Question

представьте в виде дроби:
1) sin² 80° - sin² 40°
2)cos² 10° - cos² 50°

Answer 1

Ответ:

$1)\ \ sin^280^\circ -sin^240^\circ =(sin80^\circ -sin40^\circ )(sin80^\circ -sin40^\circ )=\\\\=2sin\, 20^\circ \cdot cos60^\circ =sin\, 20^\circ =\sqrt{\dfrac{1-cos40^\circ }{2}}=\dfrac{\sqrt{2\, (1-cos40^\circ )}}{2}\\\\\\2)\ \ cos^210^\circ -cos^250^\circ =(cos10^\circ -cos50^\circ )(cos10^\circ +cos50^\circ )=\\\\=-2sin30^\circ \cdot sin\, 20^\circ =sin\, 20^\circ =\sqrt{\dfrac{1-cos40^\circ }{2}}=\dfrac{\sqrt{2\, (1-cos40^\circ )}}{2}$