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

Solution

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1.8 (5 ratings)

Answered 2 years ago

1.8 (5 ratings)

Answered 2 years ago

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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}$