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
Square both sides:


Since

is positive, you can discard the negative sign. So,

Substitute this value back into

to find


I hope this helps. =)
Tags: <em>trigonometric identity relation trig sine cosine tangent sin cos tan trigonometry precalculus</em>
We have to find what percent of the audience from the 16-25 age group gave the movie poor rating. There were 113 viewers from this group. There were 7 of them that gave the movie poor rating. We will use the proportion: 113 / 7 = 100% / x% . Then we will cross multiply: 113 x = 700; x = 700 : 113; x = 6.19%. Answer: A ) 6.19%.
Answer: see below
<u>Step-by-step explanation:</u>
The coordinates on the Unit Circle are (cos, sin). Since we are focused on cosine, we only need to focus on the left side of the coordinate. The cosine value (left side) will be the y-value of the function y = cos x
Use the quadrangles (angles on the axes) to represent the x-values of the function y = cos x.
Quadrangles are: 0°, 90°, 180°, 270°, 360° <em>(360° = 0°)</em>
Together, the coordinates will be as follow:

Answer:
b. the degree is 3
Step-by-step explanation:
degree is the highest exponent