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shubham.bagrecha
 3 years ago
Prove that:
cos(A+B)+sin(AB)=2sin(45+A)cos(45+B)
shubham.bagrecha
 3 years ago
Prove that: cos(A+B)+sin(AB)=2sin(45+A)cos(45+B)

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shubham.bagrecha
 3 years ago
Best ResponseYou've already chosen the best response.0\[ \cos(A+B)+\sin(AB)=2\sin(45+A)\cos(45+B)\]

Neo92
 3 years ago
Best ResponseYou've already chosen the best response.1first write cos(A+B) as sin[90(A+B)] then use sinC + sinD = 2sin[(C+D)/2]cos[(CD)/2] try and tell me what u get

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0ok ... for this first of all tell me what is cos(A+B)

shubham.bagrecha
 3 years ago
Best ResponseYou've already chosen the best response.0@Neo92 ok i will solve

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0cos(A+B)=sin(90AB) sin(90AB) + sin(AB) = 2sin(45+A)cos(45+B) we know that sin(x) + sin(y)=2sin([x+y]/2)cos([xy]/2)

mathslover
 3 years ago
Best ResponseYou've already chosen the best response.0so we get that sin(90AB) + sin(AB)=2sin(45B)cos(45A) but we know that sin(45B)=cos(90(45b))=cos(45+B) and cos(45A)=sin(90(45A))=sin(45+A) hence we get that cos(A+B) + sin(AB) = 2sin(45+A)cos(45+B)

klimenkov
 3 years ago
Best ResponseYou've already chosen the best response.2\[\cos(A+B)+\sin(AB)=\cos A\cos B\sin A\sin B+\sin A \cos B\sin B\cos A=\] \[=\cos A(\cos B\sin B)+\sin A (\cos B\sin B)=(\cos A+\sin A)(\cos B\sin B)\] Now use: \(\cos A+\sin A=\sqrt2\cdot(\frac1{\sqrt2}\cos A+\frac1{\sqrt2}\sin A)=\sqrt2\sin(45^\circ +A)\) and \(\cos B\sin B=\sqrt2\cdot(\frac1{\sqrt2}\cos B\frac1{\sqrt2}\sin B)=\sqrt2\cos(45^\circ +B)\). Now multiply this: \[(\cos A+\sin A)(\cos B\sin B)=\sqrt2\sin(45^\circ +A)\cdot \sqrt2\cos(45^\circ +B)=2\sin(45^\circ +A)\cdot\]\[\cdot\cos(45^\circ +B)\]
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