To find the terminal point P determined by any value of t, you we can use the following steps:
1. Find the reference number .
2. Find the terminal point Q(a, b) determined by t'.
3. The terminal point determined by t is P(±a, ±b), where the signs are chosen according to the quadrant in which this terminal point lies.
Find the terminal point determined by each given real number t.
Example 1 :
t = 5π/6
Solution :
The reference numbers associated with these values of t were found in Example 9.
The reference number is t' = π/6, which determines the terminal point (√3/2, 1/2) from the table shown above.
Because the terminal point determined by t is in quadrant II, its x-coordinate is negative and its y-coordinate is positive. Thus, the desired terminal point is
(-√3/2, 1/2)
Example 2 :
t = 7π/4
Solution :
The reference numbers associated with these values of t were found in Example 10.
The reference number is t' = π/4, which determines the terminal point (√2/2, √2/2) from the table shown above.
Because the terminal point is in quadrant IV, its x-coordinate is positive and its y-coordinate is negative. Thus, the desired terminal point is
(√2/2, -√2/2)
Example 3 :
t = -2π/3
Solution :
The reference numbers associated with these values of t were found in Example 11.
The reference number is t' = π/3, which determines the terminal point (1/2, √3/2) from the table shown above.
Because the terminal point determined by t is in quadrant III, its both x-coordinate and y-coordinate are negative. Thus, the desired terminal point is
(-1/2, -√3/2)
Because the circumference of the unit circle is 2π, the terminal point determined by t is the same as that determined by (t + 2π) or (t - 2π). In general, we can add or subtract 2π any number of times without changing the terminal point determined by t. We can use this observation in the next example to find terminal points for large t.
Example 4 :
Find the terminal point determined by t = 29π/6.
Solution :
Because
t = 29π/6 = 4π + 5π/6
we see that the terminal point of t is the same as that of 5π/6 (that is, we subtract 4π).
So by Example 13 the terminal point is
(-√3/2, 1/2)
2. Terminal Points on the Unit Circle
3. Reference Number on the Unit Circle
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