Wiedząc, że liczba sin α + cos α = √5/2 oblicz liczbę b= (sin α - cos α)²
sin α + cos α = √5/2/², podnoszę obustronnie do potęgi 2
sin²+2sinacosa+cos²a=5/4, poniewaz sin^2+cos^2=1
1+2sinacosa=5/4
2sinacosa=5/4-1=1/4
b= (sin α - coas α)²=sin²a-2sinacosa+cos²a=1-1/4=3/4
więc
b=1-3/2=-1/2
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sin α + cos α = √5/2/², podnoszę obustronnie do potęgi 2
sin²+2sinacosa+cos²a=5/4, poniewaz sin^2+cos^2=1
1+2sinacosa=5/4
2sinacosa=5/4-1=1/4
b= (sin α - coas α)²=sin²a-2sinacosa+cos²a=1-1/4=3/4
więc![2 sin\alpha cos \alpha=5/2 -1= 3/2 2 sin\alpha cos \alpha=5/2 -1= 3/2](https://tex.z-dn.net/?f=2+sin%5Calpha+cos+%5Calpha%3D5%2F2+-1%3D+3%2F2)
b=1-3/2=-1/2