Com es calcula cos (tan ^ -1 (3/4))?

Resposta:

# cos (tan ^ -1 (3/4)) = 0,8 #

Explicació:

# cos (tan ^ -1 (3/4)) =? # Deixar # tan ^ -1 (3/4) = teta

#:. tan theta = 3/4 = P / B, P i B # són perpendiculars i base

del triangle dret, doncs # H ^ 2 = P ^ 2 + B ^ 2 = 3 ^ 2 + 4 ^ 2 = 25 #

#:. H = 5;:. cos theta = B / H = 4/5 = 0,8 #

# cos (tan ^ -1 (3/4)) = cos theta = 0,8 #

#:. cos (tan ^ -1 (3/4)) = 0,8 # Ans

Resposta:

#4/5#

Explicació:

#tan (tan ^ -1 (3/4)) = 3/4 #

# "Nom" y = tan ^ -1 (3/4) #

# "Llavors tenim" #

#tan (i) = 3/4 #

# "Ara utilitzeu" sec² (x) = 1 + tan² (x) #

# => sec² (y) = 1 + tan² (y) = 1 + 9/16 = 25/16 #

# => sec (y) = 1 / cos (i) = pm 5/4 #

# => cos (i) = pm 4/5 #

# => cos (tan ^ -1 (3/4)) = pm 4/5 #

# "Hem de prendre la solució amb signar + com" #

# -pi / 2 <= arctan (x) <= pi / 2 #

# "i" #

#cos (x)> 0, si -pi / 2 <= x <= pi / 2 #

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

# "Tingueu en compte que també podríem haver usat" #

#tan (y) = sin (y) / cos (y) #

# "i" #

# sin ^ 2 (i) + cos ^ 2 (i) = 1

#tan (y) = sin (y) / cos (i) = 3/4 #

# => pm sqrt (1-cos ^ 2 (i)) / cos (i) = 3/4 #

# => 1-cos ^ 2 (y) = ((3/4) cos (i)) ^ 2 #

# => (1 + 9/16) cos ^ 2 (y) = 1

# => cos ^ 2 (y) = 16/25 #

# => cos (y) = 4/5 #