Section 7.2

The Inverse Trigonometric Functions (Continued)

We want sin θ, where θ is the angle whose tangent is ½.θ will be an angle in the first quadrant, so sin θ will be positive.

θ

1

2

And Pythagoras says to me, he says,

5 So…

5

1

2

1tansin 1

We want cos θ, where θ is the angle whose sine is –1/3.θ will be an angle in Quadrant IV, so cos θ will be positive.

θ

1

3

And Pythagoras says to me, he says,

8

So…

3

8

3

1sincos 1

We want tan θ, where θ is the angle whose cosine is –1/3.θ will be an angle in Quadrant II, so tan θ will be negative.

θ

1

3

And Pythagoras says to me, he says,

8So…

1

8

3

1costan 1

We want an angle θ, with

whose cosecant is 2. That is,

an angle θ in Q4 or Q1 whose sine equals 1/2.

62csc 1

We want sin θ, where θ is the angle (Q4 or Q1) whose tangent is u.θ will have the same sign as u, and so will sin θ.

θ

u

1

And Pythagoras says to me, he says,21 u So…

2

1

1tansin

u

uu

Notice that this has the same sign as u.