SinA + cosA = 1 Poišči vrednost cos ^ 2A + cos ^ 4A =?

Odgovor:

# rarrcos ^ 2A + cos ^ 4 (A) = 0 #

Pojasnilo:

Glede, # rarrsinA + cosA = 1 #

# rarrsin90 ^ @ + cos90 ^ @ = 1 + 0 = 1 #

To pomeni #90^@# je koren equtaion

Zdaj, # cos ^ 2A + cos ^ 4 (A) = (cos90 ^ @) ^ 2+ (cos90 ^ @) ^ 4 = 0 ^ 2 + 0 ^ 4 = 0 #

Odgovor:

0 ali 2

Pojasnilo:

#sin A + cos A = sqrt2cos (A - pi / 4) = 1 #

#cos (A - pi / 4) = 1 / sqrt2 = sqrt2 / 2 #

Trig tabela in enota kroga dajeta dve rešitvi:

#A - pi / 4 = + - pi / 4 #

a. #A = pi / 4 + pi / 4 = pi / 2 #

#cos A = cos (pi / 2) = 0 # --> # cos ^ 2 A = cos ^ 4 A = 0 #

# cos ^ 2 A + cos ^ 4 A = 0 #

b. #A - pi / 4 = - pi / 4 # --> #A = -pi / 4 + pi / 4 = 0 #

#cos A = 1 # --> #cos ^ 2 A = cos ^ 4 A = 1 #

# cos ^ 2 A + cos ^ 4 A = 1 + 1 = 2. #