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7
Analytic Trigonometry
7.1
Trigonometric Identities
Let’s begin by listing the identities we already know.
Reciprocal Identities:
csc θ =
1
sin θ
sec θ =
1
cos θ
cot θ =
tan θ =
sin θ
cos θ
cot θ =
cos θ
sin θ
1
tan θ
Pythagorean Identities:
sin2 θ + cos2 θ = 1
tan2 θ + 1 = sec2 θ
cot2 θ + 1 = csc2 θ
Even/Odd Identities:
sin(−θ) = − sin θ
cos(−θ) = cos θ
tan(−θ) = − tan θ
Cofunction Identities:
sin
tan
π
2
π
2
π
sec
2
− θ = cos θ
− θ = cot θ
− θ = csc θ
cos
cot
csc
π
π2
π2
2
− θ = sin θ
− θ = tan θ
− θ = sec θ
Examples:
1. Simplify cos(t) + tan(t) sin(t)
solution:
sin(t)
sin(t)
cos(t)
sin2 (t)
cos(t) +
cos(t)
2
cos (t) + sin2 (t)
cos(t)
1
cos(t)
sec(t)
cos(t) + tan(t) sin(t) = cos(t) +
=
=
=
=
2. Simplify
sin θ
cos θ
+
.
cos θ 1 + sin θ
solution:
sin θ
cos θ
sin θ(1 + sin θ) + cos2 θ
+
=
cos θ 1 + sin θ
cos θ(1 + sin θ)
sin θ + sin2 θ + cos2 θ
=
cos θ(1 + sin θ)
sin θ + 1
=
cos θ(1 + sin θ)
1
= sec θ
=
cos θ
3. Verify the identity
cos θ(sec θ − cos θ) = sin2 θ
solution: To verify identities we have to transform one side until it looks like the other
- do not move terms from side to side.
LHS = cos θ(sec θ − cos θ)
1
− cos θ
= cos θ
cos θ
1 − cos2 θ
= cos θ
cos θ
2
= 1 − cos θ
= sin2 θ = RHS
4. Verify the identity
2 tan θ sec θ =
1
1
−
1 − sin θ 1 + sin θ
solution: We work with the right hand side
RHS =
=
=
=
=
1
1
−
1 − sin θ 1 + sin θ
1 + sin θ − (1 − sin θ)
(1 − sin θ)(1 + sin θ)
2 sin θ
1 − sin2 θ
2 sin θ
cos2 θ
2 tan θ sec θ = LHS
5. Verify the identity
cos u
= sec u + tan u
1 − sin u
solution: This one is a little trickier. We start with the right hand side:
RHS = sec u + tan u
sin u
1
+
=
cos u cos u
1 + sin u
=
cos u
Now we appear to be stuck. We want this to look like the LHS which has 1 − sin u in
the denominator, so we could try multiplying the RHS by this on the top and bottom
to see what happens:
RHS =
=
=
=
=
1 + sin u 1 − sin u
·
cos u
1 − sin u
(1 + sin u)(1 − sin u)
cos u(1 − sin u)
1 − sin2 u
cos u(1 − sin u)
cos2 u
cos u(1 − sin u)
cos u
= LHS
1 − sin u
6. Verify the identity
tan2 θ
1 + cos θ
=
cos θ
sec θ − 1
solution: We should start on the right hand side and convert to sines and coses:
RHS =
=
=
tan2 θ
sec θ − 1
sin2 θ
cos2 θ
1
−
cos θ
2
sin θ
cos2 θ
1−cos θ
cos θ
2
1
sin θ
cos θ
·
2
cos θ 1 − cos θ
(1 − cos2 θ) cos θ
=
cos2 θ(1 − cosθ )
(1 − cos θ)(1 + cos θ)
=
cos θ(1 − cosθ )
1 + cos θ
= RHS
=
cos θ
=
7.2
Addition and Subtraction Formulas
Formulas for sine:
sin(s + t) = sin s cos t + cos s sin t
sin(s − t) = sin s cos t − cos s sin t
Formulas for cosine:
cos(s + t) = cos s cos t − sin s sin t
cos(s − t) = cos s cos t + sin s sin t
Formulas for tangent:
tan s + tan t
1 − tan s tan t
tan s − tan t
tan(s − t) =
1 + tan s tan t
π
Example: Find cos(75◦ ) and cos 12
tan(s + t) =
solution:
cos(75◦ ) = cos(45◦ + 30◦ )
= cos(45◦ ) cos(30◦ ) − sin(45◦ ) sin(30◦ )
√ √
√
2 3
21
=
−
√2 2 √ 2 2
6− 2
=
4
cos
π
12
=
=
=
=
π
π
cos
−
3 4 π π π
π
cos
+ sin
sin
cos
3
4
√ 3 √ √4
1 2
3 2
+
2√ 2 √ 2 2
2+ 6
4
Expressions of the form A sin x + B cos x
We can use the sum and difference identities to rewrite expressions of the form A sin x +
B cos x as something simpler. The trick is to simultaneously transform the A into something
that looks like sin φ and transform the B into something that looks like cos φ. Then we
can use a sum identity. We do this by imagining a point in the xy-plane with coordinates
(A, B). If φ is the angle between the line connecting (A, B) with the origin and the x-axis,
then
B
A
sin φ = √
cos φ = √
2
2
2
A +B
A + B2
Thus we can rewrite, using the sum identity:
√
A
B
2
2
A sin x + B cos x =
A +B √
sin x + √
cos x
A2 + B 2
A2 + B 2
√
=
A2 + B 2 (cos φ sin x + sin φ cos x)
√
=
A2 + B 2 sin(x + φ)
Example: Express 3 sin x + 4 cos x in the form k sin(x + φ)
√
√
solution: Here, k = A2 + B 2 = 32 + 42 = 5 also, sin φ = 45 , cos φ =
radians.
Thus,
3 sin x + 4 cos x = 5 sin(x + 0.927)
7.3
Double-Angle, Half-Angle, and Product-sum Formulas
We state the double-angle identites:
sin 2x = 2 sin x cos x
3
5
so φ = 0.927
cos 2x = cos2 x − sin2 x
= 1 − 2 sin2 x
= 2 cos2 x − 1
tan 2x =
2 tan x
1 − tan2 x
Example: If cos x = − 23 and x is in quadrant II, find cos 2x and sin 2x.
solution:
2
2
8
1
cos 2x = 2 cos x − 1 = 2 −
−1= −1=−
3
9
9
2
sin 2x = 2 sin x cos x
√
Since x is in quadrant II, we can sub in sin x = 1 − cos2 x:
r
r
√
s
2
2
4 5
4 5
2
4
4
=−
sin 2x = 2 −
1− −
=−
1− =−
3
3
3
9
3 9
9
Formulas for Lowering Powers
sin2 x =
1 − cos 2x
2
tan2 x =
cos2 x =
1 + cos 2x
2
1 − cos 2x
1 + cos 2x
Half-angle formulas
sin
u
2
r
=±
r
u
1 − cos u
1 + cos u
cos
=±
2
2
2
u 1 − cos u
tan
=
2
sin u
The ± is chosen in the first two depending on what quadrant
u
2
is in.
Example: Find the exact value of sin 22.5◦ .
solution: Since 22.5◦ is half of 45◦ , we use the half-angle formula for sin with u = 45◦ .
Since 22.5◦ is in the first quadrant, we choose the + sign.
◦
45
◦
sin 22.5 = sin
2
r
1 − cos 45◦
=
2
s
√
1 − 22
=
2
s
√
2− 2
=
4
Example: Fine tan
u
2
if sin u =
2
5
and u is in quadrant II.
solution: From the formula,
tan
u
2
=
1 − cos u
sin u
We know sin u = 25 , and from the usual pythagorean identity we know
p
p
2
cos u = ± 1 − sin u = − 1 − sin2 u
We choose the negative because cos is negative in the second quadrant. Hence
u
1 − cos u
=
tan
2
sinpu
1 + 1 − sin2 u
=
qsin u
1 − ( 52 )2
1+
=
1+
=
21
25
2/5
5
=
2
=
2/5
q
5+
5+
√ !
21
5
√
21
2
Product-to-Sum Formulas
1
[sin(u + v) + sin(u − v)]
2
1
cos u sin v =
[sin(u + v) − sin(u − v)]
2
1
cos u cos v =
[cos(u + v) + cos(u − v)]
2
1
sin u sin v =
[sin(u − v) − sin(u + v)]
2
sin u cos v =
Example: Express sin 3x sin 5x as a sum of trig functions.
solution: We use the fourth product-to-sum formula:
1
1
1
sin 3x sin 5x = [sin(3x − 5x) − sin(3x + 5x)] = (cos(−2x) − cos 8x) = (cos(2x) − cos 8x)
2
2
2
Sum-to-Product Formulas
sin x + sin y =
sin x − sin y =
cos x + cos y =
cos x − cos y =
x+y
x−y
2 sin
cos
2
2
x+y
x−y
2 cos
sin
2
2
x+y
x−y
2 cos
cos
2
2
x−y
x+y
sin
−2 sin
2
2
Example: Write sin 7x + sin 3x as a product.
solution: We use the first formula:
7x + 3x
7x − 3x
sin 7x + sin 3x = 2 sin
cos
= 2 sin 5x cos 2x
2
2
7.4
Inverse Trigonometric Functions
From looking at the graphs of the trig functions, we see that they fail the horizontal line
test spectacularly. However, if you restrict their domain, you can find an inverse for these
functions on this domain only.
Inverse Sine
The function sin is one-to-one when restricted to sin : − π2 , π2 → [−1, 1]. Thus there exists
a function sin−1 or arcsin
h π πi
sin−1 : [−1, 1] → − ,
2 2
arcsin x(= sin−1 x)
sin x
The arcsin function satisfies
arcsin(x) = y ⇔ sin(y) = x
π
π
≤x≤
2
2
sin(arcsin x) = x if − 1 ≤ x ≤ 1
arcsin(sin x) = x if −
Examples: Find (a)sin−1 ( 12 ), (b)arcsin(− 21 ), (c)sin−1 ( 32 ) and (d)cos(sin−1 ( 35 )).
solution:
(a) We know that sin( π6 ) = 12 , so sin−1 ( 12 ) = π6 .
(b) Likewise, sin(− π6 ) = − 21 , so arcsin(− 12 ) = − π6 .
(c) We know sin x is never 32 (it is never greater than 1), so sin−1 ( 32 ) is undefined.
p
(d) For u ∈ − π2 , π2 , we have cos(u) = 1 − sin2 u, so
s
3
3
−1
2
−1
cos sin
=
1 − sin sin
5
5
s
2
3
=
1−
5
r
9
1−
=
25
r
16
=
25
4
=
5
Inverse Cosine
The function cos is one-to-one when restricted to cos : [0, π] → [−1, 1]. Thus there exists a
function cos−1 or arccos
cos−1 : [−1, 1] → [0, π]
cos x
arccos x(= cos−1 x)
The arccos function satisfies
arccos(x) = y ⇔ cos(y) = x
arccos(cos x) = x if 0 ≤ x ≤ π
cos(arccos x) = x if − 1 ≤ x ≤ 1
√ 3
−1
and (b) cos−1 (0).
Examples: Find (a) cos
2
solution:
(a) cos( π6 ) =
√
3
,
2
thus cos−1
√ 3
2
(b) cos( π2 ) = 0, thus cos−1 (0) =
= π6 .
π
2
Examples:
√
1. Show that sin(cos−1 (x)) = 1 − x2 .
p
solution: sin u = 1 − cos2 (u) for u ∈ [0, π]. Thus
p
√
sin(cos−1 (x)) = 1 − cos2 (cos−1 (x)) = 1 − x2
2. Show that tan(cos−1 (x)) =
solution: tan(u) =
sin u
,
cos u
√
1−x2
.
x
so
sin(cos−1 (x))
tan(cos (x)) =
=
cos(cos−1 (x))
−1
√
1 − x2
x
√
3. Show that sin(2 cos−1 (x)) = 2x 1 − x2
solution: sin 2u = 2 sin u cos u, so
√
√
sin(2 cos−1 (x) = 2 sin(cos−1 (x)) cos(cos−1 (x)) = 2 1 − x2 x = 2x 1 − x2
Inverse Tangent
The function tan is one-to-one when restricted to tan : − π2 , π2 → R. Thus there exists a
function tan−1 or arctan
h π πi
−1
tan : R → − ,
2 2
arctan x(= tan−1 x)
tan x
The arctan function satisfies
arctan(x) = y ⇔ tan(y) = x
π
π
≤x≤
2
2
tan(arctan x) = x if x ∈ R
arctan(tan x) = x if −
Examples: Find (a) tan−1 (1), (b) tan−1
√
3 and (c)arctan(−20)
solution:
(a) tan( π4 ) = 1, thus tan−1 (1) = π4 .
√
√
(b) tan( π3 ) = 3, thus tan−1 ( 3) =
π
3
(c) Using a calculator, we find arctan(−20) ≈ 1.52084
Other Inverse Trig Functions
The other trig functions cot, csc and sec also have inverses when restricted to suitable
domains, namely cot−1 , csc−1 and sec−1 . You don’t need to worry about graphing these.
Just keep in mind:
1
cot−1 6=
tan−1
1
csc−1 6=
sin−1
1
sec−1 6=
cos−1
7.5
Trigonometric Equations
One frequently has to solve equations involving trig functions. Sometimes the values of x
you look at are restricted, while others you are asked to find all the values of x that make
a given equation true. In the latter case, there are usually an infinite number of solutions
whose form depends on the period of the trig functions involved.
Examples:
1. Solve
(a) tan2 (x) − 3 = 0
(b) sin(x) = cos(x)
(c) 1 + sin x = 2 cos2 (x)
(d) sin 2x − cos x = 0
(e) cos(x) + 1 = sin x with t ∈ [0, 2π]
solution:
√
(a) tan2 (x) − 3 = 0 ⇒ tan(x) = ± 3.
√
tan(x) = 3 when x = π3 . But tan has period π, so
√
π
tan(x) = 3 ⇒ x = + kπ k ∈ Z
3
Likewise,
√
π
tan(x) = − 3 ⇒ x = − + kπ k ∈ Z
3
Thus the full set of solutions is
n π
o
π
S = − + kπ, x = + kπ | k ∈ Z
3
3
(b) If sin(x) = cos(x), cos(x) 6= 0 because sin and cos are not 0 in the same places.
Thus we can divide both sides by cos and get
sin(x)
= 1 ⇒ tan(x) = 1
cos(x)
This happens when x = π4 . Once again, tan has a period of π so
π
x = + kπ k ∈ Z
4
(c) We can get the equation 1 + sin x = 2 cos2 (x) into an easier form to deal with by
subbing in 1 − sin2 x for cos2 x:
1 + sin x = 2 cos2 (x)
1 + sin x = 2 − 2 sin2 x
2 sin2 x + sin x − 1 = 0
This is a quadratic in sin x that factors as
(2 sin x − 1)(sin x + 1) = 0
Which means either 2 sin x − 1 = 0 or sin x + 1 = 0. The first gives us that
sin x = 12 , and the second gives sin x = −1.
• sin x = −1 ⇒ x =
• sin x =
1
2
⇒x=
π
6
3π
2
+ 2kπ, k ∈ Z
+ 2kπ, k ∈ Z or x =
5π
6
+ 2kπ, k ∈ Z
(d) We use a double angle identity on sin 2x − cos x = 0 to get
2 sin x cos x − cos x = 0
cos x(2 sin x − 1) = 0
Hence either
• cos x = 0 ⇒ x =
• sin x =
1
2
⇒x=
π
2
π
6
+ 2kπ, k ∈ Z or x =
+ 2kπ, k ∈ Z or x =
3π
2
5π
6
+ 2kπ, k ∈ Z.
+ 2kπ, k ∈ Z.
(e) Here we have to be a little trickier and square both sides
cos x + 1 = sin x
(cos x + 1)2 = 1 − cos2 x
cos2 x + 2 cos x + 1 = 1 − cos2 x
2 cos2 x + 2 cos x = 0
2 cos x(cos x − 1) = 0
Hence either cos x = 0 or cos x = −1. Between 0 and 2π, the first only happens
at π2 and 3π
and the second happens at −π. Hence the three solutions are
2
x=
π 3π
, ,π
2 2
2. Consider the equation 2 sin(3x) − 1 = 0.
(a) Find all the solutions to the equation.
(b) Find all the solutions to the equation in the interval [0, 2π).
solution:
(a) The equation rearranges to sin(3x) = 12 . As we’ve seen before, this means that
π
5π
+ 2kπ, k ∈ Z or 3x =
+ 2kπ, k ∈ Z.
6
6
π
2kπ
5π 2kπ
x=
+
, k ∈ Z or x =
+
, k ∈ Z.
18
3
18
3
3x =
(b) We solve the inequalities
π
2kπ
+
< 2π
18
3
1
2k
0≤
+
<2
18
3
2k
1
35
1
<2−
=
− ≤
18
3
18
18
1
35
− ≤ 2k <
6
6
35
1
≈ 2.91
− ≤k<
12
12
The ks in this range are k = 0, k = 1 and k = 2. Hence
0≤
x=
π 13π 25π
,
,
18 18 18
For the other solutions we have
0≤
5π 2kπ
+
< 2π
18
3
5
2k
+
<2
18
3
5
2k
5
31
− ≤
<2−
=
18
3
18
18
31
5
− ≤ 2k <
6
6
31
5
≈ 2.58
− ≤k<
12
12
The ks in this range are k = 0, k = 1 and k = 2. Hence
0≤
x=
3. Consider the equation
√
5π 17π 29π
,
,
18 18 18
3 tan( x2 ) − 1 = 0.
(a) Find all the solutions of the equation.
(b) Find all the solutions in the interval [0, 4π).
solution:
(a) This rearranges to tan( x2 ) =
√1
3
This gives us
x
π
= + kπ, k ∈ Z
2
6
π
x = + 2kπ, k ∈ Z
3
(b) As before, we find the ks we need:
0≤
π
+ 2kπ < 4π
3
1
+ 2k < 4
3
1
1
11
− ≤ 2k < 4 − =
3
3
3
1
11
− ≤k<
≈ 1.83
6
6
The ks in this range are k = 0 and k = 1. Hence x =
0≤
π
3
or x =
7π
.
3
4. Solve the equation tan2 x − tan x − 2 = 0.
solution: This is a quadratic in tan x, so
tan2 x − tan x − 2 = 0
(tan x − 2)(tan x + 1) = 0
So tan x = 2 or tan x = −1. There isn’t a convenient angle with tan x = 2, but we can
use tan−1 to write
x = tan−1 (2) + kπ, k ∈ Z
The second one of course gives us
x=−
π
+ kπ, k ∈ Z
4
5. Solve the equation 3 sin θ − 2 = 0.
solution: This rearranges to sin θ = 32 . Once again there is no easy angle that gives
us sin θ = 23 . We know that sin is positive in the first and second quadrants, and thus
the two solutions in [0, 2π) are
2
2
−1
−1
sin
and π − sin
3
3
Hence the full solution is
−1
θ = sin
2
2
−1
+ 2kπ or θ = π − sin
+ 2kπ
3
3