3x \cos x ,dx$.

A product of trig functions is awkward to integrate directly, so convert it first using Example 1:

$$\int \sin 3x \cos x ,dx = \frac{1}{2}\int \left(\sin 4x + \sin 2x\right)dx$$

Integrate each sine term separately:

$$= \frac{1}{2}\left(-\frac{\cos 4x}{4} - \frac{\cos 2x}{2}\right) + C$$

$$= -\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C$$

Final answer: $\displaystyle\int \sin 3x \cos x ,dx = -\frac{\cos 4x}{8} - \frac{\cos 2x}{4} + C$.

Example 5

Show that $\sin A \cos A = \frac{1}{2}\sin 2A$ falls straight out of this identity.

Set $B = A$. Then $A + B = 2A$ and $A - B = 0$:

$$\sin A \cos A = \frac{1}{2}\left[\sin 2A + \sin 0\right] = \frac{1}{2}\sin 2A$$

Because $\sin 0 = 0$, the difference term vanishes and the single-angle result sin A cos A appears as the special case where the two angles are equal.

Final answer: $\sin A \cos A = \dfrac{1}{2}\sin 2A$.

Example 6

Express $\sin 2x \cos 5x$ as a sum, where the cosine angle is larger.

Keep the roles fixed: $A = 2x$ (inside sine), $B = 5x$ (inside cosine).

$$\sin 2x \cos 5x = \frac{1}{2}\left[\sin 7x + \sin(-3x)\right]$$

Sine is an odd function, so $\sin(-3x) = -\sin 3x$:

$$= \frac{1}{2}\left[\sin 7x - \sin 3x\right]$$

Final answer: $\sin 2x \cos 5x = \dfrac{1}{2}\left[\sin 7x - \sin 3x\right]$.

Where Is The Sin A Cos B Formula Used?

The identity earns its keep by solving one recurring problem: a product of trig functions is hard to work with, and a sum is easy. Why bother converting at all? Because almost every operation that stalls on a product runs smoothly on a sum.

• Integration. A product like $\sin 3x \cos x$ has no simple antiderivative as written, but the sum $\frac{1}{2}(\sin 4x + \sin 2x)$ integrates term by term, as Example 4 showed.

• Solving equations. Turning products into sums exposes shared frequencies, which is how many trigonometric equations get factored and solved.

• Signal processing. When two waves multiply (amplitude modulation in radio, the "beats" you hear when two close musical notes sound together), the product-to-sum identity is what reveals the individual frequencies actually present.

The reverse direction, going from a sum back to a product, is the job of the sum to product formula, and the two are used together whenever a trig expression needs to change shape.

Common Mistakes With Sin A Cos B

Mistake 1: Dropping The $\frac{1}{2}$ Factor

Where it slips in: Applying the identity quickly and writing only $\sin(A+B) + \sin(A-B)$.

Don't do this: $\sin A \cos B = \sin(A+B) + \sin(A-B)$.

The correct way: The whole bracket is halved: $\sin A \cos B = \frac{1}{2}\left[\sin(A+B) + \sin(A-B)\right]$. The factor of $\frac{1}{2}$ is what balances the $2\sin A \cos B$ that came out of the derivation.

The habit that fixes this is deriving the identity once by hand before using it. A student who has seen the $2$ appear and cancel remembers the $\frac{1}{2}$; one who only memorised the right-hand side often loses it.

Mistake 2: Confusing The Order Of A And B

Where it slips in: Reaching for the formula when the product is actually $\cos A \sin B$, not $\sin A \cos B$.

Don't do this: Treating $\cos 5x \sin 2x$ as if it gives $\frac{1}{2}[\sin 7x + \sin 3x]$.

The correct way: $\cos A \sin B$ uses a minus sign: $\cos A \sin B = \frac{1}{2}\left[\sin(A+B) - \sin(A-B)\right]$. Which angle sits in the sine and which in the cosine decides the sign inside the bracket, so read the product carefully before choosing the identity.

Mistake 3: Mishandling The Negative Difference Angle

Where it slips in: When $B > A$, so that $A - B$ comes out negative.

Don't do this: Writing $\sin(A-B)$ as $\sin(B-A)$ and keeping it positive.

The correct way: Sine is odd, so $\sin(A-B) = -\sin(B-A)$. In Example 6, $\sin(-3x)$ became $-\sin 3x$, and skipping that sign flip is a common source of a wrong final sign.

Key Takeaways

• The sin A cos B formula is $\frac{1}{2}\left[\sin(A+B) + \sin(A-B)\right]$, a product-to-sum identity.

• It is derived by adding the sine sum and difference identities, which cancels the cross terms.

• Keep the $\frac{1}{2}$ factor and respect the order of $A$ and $B$, since $\cos A \sin B$ carries a minus sign instead.

• Its main payoff is turning products into sums that integrate and solve cleanly.

To work through product-to-sum identities with a teacher, explore Bhanzu's trigonometry tutor sessions, a high school math tutor for exam practice, or live math classes online with peers from 20+ countries.

A Practical Next Step

Practice these to solidify your understanding: rewrite $\sin 4x \cos 2x$ as a sum, then evaluate $\sin 105° \cos 15°$ exactly using the identity. If you get stuck, come back to the derivation and rebuild the $\frac{1}{2}$ from the two sine identities. Want a live Bhanzu trainer to work these with you? Book a free demo class.

Read More

• 2 Sin A Cos A Formula — the doubled single-angle product that lands on sin 2A.

• Sin Double Angle Formula — how $\sin 2x = 2\sin x\cos x$ connects to this product.

• Sin A + Sin B Formula — the sum-to-product partner that runs the other direction.

• Cos(A − B) Formula — the difference identity used in the derivation.