Ha 2 theta + 3cos theta = 2 bizonyítja, hogy a 3sin theta - 2 cos theta = ± 3?

Válasz:

Lásd alább.

Magyarázat:

Adott # Rarr2sinx + 3cosx = 2 #

# Rarr2sinx = 2-3cosx #

#rarr (2sinx) ^ 2 = (2-3cosx) ^ 2 #

# Rarr4sin ^ 2x = 4-6cosx + 9cos ^ 2x #

#rarrcancel (4) -4cos ^ 2x = megszünteti (4) -6cosx + 9cos ^ 2x #

# Rarr13cos ^ 2x-6cosx = 0 #

#rarrcosx (13cosx-6) = 0 #

# Rarrcosx = 0,6 / 13 #

# Rarrx = 90 ° #

Most, # 3sinx-2cosx = 3sin90 ° -2cos90 ° = 3 #

Adott# 2 theta + 3cos theta = 2 #

Most

# (3 theta - 2 cos theta) ^ 2 #

# = (9sin ^ 2 théta-2 * 3sintheta * 2costheta + 4cos ^ 2 théta #

# = 9-9cos ^ 2 théta-2 * 3costheta * 2sintheta + 4-4sin ^ 2 théta #

# = 13 - ((3costheta) ^ 2 + 2 * 3costheta * 2sintheta + (2sintheta) ^ 2 #

# = 13- (2sintheta + 3costheta) ^ 2 #

#=13-2^2=9#

Így

# (3 theta - 2 cos theta) ^ 2 = 9 #

# => 3in theta - 2 cos theta = pmsqrt9 #

#=±3#