Wie vereinfacht man # (sinX-cosX) ^ 2 #?
Antworten:
# (sinX-cosX)^2 = 1-sin2X #
Erläuterung:
# sin^2 A + cos^2 A = 1 #
# sin 2A = 2 sin A cos A #
# (sinX-cosX)^2 = sin^2 X -2sin X cos X + cos^2 X = 1-2sin XcosX
= 1-sin2X #
# (sinX-cosX)^2 = 1-sin2X #
# sin^2 A + cos^2 A = 1 #
# sin 2A = 2 sin A cos A #
# (sinX-cosX)^2 = sin^2 X -2sin X cos X + cos^2 X = 1-2sin XcosX
= 1-sin2X #