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maxxx2000
@maxxx2000
August 2022
1
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Вычислите период функций
А)у=sin^2(x)
б)y=cos^2(x)
в)y=sin^2(x)-cos^2(x)
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sedinalana
T=2π/|k|
k=2⇒T=2π/2=π
А)у=sin^2(x)
y=1/2-1/2*cos2x
T=2
π/|k|
k=2⇒T=2π/2=π
б)y=cos^2(x)
y=1/2+1/2*cos2x
T=2π/|k|
k=2⇒T=2π/2=π
в)y=sin^2(x)-cos^2(x)
y=-cos2x
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Answers & Comments
k=2⇒T=2π/2=πА)у=sin^2(x)
y=1/2-1/2*cos2x
T=2π/|k|
k=2⇒T=2π/2=π
б)y=cos^2(x)
y=1/2+1/2*cos2x
T=2π/|k|
k=2⇒T=2π/2=π
в)y=sin^2(x)-cos^2(x)
y=-cos2x