Dokaži to ?

Odgovor:

Dokaz ispod …

Obrazloženje:

Možemo iskoristiti naše znanje o dodatnim formulama …

#cos (A + B) = cosAcosB - sinAsinB #

# cos ^ 2 (x + pi / 3) = (cosxcos (pi / 3) - sinx sin (pi / 3)) ^ 2 #

# = (1 / 2cosx - sqrt (3) / 2 sinx) ^ 2 = 1 / 4cos ^ 2x-sqrt (3) / 2 sinxcosx +3/4 sin ^ 2 x #

# cos ^ 2 (x-pi / 3) = (cosxcos (pi / 3) + sinxsin (pi / 3)) ^ 2 #

# = (1 / 2cosx + sqrt (3) / 2 sinx) ^ 2 = 1 / 4cos ^ 2x + sqrt (3) / 2 sinxcosx + 3 / 4cos ^ 2 x #

# => cos ^ 2x + cos ^ 2 (x-pi / 3) + cos ^ 2 (x + pi / 3) #

# = cos ^ 2x + 1 / 2cos ^ 2x + 3/2 sin ^ 2 x = 3 / 2cos ^ 2x + 3 / 2sin ^ 2x #

# - = 3/2 (cos ^ 2 x + sin ^ 2 x) = boja (plava) (3/2 #

Korištenje identiteta # sin ^ 2 theta + cos ^ 2 theta - = 1 #

Odgovor:

Drugi pristup.

Obrazloženje:

Koristit ćemo 1) # 2cos ^ 2x = 1 + # cos2x

2) # Cosa + = cosB 2cos ((A + B) / 2) + cos ((A-B) / 2) *

# LHS = cos ^ 2x + cos ^ 2 (x + 60 °) + cos ^ 2 (x-60 ° C) #

# = 1/2 2cos ^ 2x + 2cos ^ 2 (x + 60 °) + 2cos ^ 2 (x-60 °) #

# = 1/2 1 + 1 + cos2x + cos2 * (x + 60 ° C) 1 + + cos2 * (X-60 °) #

# = 1/2 3 + + cos2x cos (2 x + 120 °) + cos (2 x-120 ° C) #

# = 3/2 + 1/2 * cos2x + 2 * cos ((2x + 120 ° + 2x-120) / 2) * cos ((2x + 120 ° - (2x-120 ° C)) / 2) #

# = 3/2 + 1/2 * cos2x + 2cos (2x) + cos120 ° #

# = 3/2 + 1/2 cos2x + 2cos2x * (- 1/2) #

# = 3/2 + 1/2 * 0 = 3/2 = RHS #