Cos (arctan (3)) + sin (arctan (4))은 무엇을 동등하게합니까?

대답:

#cos (arctan (3)) + sin (arctan (4)) = 1 / sqrt (10) + 4 / sqrt (17) #

설명:

방해 # tan ^ -1 (3) = x #

그때 # rarrtanx = 3 #

# rarrsecx = sqrt (1 + tan ^ 2x) = sqrt (1 + 3 ^ 2) = sqrt (10) #

# rarrcosx = 1 / sqrt (10) #

# rarrx = cos ^ (-1) (1 / sqrt (10)) = tan ^ (- 1) (3) #

또한, #tan ^ (- 1) (4) = y #

그때 # rarrtany = 4 #

# rarrcoty = 1 / 4 #

# rarrcscy = sqrt (1 + cot ^ 2y) = sqrt (1+ (1/4) ^ 2) = sqrt (17) / 4 #

# rarrsiny = 4 / sqrt (17) #

# rarry = sin ^ (-1) (4 / sqrt (17)) = tan ^ (- 1) 4 #

지금, #rarrcos (tan ^ (-1) (3)) + sin (tan ^ (- 1) tan (4)) #

(17)) = 1 / sqrt (10) + 4 / sqrt (17) #rarrcos (cos ^ -1 (1 / sqrt (10)