Intuition The big pictureWhen you add two sine or cosine waves that have different frequencies, the result looks like a single wave whose amplitude slowly wobbles . Sum-to-product formulas turn a sum/difference of trig functions into a product of trig functions. This is exactly what your ear hears as beats , and what factoring needs to solve equations like sin 3 x + sin x = 0 \sin 3x + \sin x = 0 sin 3 x + sin x = 0
WHAT: Rewrite sin A ± sin B \sin A \pm \sin B sin A ± sin B cos A ± cos B \cos A \pm \cos B cos A ± cos B WHY: Products are easy to set to zero (solve equations), and reveal a slow "envelope" × \times × HOW: Run the product-to-sum identities backwards using a clever substitution.
We start from the angle addition formulas , which we take as known:
sin ( x + y ) = sin x cos y + cos x sin y \sin(x+y) = \sin x\cos y + \cos x\sin y sin ( x + y ) = sin x cos y + cos x sin y sin ( x − y ) = sin x cos y − cos x sin y \sin(x-y) = \sin x\cos y - \cos x\sin y sin ( x − y ) = sin x cos y − cos x sin y
Add them — the cos x sin y \cos x \sin y cos x sin y
\sin(x+y) + \sin(x-y) = 2\sin x\cos y \tag{1}
Subtract them — the sin x cos y \sin x\cos y sin x cos y
\sin(x+y) - \sin(x-y) = 2\cos x\sin y \tag{2}
Now the same for cosine:
cos ( x + y ) = cos x cos y − sin x sin y \cos(x+y) = \cos x\cos y - \sin x\sin y cos ( x + y ) = cos x cos y − sin x sin y cos ( x − y ) = cos x cos y + sin x sin y \cos(x-y) = \cos x\cos y + \sin x\sin y cos ( x − y ) = cos x cos y + sin x sin y
Add: cos ( x + y ) + cos ( x − y ) = 2 cos x cos y \cos(x+y)+\cos(x-y) = 2\cos x\cos y cos ( x + y ) + cos ( x − y ) = 2 cos x cos y
Subtract: cos ( x − y ) − cos ( x + y ) = 2 sin x sin y \cos(x-y)-\cos(x+y) = 2\sin x\sin y cos ( x − y ) − cos ( x + y ) = 2 sin x sin y
Intuition The key substitution — WHY it worksEquations (1)–(4) still contain x + y x+y x + y x − y x-y x − y rename them so the left side becomes a plain sum of two angles . Let
A = x + y , B = x − y . A = x+y, \qquad B = x-y. A = x + y , B = x − y . ⇒ x = A + B 2 \Rightarrow x = \dfrac{A+B}{2} ⇒ x = 2 A + B ⇒ y = A − B 2 \Rightarrow y = \dfrac{A-B}{2} ⇒ y = 2 A − B A , B A,B A , B A + B 2 \frac{A+B}{2} 2 A + B A − B 2 \frac{A-B}{2} 2 A − B
Applying the substitution to (1)–(4):
Definition Reading the pattern
Every RHS has a factor of = = 2 = = ==2== == 2 ==
The two arguments are always the half-sum A + B 2 \frac{A+B}{2} 2 A + B half-difference A − B 2 \frac{A-B}{2} 2 A − B
Sine sum → \to → sin ⋅ cos \sin\cdot\cos sin ⋅ cos sine difference → \to → cos ⋅ sin \cos\cdot\sin cos ⋅ sin Cosine stays "same type": sum → cos cos \to \cos\cos → cos cos → − sin sin \to \boxed{-}\sin\sin → − sin sin
Worked example 1) Factor and solvesin 3 x + sin x = 0 \sin 3x + \sin x = 0 sin 3 x + sin x = 0 A = 3 x A = 3x A = 3 x B = x B = x B = x
A + B 2 = 4 x 2 = 2 x \frac{A+B}{2} = \frac{4x}{2} = 2x 2 A + B = 2 4 x = 2 x Why? half-sum of the two angles.A − B 2 = 2 x 2 = x \frac{A-B}{2} = \frac{2x}{2} = x 2 A − B = 2 2 x = x Why? half-difference.
sin 3 x + sin x = 2 sin 2 x cos x = 0 \sin 3x + \sin x = 2\sin 2x\cos x = 0 sin 3 x + sin x = 2 sin 2 x cos x = 0 Why this helps: a product = 0 means one factor is 0. So sin 2 x = 0 \sin 2x = 0 sin 2 x = 0 cos x = 0 \cos x = 0 cos x = 0 sin 2 x = 0 ⇒ 2 x = n π ⇒ x = n π 2 \sin 2x = 0 \Rightarrow 2x = n\pi \Rightarrow x = \frac{n\pi}{2} sin 2 x = 0 ⇒ 2 x = nπ ⇒ x = 2 nπ cos x = 0 ⇒ x = π 2 + n π \cos x = 0 \Rightarrow x = \frac{\pi}{2}+n\pi cos x = 0 ⇒ x = 2 π + nπ Answer: x = n π 2 , n ∈ Z x = \dfrac{n\pi}{2},\ n\in\mathbb{Z} x = 2 nπ , n ∈ Z
Worked example 2) Exact value ofcos 75 ∘ + cos 15 ∘ \cos 75^\circ + \cos 15^\circ cos 7 5 ∘ + cos 1 5 ∘ A = 75 ∘ , B = 15 ∘ A=75^\circ,\ B=15^\circ A = 7 5 ∘ , B = 1 5 ∘ = 45 ∘ =45^\circ = 4 5 ∘ = 30 ∘ =30^\circ = 3 0 ∘ cos 75 ∘ + cos 15 ∘ = 2 cos 45 ∘ cos 30 ∘ = 2 ⋅ 2 2 ⋅ 3 2 = 6 2 \cos 75^\circ+\cos 15^\circ = 2\cos 45^\circ\cos 30^\circ = 2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2} = \frac{\sqrt6}{2} cos 7 5 ∘ + cos 1 5 ∘ = 2 cos 4 5 ∘ cos 3 0 ∘ = 2 ⋅ 2 2 ⋅ 2 3 = 2 6 Why this step: we picked cosine-sum → cos cos \to \cos\cos → cos cos 45 ∘ , 30 ∘ 45^\circ,30^\circ 4 5 ∘ , 3 0 ∘
Worked example 3) Beats — physical meaning ofsin ( 2 π f 1 t ) + sin ( 2 π f 2 t ) \sin(2\pi f_1 t) + \sin(2\pi f_2 t) sin ( 2 π f 1 t ) + sin ( 2 π f 2 t ) A = 2 π f 1 t A=2\pi f_1 t A = 2 π f 1 t B = 2 π f 2 t B=2\pi f_2 t B = 2 π f 2 t sin A + sin B = 2 cos ( 2 π f 1 − f 2 2 t ) ⏟ slow envelope sin ( 2 π f 1 + f 2 2 t ) ⏟ fast carrier \sin A+\sin B = 2\underbrace{\cos\!\big(2\pi\tfrac{f_1-f_2}{2}t\big)}_{\text{slow envelope}}\ \underbrace{\sin\!\big(2\pi\tfrac{f_1+f_2}{2}t\big)}_{\text{fast carrier}} sin A + sin B = 2 slow envelope cos ( 2 π 2 f 1 − f 2 t ) fast carrier sin ( 2 π 2 f 1 + f 2 t ) Why it matters: the ear hears loudness rise and fall at the beat frequency ∣ f 1 − f 2 ∣ |f_1-f_2| ∣ f 1 − f 2 ∣ explains the physics .
sin A + sin B cos A + cos B = tan A + B 2 \dfrac{\sin A + \sin B}{\cos A + \cos B} = \tan\dfrac{A+B}{2} cos A + cos B sin A + sin B = tan 2 A + B 2 sin A + B 2 cos A − B 2 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} 2 sin 2 A + B cos 2 A − B 2 cos A + B 2 cos A − B 2 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} 2 cos 2 A + B cos 2 A − B 2 2 2 cos A − B 2 \cos\frac{A-B}{2} cos 2 A − B cancel , leaving sin A + B 2 cos A + B 2 = tan A + B 2 \frac{\sin\frac{A+B}{2}}{\cos\frac{A+B}{2}} = \tan\frac{A+B}{2} c o s 2 A + B s i n 2 A + B = tan 2 A + B Why: factoring exposed a common factor invisible in the original form.
sin A + sin B = sin ( A + B ) \sin A + \sin B = \sin(A+B) sin A + sin B = sin ( A + B ) Why it feels right: addition on the outside looks like it should slide inside, and logs/exponents behave nicely with sums. Why it's wrong: sin \sin sin sin \sin sin Fix: sin A + sin B = 2 sin A + B 2 cos A − B 2 \sin A+\sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} sin A + sin B = 2 sin 2 A + B cos 2 A − B A = B = 90 ∘ A=B=90^\circ A = B = 9 0 ∘ = 2 =2 = 2 = 2 sin 90 ∘ cos 0 ∘ = 2 =2\sin90^\circ\cos0^\circ=2 = 2 sin 9 0 ∘ cos 0 ∘ = 2 sin 180 ∘ = 0 \sin180^\circ=0 sin 18 0 ∘ = 0
Common mistake Forgetting the minus sign incos A − cos B \cos A - \cos B cos A − cos B Why it feels right: all the other three formulas have a clean + 2 +2 + 2 Fix: cos A − cos B = − 2 sin A + B 2 sin A − B 2 \cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} cos A − cos B = − 2 sin 2 A + B sin 2 A − B decreasing , so a difference of cosines flips sign — that's the source of the minus.
Common mistake Swapping sin/cos inside the sine-difference formulaFix: For sin A − sin B \sin A - \sin B sin A − sin B outer factor is cos A + B 2 \cos\frac{A+B}{2} cos 2 A + B inner is sin A − B 2 \sin\frac{A-B}{2} sin 2 A − B swap compared to sin A + sin B \sin A + \sin B sin A + sin B A = 90 ∘ , B = 30 ∘ A=90^\circ,B=30^\circ A = 9 0 ∘ , B = 3 0 ∘
Recall Try before revealing
Derive cos A + cos B \cos A + \cos B cos A + cos B
What substitution converts (1)–(4) into product form?
Which formula carries a minus sign, and why physically?
Factor cos 5 x − cos x \cos 5x - \cos x cos 5 x − cos x
Answers: 1. Add cos ( x + y ) + cos ( x − y ) = 2 cos x cos y \cos(x+y)+\cos(x-y)=2\cos x\cos y cos ( x + y ) + cos ( x − y ) = 2 cos x cos y x = A + B 2 , y = A − B 2 x=\frac{A+B}{2}, y=\frac{A-B}{2} x = 2 A + B , y = 2 A − B A = x + y , B = x − y A=x+y,\ B=x-y A = x + y , B = x − y cos A − cos B = − 2 sin A + B 2 sin A − B 2 \cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2} cos A − cos B = − 2 sin 2 A + B sin 2 A − B − 2 sin 3 x sin 2 x -2\sin 3x\sin 2x − 2 sin 3 x sin 2 x
Recall Feynman: explain to a 12-year-oldImagine two friends clapping at almost the same speed. Sometimes their claps line up and it sounds LOUD; a moment later they're opposite and it sounds soft. So two steady clappers together make a slow "loud–soft–loud" pulse. Sum-to-product is the math that says: two simple wiggles added together = one fast wiggle whose loudness slowly breathes in and out. And when the total is exactly zero, that's just one of the two pieces being zero — which is why turning a sum into a multiplication makes solving so easy.
Mnemonic Remember the argument + signs"Half the SUM, half the DIFFERENCE, times 2."
Signs: S in-sum → \to → S in× \times × C os; S in-diff swaps to C os× \times × S in. C os-sum → \to → C os× \times × C os. C os-diff is the odd one out : − 2 sin sin -2\sin\sin − 2 sin sin D ifference gets the D own/minus).
What is sin A + sin B \sin A + \sin B sin A + sin B 2 sin A + B 2 cos A − B 2 2\sin\frac{A+B}{2}\cos\frac{A-B}{2} 2 sin 2 A + B cos 2 A − B What is sin A − sin B \sin A - \sin B sin A − sin B 2 cos A + B 2 sin A − B 2 2\cos\frac{A+B}{2}\sin\frac{A-B}{2} 2 cos 2 A + B sin 2 A − B What is cos A + cos B \cos A + \cos B cos A + cos B 2 cos A + B 2 cos A − B 2 2\cos\frac{A+B}{2}\cos\frac{A-B}{2} 2 cos 2 A + B cos 2 A − B What is cos A − cos B \cos A - \cos B cos A − cos B − 2 sin A + B 2 sin A − B 2 -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} − 2 sin 2 A + B sin 2 A − B Which sum-to-product formula has a minus sign? cos A − cos B \cos A - \cos B cos A − cos B , because cosine is decreasing
What substitution derives these from angle-addition? A = x + y , B = x − y ⇒ x = A + B 2 , y = A − B 2 A=x+y,\ B=x-y \Rightarrow x=\frac{A+B}{2}, y=\frac{A-B}{2} A = x + y , B = x − y ⇒ x = 2 A + B , y = 2 A − B Simplify sin A + sin B cos A + cos B \frac{\sin A+\sin B}{\cos A+\cos B} c o s A + c o s B s i n A + s i n B tan A + B 2 \tan\frac{A+B}{2} tan 2 A + B Factor sin 3 x + sin x \sin 3x + \sin x sin 3 x + sin x 2 sin 2 x cos x 2\sin 2x\cos x 2 sin 2 x cos x In beats, sin ( 2 π f 1 t ) + sin ( 2 π f 2 t ) \sin(2\pi f_1t)+\sin(2\pi f_2t) sin ( 2 π f 1 t ) + sin ( 2 π f 2 t ) ∣ f 1 − f 2 ∣ 2 \frac{|f_1-f_2|}{2} 2 ∣ f 1 − f 2 ∣ (heard as beat
∣ f 1 − f 2 ∣ |f_1-f_2| ∣ f 1 − f 2 ∣ )
Exact value of cos 75 ∘ + cos 15 ∘ \cos 75^\circ+\cos 15^\circ cos 7 5 ∘ + cos 1 5 ∘ sin sum-diff -> 2 sinx cosy
cos sum-diff -> 2 cosx cosy etc
Substitution A=x+y, B=x-y
Four sum-to-product identities
Pattern: factor 2, half-sum, half-diff
sine sum -> sin cos; diff swaps
cosine diff has minus sign
Beats: envelope x carrier
Intuition Hinglish mein samjhoDekho, sum-to-product formulas ka basic idea simple hai: jab tum do sine ya cosine ko add ya subtract karte ho, unhe ek multiplication (product) me badal do. Kyun? Kyunki agar equation product form me aa jaye, jaise 2 sin 2 x cos x = 0 2\sin 2x\cos x = 0 2 sin 2 x cos x = 0
Yeh formulas kahi se aasman se nahi aaye — inhe hum angle addition formulas se derive karte hain. sin ( x + y ) \sin(x+y) sin ( x + y ) sin ( x − y ) \sin(x-y) sin ( x − y ) 2 sin x cos y 2\sin x\cos y 2 sin x cos y A = x + y A=x+y A = x + y B = x − y B=x-y B = x − y x = A + B 2 x=\frac{A+B}{2} x = 2 A + B y = A − B 2 y=\frac{A-B}{2} y = 2 A − B half of sum, half of difference, aur aage 2 .
Physical meaning bhi mast hai — do thodi alag frequency ki sound waves jab milti hain, to beats sunai dete hain (loud-soft-loud-soft). Sum-to-product exactly yehi explain karta hai: fast "carrier" wave × \times ×
Do galtiyan avoid karo: (1) sin A + sin B \sin A+\sin B sin A + sin B sin ( A + B ) \sin(A+B) sin ( A + B ) cos A − cos B \cos A-\cos B cos A − cos B minus sign aata hai: − 2 sin A + B 2 sin A − B 2 -2\sin\frac{A+B}{2}\sin\frac{A-B}{2} − 2 sin 2 A + B sin 2 A − B + 2 +2 + 2
