Prove that sin2A-Bsin A- 2cos A-B- sin BsinA

[ sin #160; ( 2A+B ) #160; sin #160; A 2cos ...

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Integration Using Substitution

Prove that s...

Question

Prove that
$\frac{s i n (2 A + B)}{s i n A} - 2 c o s (A + B) = \frac{s i n B}{s i n A}$

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

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

\frac{s i n B}{s i n A} = \frac{s i n (2 A + B)}{s i n A} - 2 c o s (A + B)

\frac{\sin (A + B) + \sin (A - B)}{\cos (A + B) + \cos (A - B)} = \tan A

\frac{\sin (A - B)}{\cos A \cos B} + \frac{\sin (B - C)}{\cos B \cos C} + \frac{\sin (C - A)}{\cos C \cos A} = 0

\frac{\sin (A - B)}{\sin A \sin B} + \frac{\sin (B - C)}{\sin B \sin C} + \frac{\sin (C - A)}{\sin C \sin A} = 0

\frac{\tan (A + B)}{\cot (A - B)} = \frac{\tan^{2} A - \tan^{2} B}{1 - \tan^{2} A \tan^{2} B}

sin (A + B) \neq sin A + sin B

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