use the given values to evaluate (if possible) all six trig functions: cos (pi/2 - x) and cosx = 12/13
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Directrix (directrix):
Is this the question as it appears in your text?
Could you post a screen shot of the problem?
Something seems to be missing.
Directrix (directrix):
^^ @Daisyblake11
OpenStudy (daisyblake11):
yes that's how it is
OpenStudy (daisyblake11):
one sec
Directrix (directrix):
>>cos (pi/2 - x) Is this complete?
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OpenStudy (daisyblake11):
lol it isn't. my bad
Directrix (directrix):
cosx = 12/13 is partially helpful but we do not know whether we are in quadrant III or quadrant I.
If you post a screen shot of the problem, I may be able to help. Otherwise, I cannot.
OpenStudy (daisyblake11):
OpenStudy (daisyblake11):
it's cos(pi/2-x) = 5/13 and cos x = 12/13
Directrix (directrix):
cos (pi/2 - x) = sin x = 5/13
So far, we have:
cos x = 12/13
sin x = 5/13
That leaves:
sec x
csc x
tan x
cot x
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Directrix (directrix):
@Daisyblake11 Do you know the reciprocals of the trig functions?
OpenStudy (daisyblake11):
yes i do. thank yo so much
Directrix (directrix):
Good to know.
What is this equal: sec x =
OpenStudy (daisyblake11):
13/12
Directrix (directrix):
csc x =
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OpenStudy (daisyblake11):
13/5
Directrix (directrix):
tan x =
OpenStudy (daisyblake11):
sinx/cosx
Directrix (directrix):
Right. Substitute the values we already have for sin x and cos x into the tangent fraction and crank out the numerical value of the tangent.
OpenStudy (daisyblake11):
got it
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OpenStudy (daisyblake11):
5/12
Directrix (directrix):
cot x =
OpenStudy (daisyblake11):
12/5
Directrix (directrix):
We have them all. Good work.
OpenStudy (daisyblake11):
thank you. good work :)
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