Transcript Slide 1

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.
And Pythagoras says to me, he says,
5
So…
1
θ
2
 1 1  1
sin tan  
2
5

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.
And Pythagoras says to me, he says,
3
So…
1
θ
8
 1  1 
8
cossin    
 3  3

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.
And Pythagoras says to me, he says,
3
So…
8
θ
1
 1  1 
8
tancos     
1
 3 

We want an angle θ, with
whose cosecant is 2. That is,
an angle θ in Q4 or Q1 whose sine equals 1/2.
1
csc 2 

6
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.
And Pythagoras says to me, he says,
1 u2
So…
u
θ
1

1

sin tan u 
u
1 u2
Notice that this has the same sign as u.