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# Angle A is acute. If If $\tan A=\frac{1}{2}$, find cos 2 A and tan 2 A.

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Using formulas:$\\\\$$\sin 2A=2\sin A\cos A$\\$\cos 2A=1-2\sin^2A= =1-2\cdot ($\dfrac{5}{13}$)^2= =1-2\cdot $\dfrac{25}{169}$= =$\dfrac{119}{169}\\\\\\ Now try to find$\cos A$so we can finish the work for$\sin 2A$:\\\\$\cos 2A=2\cos^2A-1 $\dfrac{119}{169}$=2\cos^2A-1 \cos^2A=$\dfrac{144}{169}$ \cos A=\pm$\sqrt{\dfrac{144}{169}}$=\pm$\dfrac{12}{13}$ $\\\\ Since$\angle A$is acute,$\cos A=+$\dfrac{12}{13}.\\\\\\$\sin 2A=2\cdot $\dfrac{5}{13}$\cdot $\dfrac{12}{13}$=$$\dfrac{120}{169} Double-Angle and Half-Angle Formulas:$\\\\$ 1)$\sin 2\alpha=2\sin \alpha\cos \alpha$\\ 2)\cos 2\alpha=\cos^2\alpha-\sin^2\alpha= =1-2\sin^2\alpha=2\cos^2\alpha-1 $\\\\ 3)$\tan 2\alpha=$\dfrac{2\tan \alpha}{1-\tan^2\alpha}\\\\ 4)$\sin$\dfrac{\alpha}{2}$=\pm$\sqrt{\dfrac{1-\cos\alpha}{2}}\\\\ 5)$\cos$\dfrac{\alpha}{2}$=\pm$\sqrt{\dfrac{1+\cos\alpha}{2}}\\\\ 6)$\tan$\dfrac{\alpha}{2}$=\pm$\sqrt{\dfrac{1-\cos\alpha}{1+\cos\alpha}}$=$\dfrac{\sin\alpha}{1+\cos\alpha}$=$$\dfrac{1-\cos\alpha}{\sin\alpha}$

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