The cosine of an angle is the x-coordinate of the point where its terminal ray intersects the unit circle. So, we can draw a line at x=-1/2 and see where it intersects the unit circle. That will tell us possible values of θ/2.
We find that vertical line intersects the unit circle at points where the rays make an angle of ±120° with the positive x-axis. If you consider only positive angles, these angles are 120° = 2π/3 radians, or 240° = 4π/3 radians. Since these are values of θ/2, the corresponding values of θ are double these values.
a) The cosine values repeat every 2π, so the general form of the smallest angle will be
... θ = 2(2π/3 + 2kπ) = 4π/3 + 4kπ
b) Similarly, the values repeat for the larger angle every 2π, so the general form of that is
... θ = 2(4π/3 + 2kπ) = 8π/3 + 4kπ
c) Using these expressions with k=0, 1, 2, we get
... θ = {4π/3, 8π/3, 16π/3, 20π/3, 28π/3, 32π/3}
Y=35
Z=112
Hope this helps you and you have a good day and yes I did calculate this
So; the answer is: A. 5x-2y=1
Answer:
-1 < m ≤ - 5
Step-by-step explanation:
Your expression includes: m, -1, -5, >, ≤
m is the value being compared so it will be in the middle.
# sign m sign #
m is no less than -1: no less than means it is greater than -1, but does not include -1. Therefore, our sign is >
m > -1
m is less than or equal to -5. This means we use ≤ because it can equal -5.
m ≤ -5
Now combine the two expressions
m > -1 – This can also be written as -1 < m.
m ≤ -5
-1 < m ≤ - 5
Answer:
The given functions are not same because the domain of both functions are different.
Step-by-step explanation:
The given functions are
First find the domain of both functions. Radicand can not be negative.
Domain of f(x):
This is possible if both numerator or denominator are either positive or negative.
Case 1: Both numerator or denominator are positive.
So, the function is defined for x≥1.
Case 2: Both numerator or denominator are negative.
So, the function is defined for x≤-1.
From case 1 and 2 the domain of the function f(x) is (-∞,-1]∪[1,∞).
Domain of g(x):
So, the function is defined for x≥1.
So, domain of g(x) is [1,∞).
Therefore, the given functions are not same because the domain of both functions are different.