Sagot:
Paliwanag:
Sagot:
Isa pang diskarte …
Paliwanag:
Ibinigay: -
#sintheta cdot costheta = 1/2 #
# => 2 cdot sintheta cdot costheta = 1 #
# "Kaya," #
#sintheta + costheta #
# = sqrt ((sintheta + costheta) ^ 2) #
# = sqrt (sin ^ 2theta + 2 cdot sintheta cdot costheta + cos ^ 2theta #
# = sqrt ((sin ^ 2theta + cos ^ 2theta) +2 cdot sintheta cdot costheta #
# = sqrt (1 +1) #
# = sqrt2 # Sana makatulong ito…
Salamat…
:-)
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
Hanapin ang halaga ng theta, kung, Cos (theta) / 1 - kasalanan (theta) + cos (theta) / 1 + sin (theta) = 4?
Theta = pi / 3 or 60 ^ @ Okay. Mayroon kaming: costheta / (1-sintheta) + costheta / (1 + sintheta) = 4 Huwag pansinin ang RHS para sa ngayon. costheta / (1-sintheta) + costheta / (1 + sintheta) (costheta (1 + sintheta) + costheta (1-sintheta)) (1-sintheta) ) (1-sin ^ 2theta) (costheta (1-sintheta + 1 + sintheta)) / (1-sin ^ 2theta) (2costheta) / (1-sin ^ 2theta) Ayon sa ang Pythagorean Identity, sin ^ 2theta + cos ^ 2theta = 1. Kaya: cos ^ 2theta = 1-sin ^ 2theta Ngayon na alam natin na, maaari nating isulat: (2costheta) / cos ^ 2theta 2 / costheta = 4 costheta / 2 = 1/4 costheta = 1/2 theta = cos ^ 1 (1/2) theta = pi / 3,
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)