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

1. anonymous

negative square root of 15 square root of 15 square root of 15 over 4 negative square root of 15 over 4

2. LynFran

last option

3. LynFran

let me show u how ok

4. anonymous

ok

5. Michele_Laino

hint: |dw:1435348645865:dw|

6. LynFran

|dw:1435355888315:dw| the third option sorry

7. Michele_Laino

there are 2 angles which satisfy the condition: sin (x) = 1/4, nevertheless only one is the right one

8. LynFran

the third option is correct

9. anonymous

ohhhhh ok thank you both

10. Michele_Laino

:)

11. anonymous

1 more?

12. Michele_Laino

ok!

13. anonymous

Where are the asymptotes of f(x) = tan(2x − π) from x = pi over 2 to x = 3 pi over 2

14. Michele_Laino

we can write your function as below: $\begin{gathered} \tan \left( {2x - \pi } \right) = \frac{{\tan \left( {2x} \right) - \tan \pi }}{{1 + \tan \left( {2x} \right)\tan \pi }} = \hfill \\ \hfill \\ = \tan \left( {2x} \right) = \frac{{\sin \left( {2x} \right)}}{{\cos \left( {2x} \right)}} \hfill \\ \end{gathered}$

15. anonymous

O.O lemme get my calculator

16. Michele_Laino

now we have vertical asymptotes, at point x, such that: $\cos \left( {2x} \right) = 0$

17. anonymous

let me post my choices

18. anonymous

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

19. Michele_Laino

namely, when: $2x = \frac{\pi }{2} + k\pi ,\quad k = 0, \pm 1, \pm 2,...$

20. anonymous

D is wrong , right?

21. Michele_Laino

so dividing by 2, we get: $x = \frac{\pi }{4} + k\frac{\pi }{2},\quad k = 0, \pm 1, \pm 2,...$

22. anonymous

is it C

23. Michele_Laino

no, it is A

24. anonymous

but i thought you said one of them was 0

25. anonymous

oh i see

26. Michele_Laino

since, using my formula, we get: $\frac{\pi }{4},\quad \frac{{3\pi }}{4},\quad \frac{{5\pi }}{4},\quad \frac{{7\pi }}{4},...$

27. anonymous

yeah i just got that

28. anonymous

thank you if i could give another medal i would

29. Michele_Laino

ok! :)