Chapter 4 Trigonometric Functions

Trigonometric Functions
3. We need values for x, y, and r. Because P = (2, 3) is a
point on the terminal side of θ , x = 2 and y = 3 .
Furthermore,
2 2 2 2
r = x + y = 2 + 3 = 4+ 9 = 13
Now that we know x, y, and r, we can find the six
trigonometric functions of θ .
y 3 3 13 3 13
sinθ
= = = ⋅ =
r 13 13 13 13
x 2 2 13 2 13
cosθ
= = = ⋅ =
r 13 13 13 13
y 3
tanθ
= =
x 2
r 13
cscθ
= =
y 3
r 13
secθ
= =
x 2
x 2
cotθ
= =
y 3
4. We need values for x, y, and r, Because
P = (3, 7) is a point on the terminal side of
θ , x = 3 and y = 7. Furthermore,
2 2 2 2
r = x + y = 3 + 7 = 9+ 49 = 58
Now that we know x, y, and r, we can find the six
trigonometric functions of θ .
y 7 7 58 7 58
sinθ
= = = ⋅ =
r 58 58 58 58
x 3 3 58 3 58
cosθ
= = = ⋅ =
r 58 58 58 58
y 7
tanθ
= =
x 3
r 58
cscθ
= =
y 7
r 58
secθ
= =
x 3
x 3
cotθ
= =
y 7
5. We need values for x, y, and r. Because P = (3, –3) is
a point on the terminal side of θ , x = 3 and y =− 3 .
2 2 2 2
Furthermore, r = x + y = 3 + ( − 3) = 9+
9
= 18 = 3 2
Now that we know x, y, and r, we can find the six
trigonometric functions of θ .
y −3 −1 2 2
sinθ
= = = − ⋅ = −
r 3 2 2 2 2
x 3 1 2 2
cosθ
= = = ⋅ =
r 3 2 2 2 2
y −3
tanθ
= = = −1
x 3
r 3 2
cscθ
= = = − 2
y −3
r 3 2
secθ
= = = 2
x 3
x 3
cotθ
= = = −1
y −3
6. We need values for x, y, and r, Because P = (5, –5) is
a point on the terminal side of θ , x = 5 and y = –5 .
Furthermore,
2 2 2
r = x + y = 5 + ( − 5) = 25 + 25 = 50
= 5 2
Now that we know x, y, and r, we can find the six
trigonometric functions of θ .
y −5 −1 2 2
sinθ
= = = ⋅ =−
r 5 2 2 2 2
x 5 1 2 2
cosθ
= = = ⋅ =
r 5 2 2 2 2
y −5
tanθ
= = = −1
x 5
r 5 2
cscθ
= = = − 2
y −5
r 5 2
secθ
= = = 2
x 5
x 5
cotθ
= = =−1
y −5
520
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