לפתור cos2A = sqrt (2) (cosA-sinA)?

תשובה:

ראה את התשובה למטה …

הסבר:

# cos2A = sqrt2 (cosA-sinA) #

# => cos2A (cOSA + sinA) = sqrt2 (cos ^ 2A-sin ^ 2A) #

# => cos2A (cOSA + sinA) = sqrt2 cdot cos2A #

# => ביטול (cos2a) (cosa + sinA) = sqrt2 cdot לבטל (cos2a #

# => (cOSA + sinA) = sqrt2 #

# => חטא ^ 2A + cos ^ 2A + 2sinAcosA = 2 #רבועות משני הצדדים

# => 1 + sin2A = 2 #

# => sin2A = 1 = sin90 ^ @ #

# => 2A = 90 ^ @ #

# => A = 45 ^ @ #

HOPE תשובה תשובה …

תודה…

# cos2A = sqrt2 (cosA-sinA) #

# => cos ^ 2A-sin ^ 2A-sqrt2 (cosA-sinA) = 0 #

# => (cosA-sinA) (cosa + sinA) -sqrt2 (cosA-sinA) = 0 #

# => (cosA-sinA) (cOSA + sinA-sqrt2) = 0 #

מתי

# cosA + sinA = 0 #

# => tana = 1 = tan (pi / 4) #

# => A = npi + pi / 4 "where" n ב ZZ #

# cosA + sinA = sqrt2 #

# => 1 / sqrt2cosA + 1 / sqrt2sinA = 1 #

# = cos (pi / 4) cosa + sin (pi / 4) sinA = 1 #

# => cos (A-pi / 4) = 1 #

# => A = 2mpi + pi / 4 "שם" m ב ZZ #