College Trigonometry, 2011a

10.3 The Six Circular Functions and Fundamental Identities 751
)(
sin(θ)
cos(θ)
At this point, it is worth pausing to remind ourselves of our goal. We wish to transform
this expression into 6 sec(θ) ) tan(θ). Using a reciprocal and quotient identity, we find
. In other words, we need to get cosines in our denomina-
6 sec(θ) tan(θ) =6(
1
cos(θ)
tor. Theorem 10.8 tells us 1 − sin 2 (θ) =cos 2 (θ) soweget:
3
1 − sin(θ) − 3
1+sin(θ)
6sin(θ)
=
1 − sin 2 (θ) = 6sin(θ)
cos 2 (θ)
( )( )
1 sin(θ)
= 6
= 6 sec(θ) tan(θ)
cos(θ) cos(θ)
6. It is debatable which side of the identity is more complicated. One thing which stands out
is that the denominator on the left hand side is 1 − cos(θ), while the numerator of the right
hand side is 1 + cos(θ). This suggests the strategy of starting with the left hand side and
multiplying the numerator and denominator by the quantity 1 + cos(θ):
sin(θ)
1 − cos(θ)
=
sin(θ) (1 + cos(θ))
·
(1 − cos(θ)) (1 + cos(θ)) = sin(θ)(1 + cos(θ))
(1 − cos(θ))(1 + cos(θ))
sin(θ)(1 + cos(θ))
=
1 − cos 2 (θ)
= ✘✘ sin(θ)(1 ✘ + cos(θ))
✘sin(θ)sin(θ)
✘ ✘
=
sin(θ)(1 + cos(θ))
sin 2 (θ)
= 1+cos(θ)
sin(θ)
In Example 10.3.3 number 6 above, we see that multiplying 1 − cos(θ) by 1 + cos(θ) produces a
difference of squares that can be simplified to one term using Theorem 10.8. Thisisexactlythe
same kind of phenomenon that occurs when we multiply expressions such as 1 − √ 2by1+ √ 2
or 3 − 4i by 3 + 4i. (Can you recall instances from Algebra where we did such things?) For this
reason, the quantities (1 − cos(θ)) and (1 + cos(θ)) are called ‘Pythagorean Conjugates.’ Below is
a list of other common Pythagorean Conjugates.
Pythagorean Conjugates
ˆ 1 − cos(θ) and 1 + cos(θ): (1 − cos(θ))(1 + cos(θ)) = 1 − cos 2 (θ) =sin 2 (θ)
ˆ 1 − sin(θ) and 1 + sin(θ): (1 − sin(θ))(1 + sin(θ)) = 1 − sin 2 (θ) =cos 2 (θ)
ˆ sec(θ) − 1 and sec(θ) + 1: (sec(θ) − 1)(sec(θ) + 1) = sec 2 (θ) − 1=tan 2 (θ)
ˆ sec(θ)−tan(θ) and sec(θ)+tan(θ): (sec(θ)−tan(θ))(sec(θ)+tan(θ)) = sec 2 (θ)−tan 2 (θ) =1
ˆ csc(θ) − 1 and csc(θ) + 1: (csc(θ) − 1)(csc(θ) + 1) = csc 2 (θ) − 1=cot 2 (θ)
ˆ csc(θ)−cot(θ) and csc(θ)+cot(θ): (csc(θ)−cot(θ))(csc(θ)+cot(θ)) = csc 2 (θ)−cot 2 (θ) =1