Exercises — Product-to-sum formulas

The two factors are mixed (a sine times a cosine), and the sine comes first, so we use $\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)].$ Matching term-by-term: $A=7x$ (the argument of $\sin$), $B=2x$ (the argument of $\cos$). WHY order matters: $\sin A\cos B\ne \cos A\sin B$ — swapping would flip a sign, so always read the sine's angle as $A$.

Both factors are cosines → same kind → output is cosines. Identity: $\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)],A=40^{\circ },B=10^{\circ }.$ Prediction: a sum of cosines, specifically $\frac{1}{2}[\cos 30^{\circ }+\cos 50^{\circ }]$.

Two sines → same kind → cosines out, and the sin·sin identity uses a minus: $\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)],A=3\theta ,B=\theta .$ The sign in front of $\cos (A+B)$ is $\boxed{-}$.

Step 1 (WHAT): identify cos·cos, $A=6x,B=2x$. Step 2 (apply): $2\cos 6x\cos 2x=2\cdot \frac{1}{2}[\cos (6x-2x)+\cos (6x+2x)]=\cos 4x+\cos 8x.$ WHY the 2 vanished: the leading $2$ cancels the formula's $\frac{1}{2}$. $\boxed{\cos 4x+\cos 8x}$ This is exactly the curve pictured at the top of the page — the product on the left equals the summed cosines on the right.

sin·sin identity (minus sign!), $A=5\theta ,B=2\theta$: $\sin 5\theta \sin 2\theta =\frac{1}{2}[\cos (5\theta -2\theta )-\cos (5\theta +2\theta )]=\frac{1}{2}[\cos 3\theta -\cos 7\theta ].$ $\boxed{\frac{1}{2}(\cos 3\theta -\cos 7\theta )}$ WHY the minus in front of $\cos (A+B)$? Recall the two cosine addition formulas: $\cos (A-B)=\cos A\cos B+\sin A\sin B$ and $\cos (A+B)=\cos A\cos B-\sin A\sin B$. To keep the $\sin A\sin B$ product and cancel the $\cos A\cos B$ part, we must subtract the second from the first: $\cos (A-B)-\cos (A+B)=2\sin A\sin B$. That subtraction is where the minus sign in front of $\cos (A+B)$ comes from — it is not a memorised quirk, it is what makes the cosine terms cancel. (Contrast cos·cos, where we add the two formulas, leaving a $+$.) See the second figure for this cancellation drawn out.

cos·cos, $A=75^{\circ },B=15^{\circ }$: $\cos 75^{\circ }\cos 15^{\circ }=\frac{1}{2}[\cos (75^{\circ }-15^{\circ })+\cos (75^{\circ }+15^{\circ })]=\frac{1}{2}[\cos 60^{\circ }+\cos 90^{\circ }].$ Now $\cos 60^{\circ }=\frac{1}{2}$ and $\cos 90^{\circ }=0$: $=\frac{1}{2}\left[\frac{1}{2}+0\right]=\boxed{\frac{1}{4}}.$

WHY product-to-sum? A product of trig functions has no direct antiderivative rule; a sum does (each cosine integrates to a sine). Step 1: sin·cos, $A=3x,B=x$: $\sin 3x\cos x=\frac{1}{2}[\sin (3x+x)+\sin (3x-x)]=\frac{1}{2}[\sin 4x+\sin 2x].$ Step 2: integrate term by term (recall $\int \sin kxdx=-\frac{\cos kx}{k}$): $\int =\frac{1}{2}\left[-\frac{\cos 4x}{4}-\frac{\cos 2x}{2}\right]+C=-\frac{\cos 4x}{8}-\frac{\cos 2x}{4}+C.$ $\boxed{-\frac{\cos 4x}{8}-\frac{\cos 2x}{4}+C}$ (See Integration of Trig Functions for the antiderivative rule used.)

Expand each product: $\cos 3x\cos x=\frac{1}{2}[\cos 2x+\cos 4x],\sin 3x\sin x=\frac{1}{2}[\cos 2x-\cos 4x].$ Subtract: $\cos 3x\cos x-\sin 3x\sin x=\frac{1}{2}[\cos 2x+\cos 4x]-\frac{1}{2}[\cos 2x-\cos 4x]=\frac{1}{2}\cdot 2\cos 4x=\cos 4x.$ $\boxed{\cos 4x}$ What it reveals: this is exactly the cosine addition formula $\cos (A+B)=\cos A\cos B-\sin A\sin B$ with $A=3x,B=x$, since $3x+x=4x$. Product-to-sum and angle-addition are two views of the same truth.

Set $A=B=\theta$. Then $A-B=0$ and $A+B=2\theta$: $\sin \theta \sin \theta =\frac{1}{2}[\cos 0-\cos 2\theta ]=\frac{1}{2}[1-\cos 2\theta ].$ So ${\sin }^2\theta =\frac{1}{2}(1-\cos 2\theta )$ — the power-reduction identity, a special case of double-angle. WHY this matters: it shows the product-to-sum formulas cover the degenerate case $A=B$ gracefully; nothing breaks when the two angles coincide.

Term 1: $\sin 5x\cos 3x=\frac{1}{2}[\sin 8x+\sin 2x]$. Term 2: $\cos 5x\sin 3x=\frac{1}{2}[\sin 8x-\sin 2x]$ (cos·sin uses the minus). Subtract: $\frac{1}{2}[\sin 8x+\sin 2x]-\frac{1}{2}[\sin 8x-\sin 2x]=\frac{1}{2}\cdot 2\sin 2x=\sin 2x.$ $\boxed{\sin 2x}$ Interpret: this is the sine difference formula $\sin (A-B)=\sin A\cos B-\cos A\sin B$ with $A=5x,B=3x$, and $5x-3x=2x$. ✓

Set $A=2\pi \cdot 442t$ and $B=2\pi \cdot 2t$. Apply cos·cos: $2\cos A\cos B=\cos (A-B)+\cos (A+B).$ Compute the angles: $A+B=2\pi \cdot 444t,A-B=2\pi (442-2)t=2\pi \cdot 440t.$ Therefore $2\cos (2\pi \cdot 442t)\cos (2\pi \cdot 2t)=\cos (2\pi \cdot 440t)+\cos (2\pi \cdot 444t).$ $\boxed{\text{Recovered frequencies: }440\text{Hz and }444\text{Hz}}$ WHY the choice of which factor is $A$ matters: by naming the larger angle as $A$, the difference $A-B$ comes out positive ($440$ Hz) directly. Had we swapped, we'd get $\cos (-2\pi \cdot 440t)$; since cosine is even, $\cos (-\varphi )=\cos \varphi$, that is the same $440$ Hz tone — a negative frequency is never physically real, it's just a bookkeeping sign the evenness erases.

Step 1 — combine the last two factors (sin·sin), with $A=\frac{\pi }{3}+x,B=\frac{\pi }{3}-x$: $\sin \left(\frac{\pi }{3}-x\right)\sin \left(\frac{\pi }{3}+x\right)=\frac{1}{2}[\cos (A-B)-\cos (A+B)].$ Here $A-B=2x$ and $A+B=\frac{2\pi }{3}$, so $=\frac{1}{2}\left[\cos 2x-\cos \frac{2\pi }{3}\right]=\frac{1}{2}\left[\cos 2x+\frac{1}{2}\right],$ using $\cos \frac{2\pi }{3}=-\frac{1}{2}$. Step 2 — multiply by $4\sin x$: $4\sin x\cdot \frac{1}{2}\left[\cos 2x+\frac{1}{2}\right]=2\sin x\cos 2x+\sin x.$ Step 3 — expand $2\sin x\cos 2x$ (sin·cos), $A=x,B=2x$: $2\sin x\cos 2x=\sin (x+2x)+\sin (x-2x)=\sin 3x+\sin (-x)=\sin 3x-\sin x.$ Step 4 — add the leftover $\sin x$: $(\sin 3x-\sin x)+\sin x=\sin 3x.■$ $\boxed{4\sin x\sin \left(\frac{\pi }{3}-x\right)\sin \left(\frac{\pi }{3}+x\right)=\sin 3x}$ WHY the $\sin (-x)$ step is crucial: $\sin$ is odd, so $\sin (-x)=-\sin x$; this minus is what lets the stray $\sin x$ terms cancel perfectly.

Step 1: cos·cos, $A=2x,B=4x$: $\cos 2x\cos 4x=\frac{1}{2}[\cos (2x-4x)+\cos (2x+4x)]=\frac{1}{2}[\cos (-2x)+\cos 6x]=\frac{1}{2}[\cos 2x+\cos 6x],$ using $\cos (-2x)=\cos 2x$ (cosine is even). Step 2 — integrate over $[0,\pi ]$: ${\int }_0^{\pi }\frac{1}{2}\cos 2xdx+{\int }_0^{\pi }\frac{1}{2}\cos 6xdx=\frac{1}{2}{\left[\frac{\sin 2x}{2}\right]}_0^{\pi }+\frac{1}{2}{\left[\frac{\sin 6x}{6}\right]}_0^{\pi }.$ WHY both boundary evaluations vanish: each term is $\frac{\sin k\pi -\sin 0}{2k}$. For any integer $k$, $\sin (k\pi )=0$ (sine is zero at every integer multiple of $\pi$) and $\sin 0=0$, so both endpoints contribute nothing. Here $k=1$ and $k=3$ are integers, so $\sin 2\pi =\sin 6\pi =\sin 0=0$: $=\frac{1}{2}\cdot \frac{0-0}{2}+\frac{1}{2}\cdot \frac{0-0}{6}=0.$ $\boxed{{\int }_0^{\pi }\cos 2x\cos 4xdx=0}$ Interpretation: distinct cosine harmonics are orthogonal on a period — their integral of the product is zero. This is the engine of Fourier Series: it lets us extract one frequency's coefficient without interference from the others.