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Allright, just want say hi to everyone, and yes this is my first post so if I miss something go easy on me.
I need to find the value for six trig functions of angle ø, angle ø is in the second quadrant and cos(ø) = -1/√5
Am I suppose to use the unit circle on this because I did not see -1/√5 in the second quadrant?
cheers.
sketch a reference triangle in quad II ... the adjacent side = -1, the hypotenuse =Quote:
Originally Posted by ipatch
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use Pythagoras to find the opposite side.
use the definitions of the trig ratios to find the other five trig values.
Step1) Draw the triangle in quad II, (check)Quote:
Originally Posted by skeetersketch a reference triangle in quad II ... the adjacent side = -1, the hypotenuse =
use Pythagoras to find the opposite side.
use the definitions of the trig ratios to find the other five trig values.
Step2) Find the opposite side using Pythagorean therom,
Pythagorean therom is a^2 + b^2 = c^2
-1^2 + b^2 = √5^2
1 + b^2 = 5
b^2 = 4
√b^2 = √4
b = 2
sin = o/h; 2/√5
cos = a/h; 2/-1
tan = o/a; 2/-1
sec = h/a; √5/-1
csc = h/o; √5/2
cot = a/o; -1/2
(check)
If I missed something on this feel free to correct me.
cheers.
...Quote:
Originally Posted by ipatch