ตรีโกณค่ะ

[KnuckleS]

ถ้า $3\tan { A } =\tan { \left( A+B \right) } $ จงหาค่าของ $\frac { \sin { \left( 2A+B \right) } }{ \sin { B } } $

[กิตติ]

$3\tan { A } =\tan { \left( A+B \right) } $
วิธีแรก
$3\tan { A } =\frac{\tan A+\tan B}{1-\tan A \tan B} $
$3\tan { A }-3\tan^2 { A }\tan B =\tan A+\tan B$
$2 \tan A=\tan B(1+3\tan^2 { A })$
$\frac{2 \tan A}{\tan B}=1+3\tan^2 { A } $
$\frac { \sin { \left( 2A+B \right) } }{ \sin { B } }=\frac{\sin 2A \cos B+\sin B \cos 2A }{ \sin { B } } $
$=\frac{\sin 2A}{\tan B} +\cos 2A$
$=\left(\,\frac{2\tan A}{(1+\tan^2 A)(\tan B)} \right)+\frac{1-\tan^2A}{1+\tan^2A} $
$=\frac{1+3\tan^2A}{1+\tan^2A}+\frac{1-\tan^2A}{1+\tan^2A}$
$=\frac{2+2\tan^2A}{1+\tan^2A}$
$=2$


วิธีที่สอง
$3\frac{\sin A}{\cos A} =\frac{\sin (A+B)}{\cos (A+B)} $
$3\sin A \cos (A+B)=\sin (A+B)\cos A$
$2\sin A \cos (A+B)=\sin (A+B)\cos A-\sin A \cos (A+B)$
$2\sin A \cos (A+B)=\sin B$
$\sin (2A+B)+\sin(-B)=\sin B$
$\sin (2A+B)=2\sin B$
$\frac{\sin (2A+B)}{\sin B} =2$

[KnuckleS]

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