i need to prove sin(u+v)sin(u-v)=sin^2u-sin^2v

Let's use trigonometric identity: \[\sin(u \pm v) = \sin(u)\cos(v) \pm \cos(u)\sin(v)\] Rewrite original expression to \[(sin(u)cos(v) + cos(u)sin(v))(sin(u)cos(v) - cos(u)sin(v))\] Now, the key thing is that it is a product of sum and difference which is known from \[(a+b)(a-b) = a^2 - b^2\] So, the expression above will be just \[(\sin(u)\cos(v))^2 - (\cos(u)\sin(v))^2\] Now, we can rewrite this as \[{1 \over 4}[(1-\cos(2u))(1 + \cos(2v)) - (1-\cos(2v))(1+\cos(2u))]\] Then, rewrite this as \[{1 \over 4}[(1 - \cos^2 u + \sin^2u)(1+\cos^2v - \sin^2v) - (1-\cos^2v+\sin^2v)(1+\cos^2u - \sin^2u)]\] (as you can see, there is pretty much of work to do with this). Now, we rewrite as: \[{1 \over 4}[(2\sin^2u)(2 - 2\sin^2v) - (2\sin^2v)(2 - 2\sin^2u)]\] This in turn can be rewrited as \[{1 \over 4}[(4\sin^2u - 4(\sin^2(u)\sin^2(v))) - (4\sin^2v - 4(\sin^2(u)\sin^2(v)))]\] which equals to \[{1 \over 4}(4 \sin^2u - 4 \sin^2v)\] which finally is \[sin^2u - sin^2v\]

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