Question

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1. sen (u+v) x sen (u-v) = sen 2 u - sen 2 v

2. cos (u+v) x cos (u-v) = cos 2 u - sen 2 v

Answer 1

$A=\sin(u-v)\times \sin(u-v)$

$\cos(x+y)-\cos(x-y)=-2\sin x\cos y \\ \\ \boxed{\sin x\cos y = \frac{1}{2}[\cos(x-y)-\cos(x+y)]} \\ \\ \\ \text{Entonces}\\ \sin(u+v)\sin(u-v)=\frac{1}{2}[\cos(2v)-\cos(2u)] \\ \\ \sin(u+v)\sin(u-v)=\frac{1}{2}[(1-2\sin^2v)-(1-2\sin^2u)] \\ \\ \sin(u+v)\sin(u-v)=\frac{1}{2}(2\sin^2u-2\sin^2v)\\ \\ \boxed{\sin(u+v)\sin(u-v)=\sin^2u-\sin^2v}$

2)

$\cos(x+y)+\cos(x-y)=2\cos x\cos y\\ \\ \boxed{\cos x\cos y = \frac{1}{2}[\cos(x+y)+\cos(x-y)]} \\ \\ \\ \text{Entonces}\\ \cos(u+v)\cos(u-v)=\frac{1}{2}[\cos(2u)+\cos(2v)]\\ \\ \cos(u+v)\cos(u-v)=\frac{1}{2}[(2\cos^2u-1)+(1-2\sin^2v)]\\ \\ \cos(u+v)\cos(u-v)=\frac{1}{2}(2\cos^2u-2\sin^2v)\\ \\ \boxed{\cos(u+v)\cos(u-v)=\cos^2u-\sin^2v}$