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Dokaži: sqrt ((1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx)) = 2 / abs (sinx)?
Dokaz spodaj z uporabo konjugatov in trigonometrične različice Pitagorejeve teoreme. Barva dela 1 ((1-cosx) / (1 + cosx)) (bela) ("XXX") = sqrt (1-cosx) / sqrt (1 + cosx) barva (bela) ("XXX") = sqrt ((1-cosx)) / sqrt (1 + cosx) * sqrt (1-cosx) / sqrt (1-cosx) barva (bela) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) 2. del Podobno sqrt ((1 + cosx) / (1-cosx) barva (bela) ("XXX") = (1 + cosx) / sqrt (1-cos ^ 2x) 3. del: Združevanje izrazov sqrt ( (1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx) barva (bela) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) + (1 + cosx) / sqrt (1-cos ^ 2x
Dokaži: tan ^ 5x = ((1 / (1-sinx) ^ 2) - (1 / (1 + sinx) ^ 2)) / ((1 / (1-cosx) ^ 2) - (1 / ( 1 + cosx) ^ 2)?
Za dokazovanje tg ^ 5x = ((1 / (1-sinx) ^ 2) - (1 / (1 + sinx) ^ 2)) / ((1 / (1-cosx) ^ 2) - (1 / (1) + cosx) ^ 2) RHS = ((1 / (1-sinx) ^ 2) - (1 / (1 + sinx) ^ 2)) / ((1 / (1-cosx) ^ 2) - (1 / (1 + cosx) ^ 2) = (((1 + sinx) ^ 2- (1-sinx) ^ 2) / (1-sin ^ 2x) ^ 2) / (((1 + cosx ^ 2) - ( 1-cosx) ^ 2) / (1-cos ^ 2x) ^ 2) = ((4sinx) / cos ^ 4x) / ((4cosx) / (sin ^ 4x)) = sin ^ 5x / cos ^ 5x = tan ^ 5x = LHS Dokazano
Pokažite, da (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)?
Glej spodaj. Naj bo 1 + costheta + isintheta = r (cosalpha + isinalpha), tukaj r = sqrt ((1 + costheta) ^ 2 + sin ^ 2theta) = sqrt (2 + 2costheta) = sqrt (2 + 4cos ^ 2 (theta / 2) ) -2) = 2cos (theta / 2) in tanalpha = sintheta / (1 + costheta) == (2sin (theta / 2) cos (theta / 2)) / (2cos ^ 2 (theta / 2)) = tan (theta / 2) ali alfa = theta / 2, nato 1 + costheta-isintheta = r (cos (-alpha) + isin (-alpha)) = r (cosalpha-isinalpha) in lahko napišemo (1 + costheta + isintheta) ^ n + (1 + costheta-isintheta) ^ n z uporabo DE MOivrejevega izreka kot r ^ n (cosnalpha + isinnalpha + cosnalpha-isinnalpha) = 2r ^ ncosnalpha = 2 *