Вычислите: f (0) ; f (П/6) ; f (П/2) если f (x) = sin^2 x+sin^2 2x+sin^2 3x

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13 февраля, 07:00

Вычислите: f (0) ; f (П/6) ; f (П/2) если f (x) = sin^2 x+sin^2 2x+sin^2 3x

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Гермес · Answer 1

F (x) = (sinx) ^2 + (sin2x) ^2 + (sin3x) ^2⇒

1) f (0) = (sin0) ^2 + (sin2*0) ^2 + (sin3*0) ^2=0^2+0^2+0^2=0

2) f (π/6) = (sinπ/6) ^2 + (sin2*π/6) ^2 + (sin3*π/6) ^2=

= (sinπ/6) ^2 + (sinπ/3) ^2 + (sinπ/2) ^2 = (1/2) ^2 + (√3/2) ^2+1^2=1/4+3/4+1=2

3) f (π/2) = (sinπ/2) ^2 + (sin2*π/2) ^2 + (sin3*π/2) ^2=

= (sinπ/2) ^2 + (sinπ) ^2 + (sin3π/2) ^2=1^2 + (0) ^2 + (-1) ^2=1+0+1=2