Last updated at April 13, 2021 by Teachoo
Transcript
Misc 20 Using the fact that sinβ‘(π΄ + π΅)=sinβ‘π΄ cosβ‘π΅+cosβ‘π΄ sinβ‘π΅ and the differentiation, obtain the sum formula for cosines.Given sinβ‘(π΄ + π΅)=sinβ‘π΄ cosβ‘π΅+cosβ‘π΄ sinβ‘π΅ Consider A & B are function of π₯ Differentiating both side π€.π.π‘.π₯. π(sinβ‘(π΄ + π΅) )/ππ₯ = π(sinβ‘π΄ cosβ‘π΅ + cosβ‘π΄ sinβ‘π΅)/ππ₯ π(sinβ‘(π΄ + π΅) )/ππ₯ = π(sinβ‘π΄ . cosβ‘π΅)/ππ₯ + π(cosβ‘γπ΄ γ. sinβ‘π΅)/ππ₯ cos (π΄+π΅) . π(π΄ + π΅)/ππ₯ = π(sinβ‘π΄ . cosβ‘π΅)/ππ₯ + π(cosβ‘γπ΄ γ. sinβ‘π΅)/ππ₯ Using Product rule As (π’π£)β = π’βπ£ + π£βπ’ πππ (π¨+π©) . (π π¨/π π + π π©/π π) = (π(sinβ‘π΄ )/ππ₯. cosβ‘π΅" +" π(cosβ‘π΅ )/ππ₯ " " π ππβ‘"A" ) + (π(cosβ‘π΄ )/ππ₯. π ππβ‘π΅" +" π(sinβ‘π΅ )/ππ₯ ". " π"os A" ) = cosβ‘π΄.ππ΄/ππ₯ ". cos B "βsinβ‘π΅.ππ΅/ππ₯ " " sinβ‘π΄ β sinβ‘π΄. ππ΄/ππ₯.sinβ‘π΅+cosβ‘π΅. ππ΅/ππ₯ ". " π"os A" = cosβ‘π΄.ππ΄/ππ₯ ". cos B "βsinβ‘π΄ .ππ΄/ππ₯ " " π ππβ‘"B" β sinβ‘π΅. ππ΅/ππ₯. π ππ π΄β‘"+ cos B" . ππ΅/ππ₯ ". " π"os A" = ππ΄/ππ₯ (cosβ‘π΄ cosβ‘π΅βsinβ‘π΄ sinβ‘π΅ ) + ππ΅/ππ₯ (βsinβ‘π΅ sinβ‘π΄+cosβ‘π΅ cosβ‘π΄ ) (cosβ‘π΄ cosβ‘π΅βsinβ‘π΄ sinβ‘π΅ ) (ππ΄/ππ₯ + ππ΅/ππ₯) Thus, cos (π΄+π΅) . (ππ΄/ππ₯ + ππ΅/ππ₯) = (cosβ‘π΄ cosβ‘π΅βsinβ‘π΄ sinβ‘π΅ ) (ππ΄/ππ₯ + ππ΅/ππ₯) πππ" " (π¨+π©) = πππβ‘π¨ πππβ‘π©βπππβ‘π¨ πππβ‘π© Hence proved
Proofs
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