How do you find the values of the other five trigonometric functions of the acute angle A with #cscA=3#?

1 Answer
Mar 21, 2018

#sinA=1/3#

#cosA=(2sqrt2)/3#

#tanA=1/(2sqrt2)#

#cscA=3/1#

#secA=3/(2sqrt2)#

#cotA=2sqrt2/1#

Explanation:

#cscA=3#
#cscA="hypotenuse"/"opposite side"=3/1#

By pythagoras theorem

#"adjacent side"=sqrt(3^2-1^2)=2sqrt2#

We have

#"opposite side"=1#
#"adjacent side"=2sqrt2#
#"hypotenuse"=3#

#sinA="opposite side"/"hypotenuse"=1/3#
#cosA="adjacent side"/"hypotenuse"=(2sqrt2)/3#
#tanA="opposite side"/"adjacent side"=1/(2sqrt2)#
#cscA="hypotenuse"/"opposite side"=3/1#
#secA="hypotenuse"/"adjacent side"=3/(2sqrt2)#
#cotA="adjacent side"/"opposite side"=2sqrt2/1#