If sin θ = 1/4 and tan θ > 0, what is the value of cos θ?
HI!!
draw a triangle, like this |dw:1439233810971:dw|
there is a picture of an angle where the sine is \(\frac{1}{4}\) since the "opposite over adjacent" is \(\frac{1}{4}\)
i meant "opposite over hypotenuse"
what you need is the adjacent side, which you find via pythagoras let me know when you get it
7
oh no not 7
use pythagoras \[a^2+b^2=c^2\\ x^2+1^2=4^2\\ x^2+1=16\\ x^2=16-1=15\] so \[x=\sqrt{15}\]
okay is the answe root15 /4
@misty1212
yes
oh no!
you want the tangent right? not the cosine
can u help me with like two more plz
the cos
oh ok then yes
and i will be happy to help with two more
thanks
What are the period and phase shift for f(x) = −4 tan(x − π)?
i think its pi and pi
period of tangent is \(\pi\) so yeah
phase shift is \(\pi\) as well you are right
thanks and
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
really need the graph for this one
okay =)
ok much better got a guess?
only need to look at the amplitude to pick from your choices
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
well i thought it was -1 but thats not a chose
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
also you can see that the period is \(\pi\) since it goes up, and back down to -1 from 0 to \(\pi\)
Amplitude: 2; period: π; midline: y = 1 is that the answer
yup
omg ur amazing thanks
lol thanks\[\color\magenta\heartsuit\]
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