Ako 2sin theta + 3cos theta = 2 dokazati da 3sin theta - 2 cos theta = ± 3?

Odgovor:

Pogledajte dolje.

Obrazloženje:

dan # Rarr2sinx + 3cosx = 2 #

# Rarr2sinx = 2-3cosx #

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

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

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

# rarr13cos ^ 2x-6cosx = 0 #

#rarrcosx (13cosx-6) = 0 #

# Rarrcosx = 0,6 / 13 #

# Rarrx = 90 ° #

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

dan# 2s theta + 3cos theta = 2 #

Sada

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

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

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

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

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

#=13-2^2=9#

Tako

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

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

#=±3#