Hoe bereken je cos (tan ^ -1 (3/4))?

Antwoord:

# cos (tan ^ -1 (3/4)) = 0.8 #

Uitleg:

# cos (tan ^ -1 (3/4)) =? # Laat # tan ^ -1 (3/4) = theta #

#:. tan theta = 3/4 = P / B, P en B # zijn loodrecht en basis

van de rechter driehoek, dan # 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

Antwoord:

#4/5#

Uitleg:

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

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

#"Dan hebben we"#

#tan (y) = 3/4 #

# "Gebruik nu" sec² (x) = 1 + tan² (x) #

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

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

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

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

# "We moeten de oplossing met + teken als" # nemen

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

#"en"#

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

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

# "Let op: we hadden ook" # kunnen gebruiken

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

#"en"#

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

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

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

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

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

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

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