find the value of cos (-8pi/3)
Answers
Step-by-step explanation:
8 pi/3 = 6pi/3 + 2 pi/3
6 pi/3 = 2 pi
2 pi = 360 degrees.
8 pi/3 = 360 degrees + (360/3) degrees
For each of the six trigonometric functions,
cos(x+2 pi) = cos(x)
sin(x + 2 pi) = sin(x)
etc
Your question is equivalent to how can I find the value of each of the trigonometric functions at (360/3) degrees = 120 degrees.
Imagine drawing a half-circle of radius equal to 1 going from (x = 1, y = 0) to
(x = -1, y = 0).
Now construct a right triangle from the point (0,0) to that point on the circle which is 120 degrees counterclockwise from (1,0), and 60 degrees clockwise from (-1,0).
Construct that right triangle by drawing the hypotenuse to that point from the origin, (0,0)
and drawing a verticle line from that point on the circle down to the x-axis.
The angle between the hypotenuse and the negative x-axis will be 60 degrees.
How can we find the length of the hypotenuse and the other two sides?
Let the vertical line you drew downward from that special point on your unit circle act as a mirror.
Construct an identically sized triangle on the other side of your vertical mirror line.
The vertical mirror line is a side of both right triangles.
The two right triangles together form another larger triangle, with twice the area of your original right triangle.
Each of the three angles of that larger triangle must be 60 degrees.
The base angles are sixty degrees because one of the base angles is the 60-degree angle of your original right triangle. The other base angle is an exact copy of it and is therefore also 60 degrees.
The angle where the two hypotenuses meet the circle is 30 degrees + 30 degrees = 60 degrees.
The larger triangle is an equilateral triangle.
The length of the hypotenuse of your first right triangle is by definition the radius of your half circle, which is 1.
If your construction was sufficiently accurate, you would note that the hypotenuse of the second constructed right triangle intersected the x-axis exactly at the same place that your half circle intersected it.
This proves that the side of your original right triangle had a length of 1/2 along the x-axis.
The cosine function is defined to be the length of the adjacent side/length of the hypotenuse
Cos(60 degrees) = (1/2)/1 = 1/2
However, we wish to find the cos(120 degrees)
Our 120-degree point on the circle is to the left of the y-axis, and above the x-axis.
We count the horizontal length as negative and the vertical length as positive.
Thus cos(120 degrees) = -1/2.
What is the length of the vertical side of our right triangle?
Use the Pythagorean theorem.
(-1/2)^2 + h^2 = 1
h^2 = 1 - 1/3 = 3/4
h = sqrt(3)/2
sin(120 degrees) = sqrt(3)/2
tan(120 degrees)
= sin(120 degrees) / cos(120 degrees)
= (sqrt(3)/2) / (-1/2)
= - sqrt(3)
cotangent (120 degrees) = 1/ tangent(120 degrees) = 1/(- sqrt(3) ) = - sqrt(3) / 3
secant(120 degrees) = 1/cos(120 degrees) = 1/(-1/2) = -2
cosecant(120 degrees) = 1/sin(120 degrees) = 1/(sqrt(3)/2) = 2/sqrt(3) = 2 sqrt(3) / 3
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Answer:
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