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, So…

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, So…

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, So…

We want an angle θ, with whose cosecant is 2. That is, an angle θ in Q4 or Q1 whose sine equals 1/2.

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, So… Notice that this has the same sign as u.