College Trigonometry, 2011a

774 Foundations of Trigonometry
Example 10.4.2.
1. Find the exact value of sin ( )
19π
12
2. If α is a Quadrant II angle with sin(α) = 5
13
,andβ is a Quadrant III angle with tan(β) =2,
find sin(α − β).
3. Derive a formula for tan(α + β) intermsoftan(α) and tan(β).
Solution.
1. As in Example 10.4.1, we need to write the angle 19π
12
as a sum or difference of common angles.
The denominator of 12 suggests a combination of angles with denominators 3 and 4. One
such combination is 19π
12 = 4π 3 + π 4
. Applying Theorem 10.15, weget
( ) ( 19π
4π
sin = sin
12
3 + π )
4
( ) 4π
( π
) ( ) 4π
( π
)
= sin cos +cos sin
3 4 3 4
( √ )(√ ) (
3 2
= −
+ − 1 ) ( √ )
2
2 2 2 2
= −√ 6 − √ 2
4
2. In order to find sin(α − β) usingTheorem10.15, we need to find cos(α) and both cos(β)
and sin(β). To find cos(α), we use the Pythagorean Identity cos 2 (α) +sin 2 (α) = 1. Since
sin(α) = 5
13 ,wehavecos2 (α)+ ( )
5 2
13 = 1, or cos(α) =±
12
13
.Sinceα is a Quadrant II angle,
cos(α) =− 12
13
. We now set about finding cos(β) and sin(β). We have several ways to proceed,
but the Pythagorean Identity 1 + tan 2 (β) = sec 2 (β) is a quick way to get sec(β), and hence,
cos(β). With tan(β) = 2, we get 1 + 2 2 = sec 2 (β) so that sec(β) =± √ 5. Since β is a
Quadrant III angle, we choose sec(β) =− √ 5socos(β) = 1
sec(β) = 1
√
− √ = − 5
5 5
. We now need
to determine sin(β). We could use The Pythagorean Identity cos 2 (β)+sin 2 (β) = 1, but we
opt instead to use a quotient identity. From tan(β) = sin(β)
cos(β)
,wehavesin(β) =tan(β) cos(β)
( √ )
so we get sin(β) =(2) − 5
5
= − 2√ 5
5
. We now have all the pieces needed to find sin(α − β):
sin(α − β) = sin(α) cos(β) − cos(α)sin(β)
( ) ( √ ) (
5 5
= − − − 12 ) ( )
− 2√ 5
13 5 13 5
= − 29√ 5
65