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ponewaz sinα≤1 wtedy sin^2013(α)≤sin²α
toz cos^2013(α)+sin^2013(α)≤cos²α+sin²α=1
"=" mamy jezeli
sin^2103(α)=sin²α i cos^2013(α)=cos²α
toz mamy uklad rownan
sin^2013(α)=sin²α
cos^2013(α)=cos²α
toz
sin^2013(α)-sin²α=0
cos^2013(α)-cos²α=0
toz
sin²α(sin^2011(α)-1)=0
cos²α(cos^2011(α)-1)=0
toz
sin²α=0 lub sin^2011(α)=1
cos²α=0 lub cos^2011(α)=1
toz
sinα=0 lub sinα=1
cosα=0 lub cosα=1
ponewaz cos²α+sin²α=1 wtedy moze byc dwa sposoby
1) sinα=0 i cosα=1, toz α=2π*k, k∈Z
2) sinα=1 i cosα=0, toz α=π/2+2π*n, n∈Z
Odpowiedz: α=2π*k; k∈Z lub α=π/2+2π*n, n∈Z