Question

prove that
sinB/sinA=sin(2A+B)/sinA-2cos(A+B)

Answer 1

Answer on Question #42293-Math-Trigonometry

Prove that

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

Solution

$$\sin (2 \mathrm {A} + \mathrm {B}) = \sin (\mathrm {A} + (\mathrm {A} + \mathrm {B})) = \sin (\mathrm {A}) \cos (\mathrm {A} + \mathrm {B}) + \sin (\mathrm {A} + \mathrm {B}) \cos (\mathrm {A}).$$ $$\begin{array}{l} \frac {\sin (2 \mathrm {A} + \mathrm {B})}{\sin \mathrm {A}} - 2 \cos (\mathrm {A} + \mathrm {B}) = \frac {\sin (\mathrm {A}) \cos (\mathrm {A} + \mathrm {B}) + \sin (\mathrm {A} + \mathrm {B}) \cos (\mathrm {A})}{\sin \mathrm {A}} - 2 \cos (\mathrm {A} + \mathrm {B}) \\ = \frac {\sin (\mathrm {A}) \cos (\mathrm {A} + \mathrm {B})}{\sin \mathrm {A}} + \frac {\sin (\mathrm {A} + \mathrm {B}) \cos (\mathrm {A})}{\sin \mathrm {A}} - 2 \cos (\mathrm {A} + \mathrm {B}) \\ = \cos (\mathrm {A} + \mathrm {B}) + \frac {\sin (\mathrm {A} + \mathrm {B}) \cos (\mathrm {A})}{\sin \mathrm {A}} - 2 \cos (\mathrm {A} + \mathrm {B}) = \frac {\sin (\mathrm {A} + \mathrm {B}) \cos (\mathrm {A})}{\sin \mathrm {A}} - \cos (\mathrm {A} + \mathrm {B}) \\ = \frac {\sin (\mathrm {A} + \mathrm {B}) \cos (\mathrm {A}) - \sin \mathrm {A} \cos (\mathrm {A} + \mathrm {B})}{\sin \mathrm {A}} = \frac {\sin \left((\mathrm {A} + \mathrm {B}) - A\right)}{\sin \mathrm {A}} = \frac {\sin \mathrm {B}}{\sin \mathrm {A}}. \\ \end{array}$$

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