Odgovor:
Glej spodaj.
Pojasnilo:
Uporabljamo formule (A) - # cosA = sin (90 ^ @ - A) #, (B) - # cos ^ 2A-sin ^ 2A = cos2A #
(C) - # 2sinAcosA = sin2A #, (D) - # sinA + sinB = 2sin ((A + B) / 2) cos ((A-B) / 2) # in
(E) - # sinA-sinB = 2cos ((A + B) / 2) sin ((A-B) / 2) #
# (cos ^ 2 33 ^ @ - cos ^ 2 57 ^ @) / (sin ^ 2 10.5 ^ 2 ^ ^ ^ ^ 34,5 ^ @) #
= # (cos ^ 2 33 ^ @ - sin ^ 2 (90 ^ @ - 57 ^ @)) / ((sin10.5 ^ @ + sin34.5 ^ @) (sin10.5 ^ @ - sin34.5 ^ @)) # - uporabljeno A
= # (cos ^ 2 33 ^ @ - sin ^ 2 33 ^ @) / (- ((2sin22.5 ^ e ^ ^ ^ ^ @ @) (2cos22.5^@sin12 ^ @)) # - uporabljeno D & E
= # (cos66 ^ @) / (- (2sin22.5 ^ @ cos22.5 ^ @ xx2sin12 ^ @ cos12 ^ @) # - uporabljeno B
= # - (sin (90 ^ @ - 66 ^ @)) / (sin45 ^ @ sin24 ^ @) # - uporabljeno A & C
= # -sin24 ^ @ / (1 / sqrt2sin24 ^ @) #
= # -sqrt2 #