Question

Tg9 - tg27 - tg63 +tg81 ,чему равно заранее

Answer 1

$tg9^{\circ} - tg27^{\circ} - tg63^{\circ} + tg81^{\circ} = \\\\ \left[ \ tg(90^{\circ} - x) = ctgx, \ ctg(90^{\circ} - x) = tgx \ \ \right]\\\\ = tg9^{\circ} -(tg27^{\circ} + tg(90^{\circ} - 27^{\circ})) + tg(90^{\circ} - 9^{\circ}) =\\\\ = tg9^{\circ} + ctg9^{\circ} - (tg27^{\circ} + ctg27^{\circ}) = \\\\ \left[ \ tga + ctga = \frac{\sin a}{\cos a} + \frac{\cos a}{\sin a} = \frac{1}{\cos a \sin a} = \frac{2}{\sin2a}\ \right]\\\\ = \frac{2}{\sin18^{\circ}} - \frac{2}{\sin54^{\circ}} = \\\\$

$= \frac{2(\sin54^{\circ} - \sin18^{\circ})}{\sin18^{\circ}\sin54^{\circ}} = \\\\ \left[ \ \sin a - \sin b = 2\sin\frac{a - b}{2}\cos \frac{a + b}{2} \ \right]\\\\ = \frac{2*2\sin18^{\circ}\cos36^{\circ}}{\sin18^{\circ}\sin54^{\circ}} = \frac{4\sin18^{\circ}\cos(90^{\circ} - 54^{\circ})}{\sin18^{\circ}\sin54^{\circ}} =\frac{4\sin18^{\circ}\sin54^{\circ}}{\sin18^{\circ}\sin54^{\circ}} = \boxed{4}$