Bagaimana anda mengira dosa (cos ^ -1 (5/13) + tan ^ -1 (3/4))?

Jawapan:

#sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) = 63/65 #

Penjelasan:

Biarkan #cos ^ (- 1) (5/13) = x # kemudian

# rarrcosx = 5/13 #

# rarrsinx = sqrt (1-cos ^ 2x) = sqrt (1- (5/13) ^ 2) = 12/13 #

# rarrx = sin ^ (- 1) (12/13) = cos ^ (- 1) (5/13) #

Juga, mari #tan ^ (- 1) (3/4) = y # kemudian

# rarrtany = 3/4 #

# rarrsiny = 1 / cscy = 1 / sqrt (1 + cot ^ 2y) = 1 / sqrt (1+ (4/3) ^ 2)

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

#rarrcos ^ (- 1) (5/13) + tan ^ (- 1) (3/4) #

# = sin ^ (- 1) (12/13) + sin ^ (- 1) (3/5) #

# = sin ^ (- 1) (12/13 * sqrt (1- (3/5) ^ 2) + 3/5 * sqrt (1- (12/13) ^ 2)

# = sin ^ (- 1) (12/13 * 4/5 + 3/5 * 5/13) = 63/65 #

Sekarang, #sin (cos ^ (- 1) (5/13) + tan ^ (- 1) (3/4)) #

# = sin (sin ^ (- 1) (63/65)) = 63/65 #