Product-to-sum formulas

3.1.15 · Maths › Advanced Trigonometry

WHAT are we trying to do?

HOW to DERIVE them (scratch se — kabhi blind memorise mat karo)

Angle addition formulas se shuru karo, jo hamare ek-maatra assumed tools hain:

\cos(A+B)=\cos A\cos B-\sin A\sin B \tag{1} \cos(A-B)=\cos A\cos B+\sin A\sin B \tag{2} \sin(A+B)=\sin A\cos B+\cos A\sin B \tag{3} \sin(A-B)=\sin A\cos B-\cos A\sin B \tag{4}

Deriving $\cos A\cos B$:

• Why? $\cos A\cos B$ term dono (1) aur (2) mein $+$ sign ke saath aata hai, jabki $\sin A\sin B$ opposite signs ke saath aata hai. Toh sine part cancel karne ke liye inhe add karo. $(1)+(2):\cos (A+B)+\cos (A-B)=2\cos A\cos B$
• Why divide by 2? Product ko cleanly isolate karne ke liye. $\boxed{\cos A\cos B=\frac{1}{2}[\cos (A-B)+\cos (A+B)]}$

Deriving $\sin A\sin B$:

• Why subtract? $\sin A\sin B$ ko rakhne aur $\cos A\cos B$ ko cancel karne ke liye, $(2)-(1)$ use karo: $\cos (A-B)-\cos (A+B)=2\sin A\sin B$ $\boxed{\sin A\sin B=\frac{1}{2}[\cos (A-B)-\cos (A+B)]}$

Deriving $\sin A\cos B$:

• Why add (3) and (4)? Dono mein $\sin A\cos B$ $+$ ke saath hai; $\cos A\sin B$ parts cancel ho jaate hain. $(3)+(4):\sin (A+B)+\sin (A-B)=2\sin A\cos B$ $\boxed{\sin A\cos B=\frac{1}{2}[\sin (A+B)+\sin (A-B)]}$

Deriving $\cos A\sin B$:

• Why subtract? $(3)-(4)$ se $\cos A\sin B$ bachta hai, $\sin A\cos B$ cancel ho jaata hai: $\cos A\sin B=\frac{1}{2}[\sin (A+B)-\sin (A-B)]$

Worked examples

Common mistakes (Steel-manned)

Flashcards

Connections

• Angle Addition Formulas — parent identities jisse hum sab kuch derive karte hain.
• Sum-to-Product Formulas — bilkul inverse operation.
• Integration of Trig Functions — main application.
• Fourier Series — $\sin /\cos$ ki orthogonality inhi se prove hoti hai.
• Wave Interference and Beats — frequencies un-mix karne ka physical meaning.
• Double Angle Formulas — special case jab $A=B$ ho.

Concept Map