Question

Помогите пожалуйста с Алгеброй.
(1-tg^2t)cos^2t
Докажите тождества
1-sinα=2sin^2(45°-α/2)

sin^2(3π/4+2t)=1-sin4t/2
p.s / знак дроби

Answer 1

(1-tg²t)·cos²t=cos²t-sin²tcos²t/cos²t=cos²t-sin²t=cos2t;
1-sinα=2sin²(45⁰-α/2);
2sin²(45-α/2)=2(sin45⁰·cosα/2-cos45⁰·sinα/2)²=
=2·(√2/2·cosα/2-√2/2·sinα/2)²=2·2/4(cosα/2-sinα/2)²=
=(cosα/2-sinα/2)²=cos²α/2-2cosα/2·sinα/2+sin²α/2=1-sinα;
sin²(3π/4+2t)=1-sin4t/2;
sin²(3π/4+2t)=(sin(3π/4)·cos2t+cos(3π/4)·sin2t)²=
=(sin(π-π/4)·cos2t+cos(π-π/4)·sin2t)²=
=(sinπ/4·cos2t+(-cosπ/4)·sin2t)²=
=(√2/2cos2t-√2/2sin2t)²=2/4(cos²2t-2sin2t·cos2t+sin²2t)=
=1/2(1-sin4t);

Answer 2

Спасибо

Answer 3

Удачи!