Cos2A = sqrt (2) (cosA-sinA) lösen?

Antworten:

Siehe die Antwort unten …

Erläuterung:

# cos2A = sqrt2 (cosA-sinA) #

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

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

# => cancel (cos2A) (cosA + sinA) = sqrt2 cdot cancel (cos2A #

# => (cosA + sinA) = sqrt2 #

# => sin ^ 2A + cos ^ 2A + 2sinAcosA = 2 #beidseitig quadratisch

# => 1 + sin2A = 2 #

# => sin2A = 1 = sin90 ^ @ #

# => 2A = 90 ^ @ #

# => A = 45 ^ @ #

Hoffe, die Antwort hilft …

DANKE DIR…

# 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 #

Wann

# cosA + sinA = 0 #

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

# => A = npi + pi / 4 "wobei" n in 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 "wo" m in ZZ #