SOLUTION: Let θ be an angle in quadrant II such that secθ=-(13)/(12)
Find the exact values of cotθ and sinθ...
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Question 586488: Let θ be an angle in quadrant II such that secθ=-(13)/(12)
Find the exact values of cotθ and sinθ...
Found 2 solutions by stanbon, solver91311:
Answer by stanbon(75887) (Show Source): You can put this solution on YOUR website!
Let θ be an angle in quadrant II such that secθ=-(13)/(12)
Find the exact values of cotθ and sinθ...
----
In QII x is negative and y is positive.
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sec(theta) = r/x = -13/12 implies that r = 13 and x = -12
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Then y^2 = 13^2-12^2 = 169-144 = 25
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y = +5
---
sin = y/r = 5/13
cos = x/r = -12/13
tan = y/x = -5/12
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csc = r/y = 13/5
sec = -13/12
cot = -12/5
===============
Cheers,
Stan H.
===============
Answer by solver91311(24713) (Show Source): You can put this solution on YOUR website!
Secant is just the reciprocal of cosine, so if
\ =\ -\frac{13}{12})
then
\ =\ -\frac{12}{13})
which is just fine because cosine is negative in Quadrant II.
Then use the Pythagorean identity:
\ +\ \sin^2(\varphi)\ =\ 1)
from which we can derive:
\ =\ \pm\sqrt{1\ -\ \cos^2(\varphi)})
And you choose 
based on the quadrant.
For you, sine is positive in QII, so:
\ =\ \sqrt{1\ -\ \frac{144}{169}}\ =\ \sqrt{\frac{25}{169}}\ =\ \frac{5}{13})
John
My calculator said it, I believe it, that settles it
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