Help finding values of trig functions.

• Jan 13th 2010, 01:05 PM
ipatch
Help finding values of trig functions.
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.
• Jan 13th 2010, 01:17 PM
skeeter
Quote:

Originally Posted by ipatch
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 = $\sqrt{5}$.

use Pythagoras to find the opposite side.

use the definitions of the trig ratios to find the other five trig values.
• Jan 13th 2010, 01:53 PM
ipatch
Quote:

Originally Posted by skeeter
sketch a reference triangle in quad II ... the adjacent side = -1, the hypotenuse = $\sqrt{5}$.

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)

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.
• Jan 13th 2010, 02:59 PM
skeeter
Quote:

Originally Posted by ipatch
Step1) Draw the triangle in quad II, (check)

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 = -2
sec = h/a; √5/-1 = -√5
csc = h/o; √5/2
cot = a/o; -1/2

(check)

If I missed something on this feel free to correct me.

cheers.

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