1-sinα = 2sin² (45° - α/2) Требуется доказать тождество.

Воспользуемся формулами вычисления синуса суммы или разности углов:

sin (A ± B) = sin A ⋅ cos B ± cos A ⋅ sin B.

Получим:

2 ⋅ sin² (45° - α/2) = 2 ⋅ (sin 45° ⋅ cos α/2 - cos 45° ⋅ sin α/2) ² =

= 2 ⋅ (√2/2 ⋅ cos α/2 - √2/2 ⋅ sin α/2) ² = 2 ⋅ (√2/2) ² ⋅ (cos α/2 - sin α/2) ² =

= 2 ⋅ (2/4) ⋅ (cos α/2 - sin α/2) ² = (cos α/2 - sin α/2) ² = cos² α/2 - 2 ⋅ cos α/2 ⋅ sin α/2 + sin² α/2 =

= cos² α/2 + sin² α/2 - (cos α/2 ⋅ sin α/2 + sin α/2 ⋅ cos α/2) =

= 1 - sin (α/2 + α/2) = 1 - sin α.

Что и требовалось доказать.