Tanong # e8ab5

Sagot:

#cos (x + y) = (a ^ 2 + b ^ 2) / 2-1 #

Paliwanag:

Una, isipin kung ano #cos (x + y) # ay:

#cos (x + y) = cosxcosy + sinxsiny #

Tandaan na:

# (sinx + siny) ^ 2 = a ^ 2 #

# -> sin ^ 2x + 2sinxsiny + sin ^ 2y = a ^ 2 #

At:

# (cosx + cozy) ^ 2 = b ^ 2 #

# -> cos ^ 2x + 2cosxcosy + cos ^ 2y = b ^ 2 #

Ngayon kami ay may dalawang mga equation na ito:

# sin ^ 2x + 2sinxsiny + sin ^ 2y = a ^ 2 #

# cos ^ 2x + 2cosxcosy + cos ^ 2y = b ^ 2 #

Kung idagdag namin ang mga ito nang magkasama, mayroon kaming:

# sin ^ 2x + 2sinxsiny + sin ^ 2y + cos ^ 2x + 2cosxcosy + cos ^ 2y = a ^ 2 + b ^ 2 #

Huwag hayaan ang laki ng equation na ito magtapon ka off. Maghanap ng mga pagkakakilanlan at mga pagpapadali:

# (sin ^ 2x + cos ^ 2x) + (2sinxsiny + 2cosxcosy) + (cos ^ 2y + sin ^ 2y) = a ^ 2 + b ^ 2 #

Mula noon # sin ^ 2x + cos ^ 2x = 1 # (Pythagorean Identity) at # cos ^ 2y + sin ^ 2y = 1 # (Pythagorean Identity), maaari naming gawing simple ang equation sa:

# 1 + (2sinxsiny + 2cosxcosy) + 1 = a ^ 2 + b ^ 2 #

# -> (2sinxsiny + 2cosxcosy) + 2 = a ^ 2 + b ^ 2 #

Maaari naming saluhin ang isang #2# dalawang beses:

# 2 (sinxsiny + cosxcosy) + 2 = a ^ 2 + b ^ 2 #

# -> 2 ((sinxsiny + cosxcosy) +1) = a ^ 2 + b ^ 2 #

At hatiin ang:

# (sinxsiny + cosxcosy) + 1 = (a ^ 2 + b ^ 2) / 2 #

At ibawas:

# sinxsiny + cosxcosy = (a ^ 2 + b ^ 2) / 2-1 #

Sa wakas, dahil #cos (x + y) = cosxcosy + sinxsiny #, meron kami:

#cos (x + y) = (a ^ 2 + b ^ 2) / 2-1 #

Given

# sinx + siny = a ……. (1) #

# cosx + cozy = b ……. (2) #

Squaring and adding (1) & (2)

# (cosx + cozy) ^ 2 + (sinx + siny) ^ 2 = a ^ 2 + b ^ 2 #

# => 2 (cosxcosy + sinxsiny) + 2 = a ^ 2 + b ^ 2 #

# => 2cos (x-y) = a ^ 2 + b ^ 2-2 …. (3) #

Squaring and Subtracting (1) mula sa (2)

# (cosx + cozy) ^ 2- (sinx + siny) ^ 2 = b ^ 2-a ^ 2 #

# => 2cos (x + y) + cos ^ 2x-sin ^ 2x + cos ^ 2y-sin ^ 2y = b ^ 2-a ^ 2 #

# => 2cos (x + y) + cos2x + cos2y = b ^ 2-a ^ 2 #

# => 2cos (x + y) + 2cos (x + y) cos (x-y) = b ^ 2-a ^ 2 #

# => cos (x + y) (2 + 2cos (x-y)) = b ^ 2-a ^ 2 #

(# "Mula sa (3)" 2cos (x-y) = a ^ 2 + b ^ 2-2 #)

# => cos (x + y) (2 + b ^ 2 + a ^ 2-2) = b ^ 2-a ^ 2 #

# => cos (x + y) (b ^ 2 + a ^ 2) = b ^ 2-a ^ 2 #

# => cos (x + y) = (b ^ 2-a ^ 2) / (b ^ 2 + a ^ 2) #

Sagot:

#cos (x + y) = (b ^ 2-a ^ 2) / (b ^ 2 + a ^ 2) #.

Paliwanag:

# sinx + siny = a rArr 2sin ((x + y) / 2) cos ((x-y) / 2) = a ……… (1) #.

# cosx + cozy = b rArr 2cos ((x + y) / 2) cos ((x-y) / 2) = b ………. (2) #.

Paghahati #(1)# sa pamamagitan ng #(2)#, meron kami, #tan ((x + y) / 2) = a / b #.

Ngayon, #cos (x + y) = {1-tan ^ 2 ((x + y) / 2)} / {1 + tan ^ 2 ((x + y) / 2)

# = (1-a ^ 2 / b ^ 2) / (1 + a ^ 2 / b ^ 2) = (b ^ 2-a ^ 2) / (b ^ 2 + a ^ 2).

Tangkilikin ang Matematika.!