If sin Θ = 1 over 4 and tan Θ > 0, what is the value of cos Θ?

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If sin Θ = 1 over 4 and tan Θ > 0, what is the value of cos Θ?

Mathematics
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negative square root of 15 square root of 15 square root of 15 over 4 negative square root of 15 over 4
last option
let me show u how ok

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Other answers:

ok
hint: |dw:1435348645865:dw|
|dw:1435355888315:dw| the third option sorry
there are 2 angles which satisfy the condition: sin (x) = 1/4, nevertheless only one is the right one
the third option is correct
ohhhhh ok thank you both
:)
1 more?
ok!
Where are the asymptotes of f(x) = tan(2x − π) from x = pi over 2 to x = 3 pi over 2
we can write your function as below: \[\begin{gathered} \tan \left( {2x - \pi } \right) = \frac{{\tan \left( {2x} \right) - \tan \pi }}{{1 + \tan \left( {2x} \right)\tan \pi }} = \hfill \\ \hfill \\ = \tan \left( {2x} \right) = \frac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}} \hfill \\ \end{gathered} \]
O.O lemme get my calculator
now we have vertical asymptotes, at point x, such that: \[\cos \left( {2x} \right) = 0\]
let me post my choices
x = 3 pi over 4, x = 5 pi over 4 x = 0, x = π, x = 2π x = 0, x = pi over 4 x = pi over 2, x = 3 pi over 2
namely, when: \[2x = \frac{\pi }{2} + k\pi ,\quad k = 0, \pm 1, \pm 2,...\]
D is wrong , right?
so dividing by 2, we get: \[x = \frac{\pi }{4} + k\frac{\pi }{2},\quad k = 0, \pm 1, \pm 2,...\]
is it C
no, it is A
but i thought you said one of them was 0
oh i see
since, using my formula, we get: \[\frac{\pi }{4},\quad \frac{{3\pi }}{4},\quad \frac{{5\pi }}{4},\quad \frac{{7\pi }}{4},...\]
yeah i just got that
thank you if i could give another medal i would
ok! :)

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