Find the values of the trigonometric functions of t if tan t= 3 and the terminal point of t is in quadrant III.
Since tan t is positive and the terminal point of t is in quadrant III, we can determine the values of other trigonometric functions as follows:
First, note that tan t = 3 can be rewritten as sin t / cos t = 3.
Next, since the terminal point of t is in quadrant III, we know that sin t is negative.
Using the Pythagorean identity, we have sin^2 t + cos^2 t = 1.
Substituting sin t = -√(1 - cos^2 t), we have (-√(1 - cos^2 t))^2 + cos^2 t = 1.
Expanding and simplifying the equation, we get 1 - cos^2 t + cos^2 t = 1.
This shows that cos^2 t = 0, and thus cos t = 0.
Now, we have sin t = -√(1 - cos^2 t) = -√(1 - 0^2) = -√1 = -1.
Therefore, the values of the trigonometric functions are as follows:
sin t = -1,
cos t = 0,
tan t = 3,
csc t = -1,
sec t = undefined (since cos t = 0, the reciprocal is undefined),
cot t = 1/3.
To find the values of the trigonometric functions of t, we first need to determine the values of sin(t), cos(t), and cot(t) since we are given the value of tan(t).
Since t is in quadrant III, we know that cosine is negative and sine is also negative.
Given that tan(t) = 3, we can use the definition of tangent: tan(t) = sin(t) / cos(t).
Therefore, we have sin(t) / cos(t) = 3.
Now, let's use the Pythagorean identity to find the value of cos(t):
cos^2(t) = 1 - sin^2(t)
Since we know that sin(t) is negative, we take the negative square root of the right side:
cos(t) = -sqrt(1 - sin^2(t))
Substituting the value of cos(t) into the equation sin(t) / cos(t) = 3, we get:
sin(t) / (-sqrt(1 - sin^2(t))) = 3
To simplify the equation, we can multiply both sides by (-sqrt(1 - sin^2(t))):
sin(t) = 3 * (-sqrt(1 - sin^2(t)))
Squaring both sides of the equation removes the square root:
sin^2(t) = 9(1 - sin^2(t))
Expanding the equation:
sin^2(t) = 9 - 9sin^2(t)
Rearranging the terms:
10sin^2(t) = 9
By dividing both sides by 10:
sin^2(t) = 9/10
Taking the square root of both sides:
sin(t) = ± sqrt(9/10)
Since sin(t) is negative in quadrant III, we take the negative square root:
sin(t) = -sqrt(9/10) = -3/sqrt(10)
Now that we have the value of sin(t), we can find cos(t) using the Pythagorean identity:
1 = sin^2(t) + cos^2(t)
cos^2(t) = 1 - sin^2(t)
cos^2(t) = 1 - (-3/sqrt(10))^2
cos^2(t) = 1 - 9/10
cos^2(t) = 1/10
Taking the square root of both sides:
cos(t) = ± sqrt(1/10)
Since cos(t) is negative in quadrant III, we take the negative square root:
cos(t) = -sqrt(1/10) = -1/sqrt(10)
Now, to find cot(t) (which is the reciprocal of tan(t)):
cot(t) = 1 / tan(t) = 1 / 3
Therefore, the values of the trigonometric functions of t are:
sin(t) = -3/sqrt(10)
cos(t) = -1/sqrt(10)
tan(t) = 3
cot(t) = 1/3
in QIII, x and y are both negative
so, we have x = -1 y = -3, h=√10
sin t = y/h = -3/√10
cos t = x/h = -1/√10
and their reciprocals.