W załączniku. Proszę o całe rozwiązanie z wyjaśnieniem
(sinα-cosα)²=(1/2)²
sin²α-2sinαcosα+cos²α=1/4
sin²α+cos²α=1
1-2sinαcosα=1/4
-2sinαcosα=1/4-1 /:(-2)
sinαcosα=-3/4:(-2)= 3/8
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(sinα-cosα)²=(1/2)²
sin²α-2sinαcosα+cos²α=1/4
sin²α+cos²α=1
1-2sinαcosα=1/4
-2sinαcosα=1/4-1 /:(-2)
sinαcosα=-3/4:(-2)= 3/8