Patunayan na ang Cos ^ 6 (x) + sin ^ 6 (x) = 1/8 (5 + 3cos4x)?

Gagamitin namin

# rarra ^ 3 + b ^ 3 = (a + b) (a ^ 2-ab + b ^ 2) #

# rarra ^ 2 + b ^ 2 = (a-b) ^ 2 + 2ab #

# rarrsin ^ 2x + cos ^ 2x = 1 #

# rarr2cos ^ 2x = 1 + cos2x # at

# rarr2sin ^ 2x = 1-cos2x #

# LHS = cos ^ 6 (x) + sin ^ 6 (x) #

# = (cos ^ 2x) ^ 3 + (sin ^ 2x) ^ 3 #

# = cos ^ 2x + sin ^ 2x (cos ^ 2x) ^ 2-cos ^ 2x * sin ^ 2x + sin ^ 2x) ^ 2 #

# = 1 * (cos ^ 2x-sin ^ 2x) ^ 2 + 2cos ^ 2x * sin ^ 2x-cos ^ 2x * sin ^ 2x #

# = cos ^ 2 (2x) + cos ^ 2x * sin ^ 2x #

# = 1/4 4cos ^ 2 (2x) + 4cos ^ 2x * sin ^ 2x #

# = 1/4 2 (1 + cos4x) + sin ^ 2 (2x) #

# = 2 / (4 * 2) 2 + 2cos4x + sin ^ 2 (2x) #

# = 1/8 4 + 4cos4x + 2sin ^ 2 (2x) #

# = 1/8 4 + 4cos4x + 1-cos4x #

# = 1/8 5 + 3cos4x = RHS #

Maaari bang tumulong ang isang tao na patunayan ang pagkakakilanlan ng trigyo? (Sinx + cosx) ^ 2 / sin ^ 2x-cos ^ 2x = sin ^ 2x-cos ^ 2x / (sinx-cosx) ^ 2

Ito ay napatunayan sa ibaba: (sinx + cosx) ^ 2 / (sin ^ 2x-cos ^ 2x) = (sin ^ 2x-cos ^ 2x) / (sinx-cosx) ^ 2 => (kanselahin ((sinx + cosx) (sinx + cosx)) / (kanselahin ((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 => kulay (green) ((sin ^ 2x) / (sinx-cosx) ^ 2) = (sin ^ 2x-cos ^ 2x) / (sinx-cosx) ^ 2

Patunayan na ((cos (33 ^ @)) ^ 2- (cos (57 ^ @)) ^ 2) / ((sin (10.5 ^ @)) ^ 2- (sin (34.5 ^ -sqrt2?

Mangyaring tingnan sa ibaba. Ginagamit namin ang mga formula (A) - cosA = sin (90 ^ @ - A), (B) - cos ^ 2A-sin ^ 2A = cos2A (C) - 2sinAcosA = sin2A, (D) - sinA + sinB = 2sin (( A + B) / 2) cos ((AB) / 2) at (E) - sinA-sinB = 2cos ((A + B) / 2) sin ((AB) / 2) cos ^ 2 57 ^ @) / (sin ^ 2 10.5 ^ @ - ^^ 34.5 ^ @) = (cos ^ 2 33 ^ @ - sin ^ 2 (90 ^ @ - 57 ^ @)) / ((sin10. 5 ^ @ + sin34.5 ^ @) (sin10.5 ^ @ - sin34.5 ^ @)) - ginamit A = (cos ^ 2 33 ^ @ - sin ^ 2 33 ^ @) / (- (2sin22.5 ^ @ cos12 ^ @) (2cos22.5 ^ @ sin12 ^ @)) - ginamit D & E = (cos66 ^ @) / (- (2sin22.5 ^ @ cos22.5 ^ @ xx2sin12 ^ @ cos12 ^ ginamit B = - (kasalan

Patunayan ito: (1-sin ^ 4x-cos ^ 4x) / (1-sin ^ 6x-cos ^ 6x) = 2/3?

LHS = (1-sin ^ 4x-cos ^ 4x) / (1-sin ^ 6x-cos ^ 6x) = (1 - ((sin ^ 2x) ^ 2 + (cos ^ 2x) ^ 2) - ((sin ^ 2x) ^ 3 + (cos ^ 2x) ^ 3)) = (1 - (sin ^ 2x + cos ^ 2x) ^ 2-2sin ^ 2cos ^ 2x)) / (1 - ^ 2x + cos ^ 2x) ^ 3-3sin ^ 2xcos ^ 2x (sin ^ 2x + cos ^ 2x)) = (1- (sin ^ 2x + cos ^ 2x) ^ 2 + 2sin ^ 2cos ^ 2x) / (1 - (sin ^ 2x + cos ^ 2x) ^ 3 + 3sin ^ 2xcos ^ 2x (sin ^ 2x + cos ^ 2x)) = (1-1 ^ 2 + 2sin ^ 2cos ^ 2x) / (1-1 ^ 3 + (2sin ^ 2cos ^ 2x) = (3sin ^ 2xcos ^ 2x) = 2/3 = RHS Pinapatunayan Sa hakbang 3 ang mga sumusunod na formula ay ginagamit ng isang ^ 2 + b ^ 2 = (a + b) ^ 2-2ab at a ^ 3 + b ^ 3 = (a + b) ^ 3-3ab (a + b)