Zadanie w załączniku.
sinα=3/4
sin²α+cos²α=1
(3/4)²+cos²α=1
9/16+cos²α=1
cosα=1-9/16
cos²α=7/16
zatem 2-cos²α=2-7/16=32/16-7/16=25/16
odp: A)25/16
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sinα=3/4
sin²α+cos²α=1
(3/4)²+cos²α=1
9/16+cos²α=1
cosα=1-9/16
cos²α=7/16
zatem 2-cos²α=2-7/16=32/16-7/16=25/16
odp: A)25/16