Sagot:
Mangyaring tingnan sa ibaba.
Paliwanag:
Sagot:
Tingnan ang patunay sa ibaba
Paliwanag:
Kailangan namin
Samakatuwid,
Ipakita na cos²π / 10 + cos²4π / 10 + cos² 6π / 10 + cos²9π / 10 = 2. Ako ay medyo nalilito kung gumawa ako Cos²4π / 10 = cos² (π-6π / 10) & cos²9π / 10 = cos² (π-π / 10), ito ay magiging negatibo bilang cos (180 ° -theta) = - costheta sa ang pangalawang kuwadrante. Paano ko mapapatunayan ang tanong?
Mangyaring tingnan sa ibaba. LHS = cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) + cos ^ 2 ((6pi) / 10) + cos ^ 2 ((9pi) / 10) 10) + cos ^ 2 (4pi) / 10) + cos ^ 2 (pi- (4pi) / 10) + cos ^ 2 (pi- (pi) / 10) = cos ^ 2 (pi / 10) + cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) / 10) = 2 * [cos ^ 2 (pi / 10) + cos ^ 2 ((4pi) 10)] = 2 * [cos ^ 2 (pi / 2- (4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * [sin ^ 2 ((4pi) / 10) + cos ^ 2 ((4pi) / 10)] = 2 * 1 = 2 = RHS
Paano makahanap ng eksaktong halaga COS (SIN ^ -1 4/5 + TAN ^ -1 5/12)?
(4/5) + tan ^ (- 1) (5/12)) = 16/65 Hayaan ang kasalanan ^ (- 1) (4/5) = x pagkatapos rarrsinx = 4/5 = 1 / (sqrt (csc ^ 2x-1)) = 1 / (sqrt ((1 / sinx) ^ 2-1)) = 1 / (sqrt ((1 / (4/5) ^ 2-1)) = 4/3 rarrx = tan ^ (- 1) (4/3) = sin ^ (- 1) = (4/5) Ngayon, rarrcos (sin ^ (- 1) (4/5 ) - tan ^ (- 1) (5/12)) = cos (tan ^ (- 1) (4/3) + tan ^ (- 1) (5/12) ((4/3 + 5/12) / (1- (4/3) * (5/12)))) = cos (tan ^ (- 1) ((63/36) / (16/36) ) = cos (tan ^ (- 1) (63/16)) Hayaan tan ^ (- 1) (63/16) = A pagkatapos rarrtanA = 63/16 rarrcosA = 1 / secA = 1 / sqrt (1 + (1 / (63/16) ^ 2) = 16/65 rarrA = cos ^ (- 1) (16/65) = tan ^ (- 1) (63/16) rarr
Ipakita na, (1 + cos theta + i * sin theta) ^ n + (1 + cos theta - i * sin theta) ^ n = 2 ^ (n + 1) * (cos theta / 2) ^ n * cos n * theta / 2)?
Mangyaring tingnan sa ibaba. (1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2 ) 2) = 2cos (theta / 2) at tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2) (theta / 2) o alpha = theta / 2 pagkatapos 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) at maaari naming isulat (1 + costheta + ^ n + (1 + costheta-isintheta) ^ gamit ang teorem ng DE MOivre bilang r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha = 2 * 2 ^ ncos ^ n (theta / 2) cos ((ntheta) / 2) = 2 ^ (n + 1) cos ^ n (theta / 2) cos ((ntheta) / 2)