Hi I have to make the bottom one from the top two and have simply forgot how to get there 2sinAcosA/cos^2A-Sin^2A or alternatives of cos2A..but I am losing my way!
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$\displaystyle \tan 2A=\frac{\sin 2A}{\cos 2A}=\frac{2\sin A\cos A}{\cos^2A-\sin^2A}=$ $\displaystyle =\frac{\dfrac{2\sin A\cos A}{\cos^2A}}{\dfrac{\cos^2A-\sin^2A}{\cos^2A}}=\frac{2\tan A}{1-\tan^2A}$
I knew I would kick myself Many Thanks
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