Question

Simplify: sin(1 + cot2x)

Answer 1

Question #9394

Simplify: $\sin(1 + \mathrm{ctg}2x)$

Solution

$$\begin{array}{l} \sin (1 + \mathrm{ctg} 2 \mathrm{x}) = \sin 1 \cos (\mathrm{ctg} 2 \mathrm{x}) + \cos 1 \sin (\mathrm{ctg} 2 \mathrm{x}) = \sin 1 \cos \left(\frac {\cos 2 \mathrm{x}}{\sin 2 \mathrm{x}}\right) + \cos 1 \sin \left(\frac {\cos 2 \mathrm{x}}{\sin 2 \mathrm{x}}\right) = \\ = \sin 1 \cos \frac {1}{2} (\mathrm{ctg} x - \mathrm{tg} x) + \cos 1 \sin \frac {1}{2} (\mathrm{ctg} x - \mathrm{tg} x) = \sin 1 \cos \left(\frac {\mathrm{ctg} x}{2} - \frac {\mathrm{tg} x}{2}\right) + \cos 1 \sin \left(\frac {\mathrm{ctg} x}{2} - \frac {\mathrm{tg} x}{2}\right) = \\ = \sin 1 \left(\cos \frac {\mathrm{ctg} x}{2} \cos \frac {\mathrm{tg} x}{2} + \sin \frac {\mathrm{ctg} x}{2} \sin \frac {\mathrm{tg} x}{2}\right) + \cos 1 \left(\sin \frac {\mathrm{ctg} x}{2} \cos \frac {\mathrm{tg} x}{2} - \cos \frac {\mathrm{ctg} x}{2} \sin \frac {\mathrm{tg} x}{2}\right) = \\ = \sin 1 \cos \frac {\mathrm{ctg} x}{2} \cos \frac {\mathrm{tg} x}{2} + \sin 1 \sin \frac {\mathrm{ctg} x}{2} \sin \frac {\mathrm{tg} x}{2} + \cos 1 \sin \frac {\mathrm{ctg} x}{2} \cos \frac {\mathrm{tg} x}{2} - \cos 1 \cos \frac {\mathrm{ctg} x}{2} \sin \frac {\mathrm{tg} x}{2} = \\ = \cos \frac {\mathrm{tg} x}{2} \sin \left(1 + \frac {\mathrm{ctg} x}{2}\right) + \sin \frac {\mathrm{tg} x}{2} \cos \left(\frac {\mathrm{ctg} x}{2} - 1\right) \end{array}$$