Πώς να βρείτε την ακριβή τιμή COS (SIN ^ -1 4/5 + TAN ^ -1 5/12);

Απάντηση:

# rarrcos (sin ^ (- 1) (4/5) + tan ^ (- 1) (5/12)) = 16/65 #

Εξήγηση:

Αφήνω #sin ^ (- 1) (4/5) = x # έπειτα

# rarrsinx = 4/5 #

= 1 / (sqrt ((1 / / sinx) ^ 2-1)) = 1 / (sqrt ((1 / (4/5)) ^ 2-1)) = 4/3 #

# rarrx = tan ^ (- 1) (4/3) = sin ^ (- 1) = (4/5) #

Τώρα,

# rarrcos (sin ^ (- 1) (4/5) + tan ^ (- 1) (5/12)) #

# = cos (tan ^ (- 1) (4/3) + tan ^ (- 1) (5/12)) #

= = cos (tan ^ (- 1) ((4/3 + 5/12) / (1- (4/3) * (5/12)

= cos (tan ^ (- 1) ((63/36) / (16/36))) #

# = cos (tan ^ (- 1) (63/16)) #

Αφήνω # t ^ (- 1) (63/16) = Α # έπειτα

# rarrtanΑ = 63/16 #

# rarrcosA = 1 / δευτερόλεπτο = 1 / sqrt (1 + tan ^ 2A) = 1 / sqrt (1+ (63/16) ^ 2) = 16 /

# rarrA = cos ^ (- 1) (16/65) = tan ^ (- 1) (63/16) #

(1) (cos) (cos ^ - (1)) = cos (cos ^ (- 1)) (16/65)) = 16/65 #