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Mikä on yhtälö linjalle, joka on normaali f (x) = sec4x-cot2x: lle x = pi / 3: ssa?
"Normaali" => y = - (3x) / (8-24sqrt3) + (152sqrt3-120 + 3pi) / (24-72sqrt2) => y ~~ 0.089x-1.52 Normaali on kohtisuora viiva tangenttiin. f (x) = sek (4x) -cot (2x) f '(x) = 4 sec (4x) tan (3x) + 2csc ^ 2 (2x) f' (pi / 3) = 4 s ((4pi) / 3 ) tan ((4pi) / 3) + 2csc ^ 2 ((2pi) / 3) = (8-24sqrt3) / 3 Normaali, m = -1 / (f '(pi / 3)) = - 3 / ( 8-24sqrt3) f (pi / 3) = sek ((4pi) / 3) -cot ((2pi) / 3) = (sqrt3-6) / 3 (sqrt3-6) / 3 = -3 / (8- 24sqrt3) (pi / 3) + cc = (sqrt3-6) / 3 + pi / (8-24sqrt3) = (152sqrt3-120 + 3pi) / (24-72sqrt2) "Normaali": y = - (3x) / (8-24sqrt3) + (152sqrt3-120 + 3
Yksinkertaista: 1 / cot2x - 1 / cos2x?
Rarr1 / (cot2x) -1 / (cos2x) = (sinx-cosx) / (sinx + cosx) rarr1 / (cot2x) -1 / cos2x = (sin2x) / (cos2x) -1 / (cos2x) = - (1 -2sinx * cosx) / (cos2x) = - (cos ^ 2x-2cosx * sinx + sin ^ 2x) / (cos2x) = - (cosx-sinx) ^ 2 / ((cosx + sinx) (cosx-sinx) = (sinx-cosx) / (sinx + cosx)
Todista, että ?? (Sinx + Sin2x + Sin3x) / (cosx + cos2x + cos3x) = tan2x
LHS = (sinx + sin2x + sin3x) / (cosx + cos2x + cos3x) = (2sin ((3x + x) / 2) * cos ((3x-x) / 2) + sin2x) / (2cos ((3x +) x) / 2) * cos ((3x-x) / 2) + cos2x = (2sin2x * cosx + sin2x) / (2cos2x * cosx + cos2x) = (sin2xcancel ((1 + 2cosx))) / (cos2xcancel (( 1 + 2cosx))) = tan2x = RHS