MTH112-007, Fall 2014
Jack Wiedrick
Trigonometric Identities — Practice Problems
KEY
1. (MODERATE)
tan +
=
(Undefined when cos = 0)
The hint was: Subtract tan from both sides and cross-multiply. The reason for this is to deal efficiently
with the inconvenient 1 + sin term in the denominator.
cos 1
1
sin =
− tan =
−
1 + sin cos cos cos cos 1 − sin ⟺
=
1 + sin cos ⟺ 1 + sin 1 − sin = cos cos ⟺ 1 − sin = cos True!
tan + 1 = sec 2. (EASY)
(Undefined when cos = 0)
The hint was: Multiply both sides by cos . It should be obvious by now why this works.
tan + 1 cos = sec cos sin 1
⟺
cos + cos =
cos cos cos ⟺ sin + cos = 1
True!
3. (MODERATE)
− = 2 tan sec (Undefined when sin = ±1 ⟺ cos = 0)
The hint was: Get a common denominator (conjugates). The key is recognizing that the denominators on
the left-hand side are conjugates whose product is 1 − sin = cos .
True!
1
1 + sin 1
1 − sin 1 + sin − 1 − sin ∙
−
∙
=
1 − sin 1 + sin 1 + sin 1 − sin 1 − sin 1 + sin − 1 − sin 2 sin =
=
1 − sin cos 2 sin 1
=
∙
cos cos = 2 tan sec 4. (EASY)
tan + cot = sec csc (Undefined when cos = 0
)
sin = 0
The hint was: Get a common denominator. Note that this problem already appeared in the identities
walkthrough handout and was solved there. First write everything in terms of sine and cosine.
cos ! sin ! cos ! cos ! sin ! sin !
+
=
∙
+
∙
sin ! cos ! sin ! cos ! cos ! sin !
cos ! + sin !
1
=
=
cos ! sin !
cos ! sin !
= sec csc tan + cot =
True!
5. (EASY)
"#$ "#$ = $ $ (Undefined when cos = 0
or tan = ±1 ⟺ |sin | = |cos |
The hint was: Multiply top and bottom by cos . The idea is to cancel out the cos factors in the
denominator of the tan terms. (But the division is only valid when cos ≠ 0.)
sin 1 + tan cos ( ∙ cos = cos + sin =
'
sin 1 − tan cos cos − sin 1−
cos 1
=
cos − sin 1+
True!
6. (EASY)
tan − sin = tan sin (Undefined when cos = 0)
The hint was: Factor out a sin . Note that this problem already appeared in a quiz and was solved
there. First write everything in terms of sine and cosine.
sin 1
tan − sin =
− sin = sin ) − 1*
cos cos = sin sec − 1
= sin tan True!
7. (MODERATE)
+
= 2 csc (Undefined when sin = 0 ⟺ cos = ±1)
The hint was: Get a common denominator and expand. The tricky part here is to realize that multiplying
the left numerator on the left-hand side by 1 − cos and the right numerator by sin will create a
cos + sin = 1 term.
True!
8. (EASY)
1 − cos 1 − cos sin sin 1 − cos + sin ∙
+
∙
=
sin 1 − cos 1 − cos sin sin 1 − cos 1 − 2 cos + cos + sin =
sin 1 − cos 1 − 2 cos + 1
2 − 2 cos =
=
sin 1 − cos sin 1 − cos 21 − cos 2
=
=
sin 1 − cos sin = 2 csc + + =
(Undefined when cos = 0 or cos = −1)
The hint was: Multiply top and bottom by cos . This one should be completely obvious to you.
1
sec − 1
cos
− 1- ∙ cos = 1 − cos =,
1
sec + 1
+ 1 cos 1 + cos cos True!
9. (EASY)
1 + cot = csc (Undefined when sin = 0)
The hint was: Multiply both sides by sin . This is just another simple transformation of the usual
Pythagorean identity.
1 + cot sin = csc sin cos 1
⟺ .1 +
/ sin =
sin sin sin ⟺ sin + cos = 1
True!
10. (EASY)
$ $ = cos (Left-hand side undefined when sin = 0)
The hint was: Expand the fraction. Note that the expression on the left-hand side can be undefined for
some angles (where csc = ±∞), but the right-hand side is never undefined; this case is an example
where an "undefined" expression actually has a real value, but you need some more math to explain it.
True!
csc − 1 csc 1
=
−
= 1 − sin = cos csc csc csc 11. (EASY)
" " =
"# "# (Undefined when tan = 0 or tan = −1 ⟺ cot = −1)
The hint was: Multiply top and bottom by tan . This one should also be completely obvious to you.
1
− 1 tan 1 − tan cot − 1
tan
=,
-∙
=
1
tan
1 + tan cot + 1
+1
tan True!
12. (HARD)
sin + cos tan + cot = sec + csc (Undefined when cos = 0
)
sin = 0
The hint was: Expand, get common denominator, then rearrange and factor. This is one where it takes a
few steps before you even begin to see the way through. Many problems are like this, and it's important
to make sure that the first few steps are reasonably likely to lead to a simplification.
sin + cos tan + cot = sin tan + cos tan + sin cot + cos cot sin sin cos cos = sin + cos + sin + cos cos cos sin sin sin cos =
+ sin + cos +
cos sin After expanding, we see a sin and a cos floating around, but we can't use them yet because they
don't have the same denominator.
cos sin sin cos sin cos cos sin + sin + cos +
=
∙
+ sin + cos ∙
+
∙
cos sin cos sin cos sin sin cos 1
1
sin + cos sin + sin cos + cos =
cos sin Now it looks more complicated, but it actually isn't because we have some factoring opportunities that
didn't exist in the original expression. The key here is to rearrange the numerator and factor in such a
way that we create cos + sin terms that we can turn into 1s.
sin1 + cos sin + sin cos + cos 1 cos 1 + cos sin + sin cos + sin1 =
cos sin cos sin cos cos + sin + sin cos + sin =
cos sin cos 1 + sin 1
=
cos sin After this, expanding the fraction reveals the solution.
True!
cos + sin cos sin 1
1
=
+
=
+
= csc + sec cos sin cos sin cos sin sin cos 13. (MODERATE)
2 2 = 1 − sin cos (Undefined when cos = − sin )
The hint was: Factor the numerator and do cancellations. You may remember from basic algebra that
sums and differences of perfect cubes can always be factored. The rule is this:
31 ± 4 1 = 3 ± 43 ∓ 34 + 4 Applying this rule to the numerator of the left-hand-side immediately reveals a cancellation.
sin1 + cos 1 sin + cos sin − sin cos + cos =
sin + cos sin + cos = sin − sin cos + cos = cos + sin − sin cos = 1 − sin cos True!
14. (EASY)
2 = (Undefined when sin = 0 ⟺ cos = ±1)
The hint was: Cross-multiply. This is because the numerator of the left-hand side and the denominator
of the right-hand side are conjugates.
cos + 1 1 + cos csc =
=
1
1
sin sin 1 − cos ⟺ 1 + cos 1 − cos = sin1 csc 1
⟺ 1 − cos = sin1 = sin sin True!
15. (MODERATE)
− = 4 tan sec (Undefined when sin = ±1 ⟺ cos = 0)
The hint was: Get a common denominator (conjugates). As with some of the previous problems, the act
of getting a common denominator introduces a conjugate product that simplifies everything.
True!
1 + sin 1 + sin 1 − sin 1 − sin 1 + sin − 1 − sin ∙
−
∙
=
1 − sin 1 − sin 1 + sin 1 + sin 1 − sin 1 + 2 sin + sin − 1 − 2 sin + sin =
1 − sin 1 + 2 sin + sin − 1 + 2 sin − sin =
1 − sin 4 sin sin 1
=
=
4
∙
∙
cos cos cos = 4 tan sec 16. (EASY)
csc 7 − cot 7 = csc + cot (Undefined when cos = 0
)
sin = 0
The hint was: Factor. You should immediately think of this as the first thing to do here, as well as look for
Pythagorean identities involving csc and cot .
True!
17. (MODERATE)
csc 7 − cot 7 = csc − cot csc + cot = 1csc + cot = csc + cot $ $ 1 = (Left-hand side undefined when cos = −1)
The hint was: Factor the numerator and denominator and do cancellations. This example is a case where
the "identity" is not strictly true because the left-hand side is undefined for angles where the right-hand
side is not undefined; the mismatch comes about as a result of cancelling a zeroable factor.
1 − cos sin =
cos + 3 cos + 2 cos + 3 cos + 2
1 − cos 1 + cos =
cos + 2cos + 1
1 − cos =
2 + cos True! (With the above caveat.)
18. (HARD)
"# "# 9
" " 9
= tan tan :
(Undefined when cot : = − cot or when
cos : = 0
cos = 0
any of or ;
< is true)
sin : = 0
sin = 0
The hint was: Convert everything to sine and cosine and get common denominators for both top and
bottom (and look for a tricky cancellation!). This problem is genuinely tricky because we have to assume
that and : are distinct angles, which means we can't combine expressions the way we usually do when
we only have a single angle to worry about. The left-hand side and right-hand side are also undefined for
different sets of angle pairs , : because of cancellation of zeroable factors, so here is another case
where the "identity" is true only in cases where both sides are defined. The key to solving this is to mess
around with the left-hand side and look for any cancellation possible.
sin sin : sin cos : sin : cos tan + tan : cos + cos : cos ∙ cos : + cos : ∙ cos =
=
cot + cot : cos + cos : cos ∙ sin : + cos : ∙ sin sin sin :
sin sin : sin : sin sin cos : + cos sin :
cos cos :
=
cos sin : + sin cos :
sin sin :
Do you see it? The numerator of the numerator and the numerator of the denominator are both the
same expression! Those cancel when we perform the division.
sin cos : + cos sin :
sin cos : + cos sin :
sin sin :
cos cos :
=
∙
cos sin : + sin cos :
cos cos :
cos sin : + sin cos :
sin sin :
sin sin :
sin sin :
=
=
∙
cos cos : cos cos :
= tan tan :
True! (With the above caveat.)
19. (EASY)
"# "# = (Undefined when cos = 0 or tan = 1 ⟺ sin = cos )
The hint was: Multiply top and bottom by cos . Another one that should be completely obvious to you.
sin 1 + cos cos cos + sin 1 + tan =,
-∙
=
sin 1 − tan 1 − cos cos cos − sin True!
20. (EASY)
sin − tan cos − cot = sin − 1cos − 1
(Undefined when
cos = 0
)
sin = 0
The hint was: Expand both sides. Notice that the right-hand side is never undefined, so this is another
example where the "identity" is not strictly true for all angles; as usual, cancellation of a zeroable factor
leads to the mismatch. Let's expand each side in turn. Here's the left-hand side:
sin − tan cos − cot = sin cos − sin cot − tan cos + tan cot cos sin 1
= sin cos − sin −
cos + tan sin cos tan = sin cos − cos − sin + 1
And here's the right-hand side:
sin − 1cos − 1 = sin cos − sin − cos + 1
Obviously they're the same expression, so the identity is true!