Cosx + sinx = sqrt (cosx)? - Trigonometrija 2024

Odgovor:

# Rarrx = 2npi # gdje #n u ZZ #

Obrazloženje:

# Rarrcosx + sinx = sqrtcosx #

# Rarrcosx-sqrtcosx = -sinx #

#rarr (cosx-sqrtcosx) ^ 2 = (- sinx) ^ 2 #

# Rarrcos ^ 2x-2cosx * sqrtcosx + cosx = sin ^ 2x = 1-cos ^ 2x #

# rarr2cos ^ 2x-2cosx * sqrtcosx + cosx-1 = 0 #

pustiti # Sqrtcosx = y # zatim # Cosx = y ^ 2 #

# Rarr2 * (y ^ 2) ^ 2-2 * y ^ 2 * y ^ + y = 0 2-1 #

# Rarr2y ^ 4-2y ^ 3 + y ^ 2-1 = 0 #

# Rarr2y ^ 3 (y-1) + (y + 1) + (y-1) = 0 #

#rarr y-1 2y ^ 3 + y + 1 = 0 #

Uzimanje, # Rarry-1 = 0 #

# Rarrsqrtcosx = 1 #

# Rarrcosx = 1 = cos0 #

# Rarrx = 2npi + -0 = 2npi # gdje #n u ZZ # što je general

rješenje za #x#.

Kako dokazati (1 + sinx-cosx) / (1 + cosx + sinx) = tan (x / 2)?

Pogledajte dolje. LHS = (1-cosx + sinx) / (1 + cosx + sinx) = (2sin ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2)) / (2cos ^ 2 (x / 2) + 2sin (x / 2) * cos (x / 2) = (2sin (x / 2) [sin (x / 2) + cos (x / 2)]) / (2cos (x / 2) * [ sin (x / 2) + cos (x / 2)]) = tan (x / 2) = RHS

Može li netko pomoći potvrditi taj identitet trigonometrije? (+ Sinx cosx) ^ 2 / sin ^ 2x-cos ^ 2 x = sin ^ 2x-cos ^ 2x / (sinx-cosx) ^ 2

To se provjerava u nastavku: (sinx + cosx) ^ 2 / (sin ^ 2x-cos ^ 2x) = (sin ^ 2x-cos ^ 2x) / (sinx-cosx) ^ 2 => (poništi ((sinx + cosx)) ) (sinx + cosx)) / (poništi ((sinx + cosx)) (sinx-cosx)) = (sin ^ 2x-cos ^ 2x) / (sinx-cosx) ^ 2 => ((sinx + cosx) ( sinx-cosx)) / ((sinx-cosx) (sinx-cosx)) = (sin ^ 2x-cos ^ 2x) / (sinx-cosx) ^ 2 => boja (zelena) ((sin ^ 2x-cos ^ 2 x) / (sinx-cosx) ^ 2) = (sin ^ 2x-cos ^ 2 x) / (sinx-cosx) ^ 2

Dokazati: sqrt ((1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx)) = 2 / abs (sinx)?

Dokaz ispod pomoću konjugata i trigonometrijske verzije Pitagorine teoreme. Dio 1 sqrt ((1-cosx) / (1 + cosx)) boja (bijela) ("XXX") = sqrt (1-cosx) / sqrt (1 + cosx) boja (bijela) ("XXX") = sqrt ((1-cosx)) / sqrt (1 + cosx) * sqrt (1-cosx) / sqrt (1-cosx) boja (bijela) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) Part 2 Slično sqrt ((1 + cosx) / (1-cosx) boja (bijela) ("XXX") = (1 + cosx) / sqrt (1-cos ^ 2x) Dio 3: Kombiniranje pojmova sqrt ( (1-cosx) / (1 + cosx)) + sqrt ((1 + cosx) / (1-cosx) boja (bijela) ("XXX") = (1-cosx) / sqrt (1-cos ^ 2x) + (1 + cosx) / sqrt (1-cos ^ 2x)