Kako ocenjujete greh (cos ^ -1 (1/2)) brez kalkulatorja?

Odgovor:

#sin (cos ^ (- 1) (1/2)) = sqrt (3) / 2 #

Pojasnilo:

Let #cos ^ (- 1) (1/2) = x # potem # cosx = 1/2 #

# rarrsinx = sqrt (1-cos ^ 2x) = sqrt (1- (1/2) ^ 2) = sqrt (3) / 2 #

# rarrx = sin ^ (- 1) (sqrt (3) / 2) = cos ^ (- 1) (1/2) #

Zdaj, #sin (cos ^ (- 1) (1/2)) = sin (sin ^ (- 1) (sqrt (3) / 2)) = sqrt (3) / 2 #

Odgovor:

#sin cos ^ -1 (1/2)) = sqrt 3/2 #

Pojasnilo:

Najti vrednost #sin (cos ^ -1 (1/2)) #

Naj theta = cos ^ -1 (1/2) #

#cos theta = (1/2) #

Vemo, iz zgornje tabele, #cos 60 = 1/2 #

Zato theta = 60 ^ @ #

Zamenjava # cos ^ -1 (1/2) # z #theta = 60 ^ @ #, Znesek postane, # => sin theta = sin 60 = sqrt3 / 2 # (Kot v zgornji tabeli)