7 Analytic Trigonometry - Faculté des sciences - Faculty of...

7 Analytic Trigonometry

7.1 Trigonometric Identities

Let’s begin by listing the identities we already know.

Reciprocal Identities:

csc θ =1

sin θsec θ =

1

cos θcot θ =

1

tan θ

tan θ =sin θ

cos θcot θ =

cos θ

sin 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(π

2− θ)

= cos θ cos(π

2− θ)

= sin θ

tan(π

2− θ)

= cot θ cot(π

2− θ)

= tan θ

sec(π

2− θ)

= csc θ csc(π

2− θ)

= sec θ

Examples:

1. Simplify cos(t) + tan(t) sin(t)

solution:

cos(t) + tan(t) sin(t) = cos(t) +sin(t)

cos(t)sin(t)

= cos(t) +sin2(t)

cos(t)

=cos2(t) + sin2(t)

cos(t)

=1

cos(t)

= sec(t)

7 ANALYTIC TRIGONOMETRY 99

2. Simplifysin θ

cos θ+

cos θ

1 + sin θ.

solution:

sin θ

cos θ+

cos θ

1 + sin θ=

sin θ(1 + sin θ) + cos2 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

cos θ= sec 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)

= cos θ

(1

cos θ− cos θ

)= cos θ

(1− cos2 θ

cos θ

)= 1− cos2 θ

= sin2 θ = RHS

4. Verify the identity

2 tan θ sec θ =1

1− sin θ− 1

1 + sin θ

solution: We work with the right hand side

RHS =1

1− sin θ− 1

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 identitycosu

1− sinu= secu+ tanu

solution: This one is a little trickier. We start with the right hand side:

RHS = secu+ tanu

=1

cosu+

sinu

cosu

=1 + sinu

cosu

Now we appear to be stuck. We want this to look like the LHS which has 1− sinu in

the denominator, so we could try multiplying the RHS by this on the top and bottom

to see what happens:

RHS =1 + sinu

cosu· 1− sinu

1− sinu

=(1 + sinu)(1− sinu)

cosu(1− sinu)

=1− sin2 u

cosu(1− sinu)

=cos2 u

cosu(1− sinu)

=cosu

1− sinu= LHS

6. Verify the identity1 + cos θ

cos θ=

tan2 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 θcos2 θ1

cos θ− 1

=sin2 θcos2 θ1−cos θcos θ

=sin2 θ

cos2 θ· 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

cos θ= RHS

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+ t) =tan s+ tan t

1− tan s tan t

tan(s− t) =tan s− tan t

1 + tan s tan t

Example: Find cos(75◦) and cos(π12

)solution:

cos(75◦) = cos(45◦ + 30◦)

= cos(45◦) cos(30◦)− sin(45◦) sin(30◦)

=

√2

2

√3

2−√

2

2

1

2

=

√6−√

2

4


cos( π

12

)= cos

(π3− π

4

)= cos

(π3

)cos(π

4

)+ sin

(π3

)sin(π

4

)=

1

2

√2

2+

√3

2

√2

2

=

√2 +√

6

4

Expressions of the form A sinx+B cosx

We can use the sum and difference identities to rewrite expressions of the form A sinx +

B cosx as something simpler. The trick is to simultaneously transform the A into something

that looks like sinφ and transform the B into something that looks like cosφ. 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

cosφ =A√

A2 +B2sinφ =

B√A2 +B2

Thus we can rewrite, using the sum identity:

A sinx+B cosx =√A2 +B2

(A√

A2 +B2sinx+

B√A2 +B2

cosx

)=√A2 +B2(cosφ sinx+ sinφ cosx)

=√A2 +B2 sin(x+ φ)

Example: Express 3 sinx+ 4 cosx in the form k sin(x+ φ)

solution: Here, k =√A2 +B2 =

√32 + 42 = 5 also, sinφ = 4

5, cosφ = 3

5so φ = 0.927

radians.

Thus,

3 sinx+ 4 cosx = 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 cosx


cos 2x = cos2 x− sin2 x

= 1− 2 sin2 x

= 2 cos2 x− 1

tan 2x =2 tanx

1− tan2 x

Example: If cos x = −23

and x is in quadrant II, find cos 2x and sin 2x.

solution:

cos 2x = 2 cos2 x− 1 = 2

(−2

3

)2

− 1 =8

9− 1 = −1

9

sin 2x = 2 sin x cosx

Since x is in quadrant II, we can sub in sinx =√

1− cos2 x:

sin 2x = 2

(−2

3

)√1−

(−2

3

)2

= −4

3

√1− 4

9= −4

3

√5

9= −4

√5

9

Formulas for Lowering Powers

sin2 x =1− cos 2x

2cos2 x =

1 + cos 2x

2

tan2 x =1− cos 2x

1 + cos 2x

Half-angle formulas

sin(u

2

)= ±

√1− cosu

2cos(u

2

)= ±

√1 + cosu

2

tan(u

2

)=

1− cosu

sinu

The ± is chosen in the first two depending on what quadrant u2

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.

sin 22.5◦ = sin

(45◦

2

)=

√1− cos 45◦

2

=

√1−

√22

2

=

√2−√

2

4

Example: Find tan(u2

)if sinu = 2

5and u is in quadrant II.

solution: From the formula,

tan(u

2

)=

1− cosu

sinu

We know sinu = 25, and from the usual pythagorean identity we know

cosu = ±√

1− sin2 u = −√

1− sin2 u

We choose the negative because cos is negative in the second quadrant. Hence

tan(u

2

)=

1− cosu

sinu

=1 +

√1− sin2 u

sinu

=1 +

√1− (2

5)2

2/5

=1 +

√2125

2/5

=5

2

(5 +√

21

5

)

=5 +√

21

2


Product-to-Sum Formulas

sinu cos v =1

2[sin(u+ v) + sin(u− v)]

cosu sin v =1

2[sin(u+ v)− sin(u− v)]

cosu cos v =1

2[cos(u+ v) + cos(u− v)]

sinu sin v =1

2[cos(u− v)− cos(u+ v)]

Example: Express sin 3x sin 5x as a sum of trig functions.

solution: We use the fourth product-to-sum formula:

sin 3x sin 5x =1

2[sin(3x− 5x)− sin(3x+ 5x)] =

1

2(cos(−2x)− cos 8x) =

1

2(cos(2x)− cos 8x)

Sum-to-Product Formulas

sinx+ sin y = 2 sin

(x+ y

2

)cos

(x− y

2

)sinx− sin y = 2 cos

(x+ y

2

)sin

(x− y

2

)cosx+ cos y = 2 cos

(x+ y

2

)cos

(x− y

2

)cosx− cos y = −2 sin

(x+ y

2

)sin

(x− y

2

)Example: Write sin 7x+ sin 3x as a product.

solution: We use the first formula:

sin 7x+ sin 3x = 2 sin

(7x+ 3x

2

)cos

(7x− 3x

2

)= 2 sin 5x cos 2x


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 :[−π

2, π2

]→ [−1, 1]. Thus there exists

a function sin−1 or arcsin

sin−1 : [−1, 1]→[−π

2,π

2

]

sinx arcsinx(= sin−1 x)

The arcsin function satisfies

arcsin(x) = y ⇔ sin(y) = x

arcsin(sinx) = x if − π

2≤ x ≤ π

2

sin(arcsinx) = x if − 1 ≤ x ≤ 1

Examples: Find (a)sin−1(12), (b)arcsin(−1

2), (c)sin−1(3

2) and (d)cos(sin−1(3

5)).

solution:

(a) We know that sin(π6) = 1

2, so sin−1(1

2) = π

6.

(b) Likewise, sin(−π6) = −1

2, so arcsin(−1

2) = −π

6.


(c) We know sin x is never 32

(it is never greater than 1), so sin−1(32) is undefined.

(d) For u ∈[−π

2, π2

], we have cos(u) =

√1− sin2 u, so

cos

(sin−1

(3

5

))=

√1− sin2

(sin−1

(3

5

))

=

√1−

(3

5

)2

=

√1− 9

25

=

√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]

cosx arccosx(= cos−1 x)

The arccos function satisfies

arccos(x) = y ⇔ cos(y) = x

arccos(cosx) = x if 0 ≤ x ≤ π


cos(arccosx) = x if − 1 ≤ x ≤ 1

Examples: Find (a) cos−1(√

32

)and (b) cos−1(0).

solution:

(a) cos(π6) =

√32

, thus cos−1(√

32

)= π

6.

(b) cos(π2) = 0, thus cos−1(0) = π

2

Examples:

1. Show that sin(cos−1(x)) =√

1− x2.

solution: sinu =√

1− cos2(u) for u ∈ [0, π]. Thus

sin(cos−1(x)) =√

1− cos2(cos−1(x)) =√

1− x2

2. Show that tan(cos−1(x)) =√1−x2x

.

solution: tan(u) = sinucosu

, so

tan(cos−1(x)) =sin(cos−1(x))

cos(cos−1(x))=

√1− x2x

3. Show that sin(2 cos−1(x)) = 2x√

1− x2

solution: sin 2u = 2 sinu cosu, so

sin(2 cos−1(x) = 2 sin(cos−1(x)) cos(cos−1(x)) = 2√

1− x2x = 2x√

1− x2

Inverse Tangent

The function tan is one-to-one when restricted to tan :[−π

2, π2

]→ R. Thus there exists a

function tan−1 or arctan

tan−1 : R→[−π

2,π

2

]


tanx arctanx(= tan−1 x)

The arctan function satisfies

arctan(x) = y ⇔ tan(y) = x

arctan(tanx) = x if − π

2≤ x ≤ π

2

tan(arctanx) = x if x ∈ R

Examples: Find (a) tan−1(1), (b) tan−1√

3 and (c)arctan(−20)

solution:

(a) tan(π4) = 1, thus tan−1(1) = π

4.

(b) tan(π3) =√

3, thus tan−1(√

3) = π3

(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:

cot−1 6= 1

tan−1

csc−1 6= 1

sin−1

sec−1 6= 1

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 + sinx = 2 cos2(x)

(d) sin 2x− cosx = 0

(e) cos(x) + 1 = sinx 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

tan(x) =√

3⇒ x =π

3+ kπ k ∈ Z

Likewise,

tan(x) = −√

3⇒ x = −π3

+ kπ k ∈ Z

Thus the full set of solutions is

S ={−π

3+ kπ, x =

π

3+ kπ | k ∈ Z

}(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)

cos(x)= 1⇒ tan(x) = 1

This happens when x = π4. Once again, tan has a period of π so

x =π

4+ kπ k ∈ Z

(c) We can get the equation 1 + sinx = 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+ sinx− 1 = 0


This is a quadratic in sinx that factors as

(2 sinx− 1)(sinx+ 1) = 0

Which means either 2 sin x − 1 = 0 or sinx + 1 = 0. The first gives us that

sinx = 12, and the second gives sin x = −1.

• sinx = −1⇒ x = 3π2

+ 2kπ, k ∈ Z• sinx = 1

2⇒ x = π

6+ 2kπ, k ∈ Z or x = 5π

6+ 2kπ, k ∈ Z

(d) We use a double angle identity on sin 2x− cosx = 0 to get

2 sinx cosx− cosx = 0

cosx(2 sinx− 1) = 0

Hence either

• cosx = 0⇒ x = π2

+ 2kπ, k ∈ Z or x = 3π2

+ 2kπ, k ∈ Z.

• sinx = 12⇒ x = π

6+ 2kπ, k ∈ Z or x = 5π

6+ 2kπ, k ∈ Z.

(e) Here we have to be a little trickier and square both sides. Unfortunately, squaring

both sides may produce so-called extraneous solutions, so after we are done we

must check each solution in the original equation.

cosx+ 1 = sinx

(cosx+ 1)2 = 1− cos2 x

cos2 x+ 2 cosx+ 1 = 1− cos2 x

2 cos2 x+ 2 cosx = 0

2 cosx(cosx− 1) = 0

Hence either cosx = 0 or cos x = −1. Between 0 and 2π, the first only happens

at π2

and 3π2

and the second happens at −π. Hence the three solutions are

x =π

2,3π

2, π.

We now check each of these numbers in the original equation

cosπ

2+ 1 = sin

π

21 = 1.

This is true, so π2

is a solution.

cos3π

2+ 1 = sin

3π

21 = −1.


This is false, so 3π2

is not a solution.

cos π + 1 = sinπ

0 = 0.

This is true, so π is a solution.

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

3x =π

6+ 2kπ, k ∈ Z or 3x =

5π

6+ 2kπ, k ∈ Z.

x =π

18+

2kπ

3, k ∈ Z or x =

5π

18+

2kπ

3, k ∈ Z.

(b) We solve the inequalities

0 ≤ π

18+

2kπ

3< 2π

0 ≤ 1

18+

2k

3< 2

− 1

18≤ 2k

3< 2− 1

18=

35

18

−1

6≤ 2k <

35

6

− 1

12≤ k <

35

12≈ 2.91

The ks in this range are k = 0, k = 1 and k = 2. Hence

x =π

18,13π

18,25π

18

For the other solutions we have

0 ≤ 5π

18+

2kπ

3< 2π

0 ≤ 5

18+

2k

3< 2

− 5

18≤ 2k

3< 2− 5

18=

31

18

−5

6≤ 2k <

31

6


− 5

12≤ k <

31

12≈ 2.58

The ks in this range are k = 0, k = 1 and k = 2. Hence

x =5π

18,17π

18,29π

18

3. Consider the equation√

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√

3This gives us

x

2=π

6+ kπ, k ∈ Z

x =π

3+ 2kπ, k ∈ Z

(b) As before, we find the ks we need:

0 ≤ π

3+ 2kπ < 4π

0 ≤ 1

3+ 2k < 4

−1

3≤ 2k < 4− 1

3=

11

3

−1

6≤ k <

11

6≈ 1.83

The ks in this range are k = 0 and k = 1. Hence x = π3

or x = 7π3

.

There is another way of doing problems of this type, see class for details.

4. Solve the equation tan2 x− tanx− 2 = 0.

solution: This is a quadratic in tanx, so

tan2 x− tanx− 2 = 0

(tanx− 2)(tanx+ 1) = 0

So tanx = 2 or tanx = −1. There isn’t a convenient angle with tanx = 2, but we can

use tan−1 to write

x = tan−1(2) + kπ, k ∈ Z

The second one of course gives us

x = −π4

+ kπ, k ∈ Z


5. Solve the equation 3 sin θ − 2 = 0.

solution: This rearranges to sin θ = 23. 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

sin−1(

2

3

)and π − sin−1

(2

3

)Hence the full solution is

θ = sin−1(

2

3

)+ 2kπ or θ = π − sin−1

(2

3

)+ 2kπ

7 Analytic Trigonometry - Faculté des sciences - Faculty of...

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