itsmichelle29
  • itsmichelle29
If sin θ = 1/4 and tan θ > 0, what is the value of cos θ?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
draw a triangle, like this |dw:1439233810971:dw|
misty1212
  • misty1212
there is a picture of an angle where the sine is \(\frac{1}{4}\) since the "opposite over adjacent" is \(\frac{1}{4}\)

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misty1212
  • misty1212
i meant "opposite over hypotenuse"
misty1212
  • misty1212
what you need is the adjacent side, which you find via pythagoras let me know when you get it
itsmichelle29
  • itsmichelle29
7
misty1212
  • misty1212
oh no not 7
misty1212
  • misty1212
use pythagoras \[a^2+b^2=c^2\\ x^2+1^2=4^2\\ x^2+1=16\\ x^2=16-1=15\] so \[x=\sqrt{15}\]
itsmichelle29
  • itsmichelle29
okay is the answe root15 /4
itsmichelle29
  • itsmichelle29
@misty1212
misty1212
  • misty1212
yes
misty1212
  • misty1212
oh no!
misty1212
  • misty1212
you want the tangent right? not the cosine
itsmichelle29
  • itsmichelle29
can u help me with like two more plz
itsmichelle29
  • itsmichelle29
the cos
misty1212
  • misty1212
oh ok then yes
misty1212
  • misty1212
and i will be happy to help with two more
itsmichelle29
  • itsmichelle29
thanks
itsmichelle29
  • itsmichelle29
What are the period and phase shift for f(x) = −4 tan(x − π)?
itsmichelle29
  • itsmichelle29
i think its pi and pi
misty1212
  • misty1212
period of tangent is \(\pi\) so yeah
misty1212
  • misty1212
phase shift is \(\pi\) as well you are right
itsmichelle29
  • itsmichelle29
thanks and
itsmichelle29
  • itsmichelle29
Use the graph below to answer the question that follows: cosine graph with points at 0, negative 1 and pi over 2, 3 and pi, negative 1 What are the amplitude, period, and midline of the function? Amplitude: 8; period: π; midline: y = 1 Amplitude: 8; period: 2π; midline: y = 5 Amplitude: 4; period: 2π; midline: y = 5 Amplitude: 2; period: π; midline: y = 1
misty1212
  • misty1212
really need the graph for this one
itsmichelle29
  • itsmichelle29
1 Attachment
itsmichelle29
  • itsmichelle29
okay =)
misty1212
  • misty1212
ok much better got a guess?
misty1212
  • misty1212
only need to look at the amplitude to pick from your choices
itsmichelle29
  • itsmichelle29
Amplitude: 8; period: π; midline: y = 1 Amplitude: 8; period: 2π; midline: y = 5 Amplitude: 4; period: 2π; midline: y = 5 Amplitude: 2; period: π; midline: y = 1
itsmichelle29
  • itsmichelle29
well i thought it was -1 but thats not a chose
misty1212
  • misty1212
from -1 to 3, that is from the lowest point to the highest, is a distance of 4 units the amplitude is half of that i.e. 2
misty1212
  • misty1212
also you can see that the period is \(\pi\) since it goes up, and back down to -1 from 0 to \(\pi\)
itsmichelle29
  • itsmichelle29
Amplitude: 2; period: π; midline: y = 1 is that the answer
misty1212
  • misty1212
yup
itsmichelle29
  • itsmichelle29
omg ur amazing thanks
misty1212
  • misty1212
lol thanks\[\color\magenta\heartsuit\]

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