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Last updated at March 22, 2023 by Teachoo
Misc 20 Using the fact that sin(𝐴 + 𝐵)=sin𝐴 cos𝐵+cos𝐴 sin𝐵 and the differentiation, obtain the sum formula for cosines.Given sin(𝐴 + 𝐵)=sin𝐴 cos𝐵+cos𝐴 sin𝐵 Consider A & B are function of 𝑥 Differentiating both side 𝑤.𝑟.𝑡.𝑥. 𝑑(sin(𝐴 + 𝐵) )/𝑑𝑥 = 𝑑(sin𝐴 cos𝐵 + cos𝐴 sin𝐵)/𝑑𝑥 𝑑(sin(𝐴 + 𝐵) )/𝑑𝑥 = 𝑑(sin𝐴 . cos𝐵)/𝑑𝑥 + 𝑑(cos〖𝐴 〗. sin𝐵)/𝑑𝑥 cos (𝐴+𝐵) . 𝑑(𝐴 + 𝐵)/𝑑𝑥 = 𝑑(sin𝐴 . cos𝐵)/𝑑𝑥 + 𝑑(cos〖𝐴 〗. sin𝐵)/𝑑𝑥 Using Product rule As (𝑢𝑣)’ = 𝑢’𝑣 + 𝑣’𝑢 𝒄𝒐𝒔 (𝑨+𝑩) . (𝒅𝑨/𝒅𝒙 + 𝒅𝑩/𝒅𝒙) = (𝑑(sin𝐴 )/𝑑𝑥. cos𝐵" +" 𝑑(cos𝐵 )/𝑑𝑥 " " 𝑠𝑖𝑛"A" ) + (𝑑(cos𝐴 )/𝑑𝑥. 𝑠𝑖𝑛𝐵" +" 𝑑(sin𝐵 )/𝑑𝑥 ". " 𝑐"os A" ) = cos𝐴.𝑑𝐴/𝑑𝑥 ". cos B "−sin𝐵.𝑑𝐵/𝑑𝑥 " " sin𝐴 − sin𝐴. 𝑑𝐴/𝑑𝑥.sin𝐵+cos𝐵. 𝑑𝐵/𝑑𝑥 ". " 𝑐"os A" = cos𝐴.𝑑𝐴/𝑑𝑥 ". cos B "−sin𝐴 .𝑑𝐴/𝑑𝑥 " " 𝑠𝑖𝑛"B" − sin𝐵. 𝑑𝐵/𝑑𝑥. 𝑠𝑖𝑛 𝐴"+ cos B" . 𝑑𝐵/𝑑𝑥 ". " 𝑐"os A" = 𝑑𝐴/𝑑𝑥 (cos𝐴 cos𝐵−sin𝐴 sin𝐵 ) + 𝑑𝐵/𝑑𝑥 (−sin𝐵 sin𝐴+cos𝐵 cos𝐴 ) (cos𝐴 cos𝐵−sin𝐴 sin𝐵 ) (𝑑𝐴/𝑑𝑥 + 𝑑𝐵/𝑑𝑥) Thus, cos (𝐴+𝐵) . (𝑑𝐴/𝑑𝑥 + 𝑑𝐵/𝑑𝑥) = (cos𝐴 cos𝐵−sin𝐴 sin𝐵 ) (𝑑𝐴/𝑑𝑥 + 𝑑𝐵/𝑑𝑥) 𝒄𝒐𝒔" " (𝑨+𝑩) = 𝒄𝒐𝒔𝑨 𝒄𝒐𝒔𝑩−𝒔𝒊𝒏𝑨 𝒔𝒊𝒏𝑩 Hence proved