W ORKED E XERCISE:

(a) Given that sin θ = 1

5, find cos θ.

(b) Repeat if it is also known that tan θ is negative.

SOLUTION:

(a) First, the angle must be in quadrant 1 or 2. Since sin θ = y

r = 1

5, we can take y = 1 and r = 5, so by Pythagoras’ theorem, x =√24 or −√24 = 2√6 or −2√6 , x y 1 5 5 1 θ θ −2 6 2 6 so cos θ = 2 √ 6 5 or − 2√6 5 .

(b) Since tan θ is negative, θ must be in quadrant 2,

so cos θ = −2

5

√ 6 .

Exercise

4E

1. (a) Given that cos θ = 3

5 and θ is acute, find sin θ and tan θ.

(b) Given that tan θ = −5

12 and θ is obtuse, find sin θ and sec θ.

2. (a) Given that sin α = 8

17, find the possible values of cos α and cot α.

(b) Given that cos x = −3

4 and 90◦< x < 180◦, find tan x and cosec x.

3. (a) Given that cot β = 3

2 and sin β < 0, find cos β.

(b) If cosec α = −5

2 and cos α > 0, find cot α.

(c) If tan θ = 2, find the possible values of cosec θ.

(d) Suppose that sin A = 1. Find sec A.

4. (a) Given that sec P = −3 and 180◦_{< P < 360}◦_{, find cosec P .}

(b) If cos θ = −1, find tan θ.

(c) Suppose that cos α = 2

3. Find the possible values of sin α and cot α.

(d) Given that cot x = −3

5, find the possible values of cosec x and sec x.

D E V E L O P M E N T

5. Given that sin θ = p

q, with θ obtuse and p and q both positive, find expressions for cos θ and tan θ.

6. If tan α = k, where k > 0, find the possible values of sin α and sec α.

7. (a) Prove the algebraic identity (1 − t2₎2_{+ 4t}2 _{= (1 + t}2₎2_.

(b) If cos x =1 − t2

1 + t2 and x is acute, find expressions for sin x and tan x.

E X T E N S I O N

8. If sin θ = k and θ is obtuse, find an expression for tan(θ + 90◦_).

9. If sec θ = a + 1

4a, prove that sec θ + tan θ = 2a or 1 2a.

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CHAPTER4:Trigonometry 4F Trigonometric Identities and Elimination 129

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4 F

Trigonometric Identities and Elimination

Working with the trigonometric functions requires knowledge of a number of formulae called trigonometric identities, which relate trigonometric functions to each other. This section introduces eleven of these in four groups: the three reciprocal identities, the two ratio identities, the three Pythagorean identities, and the three identities concerning complementary angles.

The Three Reciprocal Identities: It follows immediately from the definitions of the trigonometric functions in terms of x, y and r that:

17

THE RECIPROCAL IDENTITIES: For all angles θ:

cosec θ = 1

sin θ (provided sin θ ̸= 0) sec θ = 1

cos θ (provided cos θ ̸= 0) cot θ = 1

tan θ (provided tan θ ̸= 0 and cot θ ̸= 0)

Note: The last identity needs attention. One cannot use the calculator to find cot 90◦ _{or cot 270}◦ _{by first finding tan 90}◦ _{or tan 270}◦_{, because both of these are}

undefined. We already know, however, that cot 90◦_{= cot 270}◦_{= 0.}

The Two Ratio Identities: Again using the definitions of the trigonometric functions:

18

THE RATIO IDENTITIES: For any angle θ:

tan θ = sin θ

cos θ (provided cos θ ̸= 0) cot θ = cos θ

sin θ (provided sin θ ̸= 0)

The Three Pythagorean Identities: Since the point P (x, y) lies on the circle with centre O and radius r, its coordinates satisfy

x2+ y2 = r2. Dividing through by r2 _gives x2

r2 + y2 r2 = 1, θ P x y( , ) r x y

then by the definitions, sin2_{θ + cos}2_{θ = 1.}

Dividing through by cos2_{θ and using the ratio and reciprocal identities,}

tan2_{θ + 1 = sec}2_{θ, provided cos θ}

̸= 0. Dividing through instead by sin2_{θ, 1 + cot}2_{θ = cosec}2_{θ, provided sin θ}

̸= 0. These identities are called the Pythagorean identities because they rely on the circle equation x2 _{+ y}2 _{= r}2_{, which is really just a restatement of Pythagoras’}

theorem.

19

THE PYTHAGOREAN IDENTITIES: For any angle θ:

sin2_{θ + cos}2_{θ = 1}

tan2_{θ + 1 = sec}2_{θ (provided cos θ ̸= 0)}

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The Three Identities for Complementary Angles: These identities relate the values of the trigonometric functions at any angle θ to their values at the complementary angle 90◦_{− θ.}

20

THE COMPLEMENTARY IDENTITIES: For any angle θ:

cos(90◦_{− θ) = sin θ}

cot(90◦_{− θ) = tan θ (provided tan θ is defined)}

cosec(90◦_{− θ) = sec θ (provided sec θ is defined)}

Proof:

A. [Acute angles] The triangle on the right shows that when θ is acute, viewing the right triangle from 90◦_{− θ instead of from θ exchanges the opposite side}

and the adjacent side, so: cos(90◦_{− θ) =} a c = sin θ, cot(90◦_{− θ) =} a b = tan θ, b a 90º− θ c θ cosec(90◦_{− θ) =} c b = sec θ.

B. [General angles] For general angles, we take the full circle diagram, and reflect it in the diagonal line y = x. Let P′ _{be the image of P under this reflection.}

1. The image OP′ _{of the ray OP corresponds with the}

angle 90◦_{− θ.} θ P x y( , ) P' y x( , ) x y 90º− θ y x=

2. The image P′ _{of P (x, y) has coordinates P}′_{(y, x).}

We have seen before that reflection in the line y = x reverses the coordinates of each point. So x and y are interchanged in passing from P to P′_.

Applying the definitions of the trigonometric functions to the angle 90◦_{− θ:}

cos(90◦_{− θ) =} y r = sin θ, cot(90◦_{− θ) =} y x = tan θ, provided x ̸= 0, cosec(90◦_{− θ) =} r x = sec θ, provided x ̸= 0.

Cosine, Cosecant and Cotangent: The complementary identities are the origin of the ‘co-’ prefix of cosine, cosecant and cotangent — the prefix is an abbreviation of the prefix ‘com-’ of complementary angle. The various identities can be easily remembered as:

21 CO-FUNCTIONS: The co-function of a complement is the function of an angle. The co-function of an angle is the function of the complement.

Proving Identities: An identity is a statement that needs to be proven true for all values of θ for which both sides are defined. It is quite different from an equation, which needs to be solved and to have its solutions listed.

!

CHAPTER4:Trigonometry 4F Trigonometric Identities and Elimination 131

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WORKEDEXERCISE: Prove that sin A sec A = tan A. Note: The necessary restric-

tion to angles for which sec A and tan A are defined is implied by the statement.

SOLUTION: LHS = sin A × 1

cos A (reciprocal identity) = tan A (ratio identity)

= RHS

WORKEDEXERCISE: Prove that 1

sin2_θ+

1

cos2_θ = sec2θ cosec2θ.

Proof: LHS = 1 sin2θ+

1 cos2_θ

= cos2θ + sin2θ

sin2θ cos2θ (common denominator)

= 1

sin2_{θ cos}2_θ (Pythagorean identity)

= sec2_{θ cosec}2_{θ (reciprocal identities)}

= RHS

Elimination: If x and y are given as functions of θ, then using the techniques of si- multaneous equations, the θ can often be eliminated to give a relation (rarely a function) between x and y.

WORKEDEXERCISE: Eliminate θ from the following pair, and describe the graph

of the relation:

x = 4 + 5 cos θ y = 3− 5 sin θ

SOLUTION: From the first equation, 5 cos θ = x − 4,

and from the second equation, 5 sin θ = 3 − y. Squaring and adding, 25 cos2_{θ + 25 sin}2_{θ = (x}

− 4)2+ (3 − y)2 and since cos2_{θ + sin}2_{θ = 1, (x}

− 4)2+ (y − 3)2= 25, y x 4 ₈ 3 6

which is a circle of radius 5 and centre (4, 3).

Exercise

4F

1. Use your calculator to verify that:

(a) sin 16◦_{= cos 74}◦

(b) tan 63◦_{= cot 27}◦

(c) sec 7◦_{= cosec 83}◦

(d) sin2₂₃◦_{+ cos}2₂₃◦_{= 1}

(e) 1 + tan2_{55◦= sec}2_55◦

(f) cosec2₃₂◦_{− 1 = cot}2₃₂◦ 2. Simplify: (a) 1 sin θ (b) 1 tan α (c) sin β cos β (d) cos φ sin φ

3. Simplify: (a) sin α cosec α (b) cot β tan β (c) cos θ sec θ

4. Prove: (a) tan θ cos θ = sin θ (b) cot α sin α = cos α (c) sin β sec β = tan β

5. Prove: (a) cos A cosec A = cot A (b) cosec x cos x tan x = 1 (c) sin y cot y sec y = 1

6. Simplify: (a) cos α

sec α (b) sin α cosec α (c) tan A sec A (d) cot A cosec A

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132 CHAPTER4:Trigonometry CAMBRIDGEMATHEMATICS3 UNITYEAR11

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7. Simplify: (a) 1

sec2_θ (b) sin2α cosec2α (c) sin 2_β

cos2_β (d)

cos2_A

sin2A

8. Simplify: (a) sin(90◦_{− θ)(b) sec(90}◦_{− α) (c)} 1

cot(90◦_{− β)} (d)

cos(90◦_{− φ)}

sin(90◦_{− φ)}

9. Simplify: (a) sin2_{α + cos}2_{α (b) 1 − cos}2_{β (c) 1 + tan}2_{φ (d) sec}2_{x− tan}2_x

10. Simplify: (a) 1 − sin2_{β (b) 1 + cot}2_{φ (c) cosec}2_{A− 1 (d) cot}2_{θ− cosec}2_θ

D E V E L O P M E N T

11. Prove the identities:

(a) (1 − sin θ)(1 + sin θ) = cos2_θ

(b) (1 + tan2_{α) cos}2_{α = 1}

(c) (sin A + cos A)2 _{= 1 + 2 sin A cos A}

(d) cos2_{x− sin}2_{x = 1− 2 sin}2_x

(e) tan2_{φ cos}2_{φ + cot}2_{φ sin}2_{φ = 1}

(f) 3 cos2_{θ− 2 = 1 − 3 sin}2_θ

(g) 2 tan2_{A− 1 = 2 sec}2_{A− 3}

(h) 1 − tan2_{α + sec}2_{α = 2}

(i) cos4_{x + cos}2_{x sin}2_{x = cos}2_x

(j) cot θ(sec2_θ

− 1) = tan θ

12. Prove the identities:

(a) sin θ cos θ cosec2_{θ = cot θ}

(b) (cos φ + cot φ) sec φ = 1 + cosec φ

(c) sin α

cos α + cos α

sin α = sec α cosec α

(d) 1 + tan2x

1 + cot2_x = tan2x

(e) sin4_{A− cos}4_{A = sin}2_{A− cos}2_A

(f) 1

1 + sin θ+ 1

1 − sin θ = 2 sec2θ

(g) sin β + cot β cos β = cosec β

(h) 1

sec φ − tan φ−

1

sec φ + tan φ = 2 tan φ

(i) 1 + cot x

1 + tan x = cot x

(j) cos α

1 + sin α = sec α(1 − sin α)

13. (a) If x = a cos α and y = a sin α, show that x2_{+ y}2 _{= a}2_.

(b) If x = a sec θ and y = b tan θ, show that x2 a2 −

y2 b2 = 1.

(c) If x = r cos θ sin φ, y = r sin θ sin φ and z = r cos φ, show that x2_{+ y}2_{+ z}2 _{= r}2_.

(d) If x = a cos θ − b sin θ and y = a sin θ + b cos θ, show that x2_{+ y}2 _{= a}2_{+ b}2_.

14. Eliminate θ from each pair of equations:

(a) x = a cos θ and y = b sin θ

(b) x = a tan θ and y = b sec θ

(c) x = 2 + cos θ and y = 1 + sin θ

(d) x = sin θ + cos θ and y = sin θ− cos θ

15. Prove that each expression is independent of θ:

(a) cos2θ

1 + sin θ +

cos2_θ

1 − sin θ

(b) tan θ(1 − cot2_{θ) + cot θ(1− tan}2_θ)

(c) tan θ + cot θ sec θ cosec θ (d) tan θ + 1 sec θ − cot θ + 1 cosec θ

16. Prove the identities:

(a) 2 cos3θ− cos θ

sin θ cos2_{θ− sin}3_θ = cot θ (b) sec y + tan y + cot y =

1 + sin y sin y cos y

(c) cos A − tan A sin A

cos A + tan A sin A = 1 − 2 sin2A

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CHAPTER4:Trigonometry 4G Trigonometric Equations 133

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(e) 1 1 + tan2_x − 1 1 + sec2_x = cos4_x 1 + cos2_x (f) cos θ 1 − tan θ + sin θ

1 − cot θ = sin θ + cos θ

(g) (tan α + cot α − 1)(sin α + cos α) = _cosecsec α₂ α +

cosec α sec2_α

(h) 1

sec θ + tan θ = sec θ − tan θ =

cos θ 1 + sin θ (i) 1 cot θ − cos θ = tan θ 1 − sin θ = sin θ + sin2_θ cos3_θ

(j) sin2x(1 + n cot2x) + cos2x(1 + n tan2x) = n + 1 = sin2_{x(n + cot}2_{x) + cos}2_{x(n + tan}2_x)

(k) (sin2α− cos2α)(1− sin α cos α)

cos α(sec α − cosec α)(sin3_{α + cos}3_α) = sin α

(l) 1 + cosec2A tan2C

1 + cosec2_{B tan}2_C =

1 + cot2_{A sin}2_C

1 + cot2_{B sin}2_C

E X T E N S I O N

17. Eliminate θ from each pair of equations:

(a) x = cosec2θ + 2 cot2θ and y = 2 cosec2θ + cot2θ

(b) x = sin θ_{− 3 cos θ and y = sin θ + 2 cos θ}

(c) x = sin θ + cos θ and y = tan θ + cot θ [Hint: Find x2y.]

18. (a) If a

sin A = b

cos A, show that sin A cos A = ab a2+ b2 .

(b) If a + b cosec x =

a− b

cot x, show that cosec x cot x =

a2− b2 4ab .

(c) If tan θ + sin θ = x and tan θ − sin θ = y, prove that x4_{+ y}4 _{= 2xy(8 + xy).}

4 G

Trigonometric Equations

This piece of work is absolutely vital, because so many problems in later work end up with a trigonometric equation that has to be solved. There are many small details and qualifications in the methods, and the subject needs a great deal of careful study.

Pay Attention to the Domain: To begin with a warning, before any other details:

23 THE DOMAIN: Always pay attention to the domain in which the angle can lie.

Equations Involving Boundary Angles: The usual quadrants-and-related-angle method described below doesn’t apply to boundary angles, which do not lie in any quad- rant.

24

THE BOUNDARY ANGLES: If a trigonometric equation

involves boundary angles, read the solutions off a sketch of the graph.

WORKEDEXERCISE: Solve sin x = −1, for 0◦_{≤ x ≤ 720}◦. SOLUTION: The graph of y = sin x is drawn on the right.

x y 360º 270º 1 −1 720º 630º

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134 CHAPTER4:Trigonometry CAMBRIDGEMATHEMATICS3 UNITYEAR11

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The Standard Method — Quadrants and Related Angle: Nearly all our trigonometric

equations will eventually come down to something like sin x = −1

2, where −180◦≤ x ≤ 360◦.

As long as the angle is not a boundary angle, the method is:

25

THE QUADRANTS-AND-RELATED-ANGLE METHOD:

1. Draw a quadrant diagram, then draw a ray in each quadrant that the angle could be in.

2. Find the related angle (only work with positive numbers here): (a) using special angles, or

(b) using the calculator to find an approximation.

3. Mark the angles on the ends of the rays, taking account of any restrictions on x, and write a conclusion.

WORKEDEXERCISE: Solve each equation. Give the solution exactly if possible, or

else to the nearest degree:

(a) sin x = −1 2, −180◦≤ x ≤ 360◦ (b) tan x = −3, 0◦≤ x ≤ 360◦ SOLUTION: (a) sin x = −1 2, where−180◦≤ x ≤ 360◦ x =−150◦, −30◦, 210◦ or 330◦

(Since sin x is negative, x is in quadrants 3 or 4,

−30º,330º −150º,210º

30º 30º

the sine of the related angle is +1 2,

so the related angle is 30◦_.)

(b) tan x = −3, where 0◦_{≤ x ≤ 360}◦

x =.. 108◦ or 288◦

(Since tan x is negative, x is in quadrants 2 or 4,

288º 108º

72º 72º

the tangent of the related angle is +3, so the related angle is about 72◦_.)

Note: When using the calculator, never enter a negative number and take an inverse trigonometric function of it. In the example above, the calculator was used to find the acute angle whose tan was 3, that is, 71◦₃₄′_{. The positive}

number 3 was entered, not −3.

The Three Reciprocal Functions: Because they are unfamiliar, and also because the calculator doesn’t have specific keys for them:

26

THE RECIPROCAL FUNCTIONS: Try to change any of the three reciprocal functions

secant, cosecant and cotangent to the three more common functions by taking reciprocals.

WORKEDEXERCISE: Suppose we are given that cosec x = −2.

Taking reciprocals of both sides gives sin x = −1 2,

!

CHAPTER4:Trigonometry 4G Trigonometric Equations 135

!

Equations with Compound Angles: These can cause trouble. Equations like

tan 2x =√3, where 0◦_{≤ x ≤ 360}◦_{, or}

sin(x − 250◦_{) =}

√ 3

2 , where 0◦≤ x ≤ 360◦,

are really trigonometric equations in the compound angles 2x and (x − 250◦₎

respectively. The secret lies in solving for the compound angle, and in calculating first the domain for that compound angle.

27

EQUATIONS WITH COMPOUND ANGLES:

1. Let u be the compound angle.

2. Find the restrictions on u from the given restrictions on x. 3. Solve the trigonometric equation for u.

4. Hence solve for x.

WORKEDEXERCISE: Solve tan 2x =√3, where 0◦_{≤ x ≤ 360}◦.

SOLUTION: Let u = 2x.

Then tan u =√3 , where 0◦_{≤ u ≤ 720}◦_,

(the restriction on u is the key step here),

so from the diagram, u = 60◦, 240◦, 420◦ or 600◦.

60º,420º 240º,600º 60º 60º Since x = 1 2u, x = 30◦, 120◦, 210◦ or 300◦.

WORKEDEXERCISE: Solve sin(x − 250◦) =

√ 3 2 , where 0◦≤ x ≤ 360◦. SOLUTION: Let u = x_{− 250}◦. Then sin u = √ 3 2 , where −250◦≤ u ≤ 110◦, (again, the restriction on u is the key step here),

so from the diagram, u =₋₂₄₀◦ or 60◦.

60º −240º

60º 60º

Since x = u + 250◦_{, x = 10}◦ _{or 310}◦_.

Equations Requiring Algebraic Substitutions: If there are powers or reciprocals of the trigonometric function present, as in

5 sin2_{x = sin x, for 0}◦_{≤ x ≤ 360}◦_{, or}

4

cos x −cos x = 0, for − 180◦≤ x ≤ 180◦,

but still only the one trigonometric function, then it is probably better to make a substitution so that the algebra can be done without interference by the trigono- metric notation.

28

ALGEBRAIC SUBSTITUTION:

1. Substitute u to obtain a purely algebraic equation.

2. Solve the algebraic equation — it may have more than one solution. 3. Solve each of the resulting trigonometric equations.

!

136 CHAPTER4:Trigonometry CAMBRIDGEMATHEMATICS3 UNITYEAR11

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WORKEDEXERCISE: Solve 5 sin2x = sin x, for 0◦ ≤ x ≤ 360◦. Give the exact value of the solutions if possible, otherwise approximate to the nearest minute.

SOLUTION: Let u = sin x. Then 5u2 _{= u} − u 5u2− u = 0 u(5u_{− 1) = 0} 11º32' 11º32' 168º28' u = 0 or u = 1₅, so sin x = 0 or sin x = 1 5.

Using the graph of y = sin x to solve sin x = 0 and the quadrant-diagram method to solve sin x = 1

5,

x = 0◦, 180◦ or 360◦, or x =.. 11◦32′ or 168◦28′.

WORKEDEXERCISE: Solve 4

cos x −cos x = 0, for −180◦≤ x ≤ 180◦.

SOLUTION: Let u = cos x. Then 4

u − u = 0 × u 4 − u2 = 0

u = 2 or u =−2, so cos x = 2 or cos x = −2.

Neither equation has a solution, because cos x lies between −1 and 1, so there are no solutions.

Equations with More than One Trigonometric Function: Often a trigonometric equation will involve more than one trigonometric function, as, for example,

sec2_{x + tan x = 1, where 180}◦_{≤ x ≤ 360}◦_.

29

EQUATIONS WITH MORE THAN ONE TRIGONOMETRIC FUNCTION: Usually use trigonomet-

ric identities to produce an equation in only one trigonometric function, then proceed by substitution as before.

If all else fails, reduce everything to sines and cosines, and hope for the best!

WORKEDEXERCISE: Solve sec2x + tan x = 1, where 180◦_{≤ x ≤ 360}◦ (as above).

SOLUTION: Recognizing that sec2x = 1 + tan2x, the equation becomes 1 + tan2_{x + tan x = 1, where 180}◦_{≤ x ≤ 360}◦

− 1 tan2x + tan x = 0, tan x(tan x + 1) = 0,

so tan x = 0 or tan x = −1. _45º

315º

Using the graph of y = tan x to solve tan x = 0,

and the quadrant-diagram method to solve tan x = −1, x = 180◦, 360◦ or 315◦.

!

CHAPTER4:Trigonometry 4G Trigonometric Equations 137

!

Homogeneous Equations: One special sort of equation which occurs quite often is

called homogeneous in sin x and cos x because the sum of the indices of sin x and cos x in each term is the same. For example, the following equation is homoge- neous of degree 2 in sin x and cos x:

sin2_x

− 3 sin x cos x + 2 cos2x = 0, for 0◦_{≤ x ≤ 180}◦.

30 HOMOGENEOUS EQUATIONS: To solve a homogeneous equation in sin x and cos x, divide through by a power of cos x to produce an equation in tan x.

WORKEDEXERCISE: Continuing with the example above,

÷ cos2x tan2x− 3 tan x + 2 = 0.

Let u = tan x, then u2_{− 3u + 2 = 0} (u − 2)(u − 1) = 0 u = 2 or u = 1 tan x = 2 or tan x = 1. So x =.. 63◦26′ or x = 45◦.

Exercise

4G

1. Solve each of these equations for 0◦_{≤ θ ≤ 360}◦ _{(each related angle is 30}◦_{, 45}◦ _{or 60}◦_):

(a) sin θ = √ 3 2 (b) tan θ = 1 (c) cos θ = −√1 2 (d) tan θ = −√3 (e) cosec θ = −2 (f) sec θ = −√2 3

2. Solve each of these equations for 0◦ _{≤ θ ≤ 360}◦ _{(the trigonometric graphs are helpful}

here): (a) sin θ = 1 (b) cos θ = −1 (c) cos θ = 0 (d) sec θ = 1 (e) tan θ = 0 (f) cot θ = 0

3. Solve each of these equations for 0◦ _{≤ x ≤ 360}◦_{. Use your calculator to find the related}

angle in each case, and give solutions correct to the nearest degree.

(a) cos x = 3 7 (b) sin x = 0·1234 (c) tan x = −7 (d) cot x = −0·45 (e) cosec x = −3 2 (f) sec x = 6

4. Solve each of these equations for α in the given domain. Give solutions correct to the nearest minute where necessary:

(a) sin α = 0·1, 0◦_{≤ α ≤ 360}◦ (b) cos α = −0·1, 0◦_{≤ α ≤ 360}◦ (c) tan α = −1, −180◦_{≤ α ≤ 180}◦ (d) cosec α = −1, 0◦_{≤ α ≤ 360}◦ (e) sin α = 3, 0◦_{≤ α ≤ 360}◦ (f) sec α =√2, 0◦_{≤ α ≤ 360}◦ (g) cos α = 0, −180◦_{≤ α ≤ 180}◦ (h) cot α = 1 2, α reflex

(i) √3 tan α + 1 = 0, α obtuse

(j) cosec α + 2 = 0, α reflex (k) 2 cos α − 1 = 0, 0◦_{≤ α ≤ 360}◦ (l) cot α = 3, 0◦_{≤ α ≤ 360}◦ (m) tan α = 0, −360◦_{≤ α ≤ 360}◦ (n) tan α = −0·3, −180◦_{≤ α ≤ 180}◦ (o) sin α = −0·7, 0◦_{≤ α ≤ 720}◦ (p) tan α = 1 −√2, 0◦_{≤ α ≤ 360}◦

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138 CHAPTER4:Trigonometry CAMBRIDGEMATHEMATICS3 UNITYEAR11

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D E V E L O P M E N T

5. Solve for 0◦_{≤ θ ≤ 360}◦_{, giving solutions correct to the nearest degree where necessary:}

(a) cos2_{θ = 1 (b) sec}2_{θ =} 4

3 (c) tan2θ = 9 (d) cosec2θ = 2

6. Solve for 0◦_{≤ x ≤ 360}◦_{(let u be the compound angle):}

(a) sin 2x =1

2 (b) cos 2x = −√12 (c) tan 3x =

√

3 (d) sec 3x = 0

7. Solve for 0◦_{≤ α ≤ 360}◦ _{(let u be the compound angle):}

(a) tan(α − 45◦_{) = √}1 3 (b) sin(α + 30◦_{) = −}√3 2 (c) cot(α + 60◦_{) = 1} (d) cosec(α − 75◦_{) = −2} 8. Solve for 0◦_{≤ θ ≤ 360}◦_:

(a) sin θ = cos θ

(b) √3 sin θ + cos θ = 0

(c) 4 sin θ = 3 cosec θ

(d) sec θ − 2 cos θ = 0

9. Solve for 0◦_{≤ θ ≤ 360}◦_{, giving solutions correct to the nearest minute where necessary:}

(a) cos2_{θ− cos θ = 0}

(b) cot2_{θ =}√_{3 cot θ}

(c) 2 sin θ cos θ = sin θ

(d) tan2_{θ− tan θ − 2 = 0}

(e) 2 sin2_{θ− sin θ = 1}

(f) sec2_{θ + 2 sec θ = 8}

(g) 3 cos2_{θ + 5 cos θ = 2}

(h) 4 cosec2_{θ− 4 cosec θ − 15 = 0}

(i) 4 sin3θ = 3 sin θ

10. Solve for 0◦_{≤ x ≤ 360}◦_{, giving solutions correct to the nearest minute where necessary:}

(a) 2 sin2_{x + cos x = 2}

(b) sec2_x

− 2 tan x − 4 = 0

(c) 8 cos2_{x = 2 sin x + 7}

(d) 6 tan2_{x = 5 sec x}

(e) 6 cosec2_{x = cot x + 8}

11. Solve for 0◦_{≤ α ≤ 360}◦_{, giving solutions correct to the nearest minute where necessary:}

(a) 3 sin α = cosec α + 2 (b) 3 tan α − 2 cot α = 5

12. Solve for 0◦_{≤ A ≤ 360}◦_{, giving solutions correct to the nearest minute where necessary:}

(a) cot A + 4 tan A = 4 cosec A (b) 3(tan A + sec A) = 2 cot A

13. Solve for 0◦_{≤ x ≤ 360}◦_{, giving solutions correct to the nearest minute where necessary:}

(a) cos x tan x + tan x = cos x + 1 (b) 6 sin x cos x + 3 sin x = 2 cos x + 1

14. Solve each of these homogeneous equations for 0◦_{≤ x ≤ 360}◦ _{by dividing both sides by a}

suitable power of cos x. Give solutions to the nearest minute where necessary.

(a) sin x = 3 cos x

(b) sin2_{x− 2 sin x cos x − 8 cos}2_{x = 0} (c) 5 sin

2_{x + 8 sin x cos x = 4 cos}2_x

(d) sin3_{x + 2 sin}2_{x cos x + sin x cos}2_{x = 0}

E X T E N S I O N

15. Solve for 0◦_{≤ θ ≤ 360}◦_{, giving solutions correct to the nearest minute where necessary:}

(a) 4 cos2_{θ + 2 sin θ = 3}

(b) 5 sec2_{θ + 7 tan θ = 7}

(c) cos2_{θ− 8 sin θ cos θ + 3 = 0}

(d) 5 sin2_{θ− 4 sin θ cos θ + 3 cos}2_{θ = 2}

(e) 8 cos4_{θ− 10 cos}2_{θ + 3 = 0}

(f) √6 cos θ+√2 sin θ+√3 cot θ+1 = 0

(g) 20 cot θ + 15 cot θ cosec θ − 4 cosec θ = 3(1 + cot2_θ)

(h) 1 − tan2θ

1 + tan2_θ + cos θ = 0

(i) (√3 + 1) cos2_{θ− 1 = (}√_{3 − 1) sin θ cos θ}

(j) 1 + 2 sin2θ

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CHAPTER4:Trigonometry 4H The Sine Rule and the Area Formula 139

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4 H

The Sine Rule and the Area Formula

The sine rule, the area rule and the cosine rule belong both to trigonometry and to geometry. On the one hand, they extend the elementary trigonometry of Section 4A to non-right-angled triangles. On the other hand, they generalise Pythagoras’ theorem, the isosceles triangle theorem, and some results about alti- tudes of triangles. They are also closely related to the four standard congruence tests, and the sine rule can be restated as a theorem about the diameter of the circumcircle of a triangle. These last three sections review the rules and their ap- plications. Their proofs should now be given more attention, particularly because they involve connections between trigonometry and Euclidean geometry.

a c b

C B

A

Statement of the Sine Rule: We will often use the convention that each side of a triangle is given the lower-case letter of the opposite vertex, as in the diagram on the right. Using that convention, here are the verbal and symbolic statements of the sine rule.

31

THEOREM — THE SINE RULE: In any triangle, the ratio of each side to the sine of the

opposite angle is constant. That is, in any triangle △ABC, a sin A = b sin B = c sin C .

Proving the Sine Rule by Constructing an Altitude: So far we can only handle right tri- angles, so any proof of the sine rule must involve a construction with a right angle. The obvious approach is to construct an altitude, which is the perpendicular from one vertex to the opposite side.

Given: Let ABC be any triangle. There are three cases, depending on whether

̸ A is an acute angle, a right angle, or an obtuse angle.

a c b C B A a c b C B A h M a c b C B A h M

Case 1: ̸ A is acute Case 2: ̸ A = 90◦ Case 3: ̸ A is obtuse

Aim: To prove that a sin A =

b sin B.

In case 2, sin A = sin 90◦_{= 1, and sin B =} b

a, so the result is clear.

Construction: In the remaining cases 1 and 3, construct the altitude from C, meeting AB, produced if necessary, at M. Let h be the length of CM.

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140 CHAPTER4:Trigonometry CAMBRIDGEMATHEMATICS3 UNITYEAR11

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Proof:

Case 1 — Suppose that̸ A is acute. In the triangle ACM, h

b = sin A × b h = b sin A. In the triangle BCM, h

a = sin B × a h = a sin B. Equating these, b sin A = a sin B

b sin B =

a sin A.

Case 3 — Suppose that ̸ A is obtuse. In the triangle ACM, h

b = sin(180◦− A), and since sin(180◦_{− A) = sin A,}

× b h = b sin A. In the triangle BCM, h

a = sin B × a h = a sin B. Equating these, b sin A = a sin B

b sin B =

a sin A.

The Area Formula: The well-known formula area = 1

2×base×height can be generalised

to a formula involving two sides and the included angle.

32

THEOREM — THE AREA FORMULA: The area of a triangle is half the product of any

two sides and the sine of the included angle. That is, area △ABC = 1

2bc sin A = 12ca sin B = 12ab sin C .

Proof: We use the same diagrams as in the proof of the sine rule. In case 2,̸ A = 90◦ and sin A = 1, so area = 1₂bc = 1₂bc sin A, as required. Otherwise, area = 1

2 × base × height

= 1

2 × AB × h

= 1

2 × c × b sin A, since we proved before that h = b sin A.

Using the Sine Rule to Find a Side — The AAS Congruence Situation: For the sine rule to be applied to the problem of finding a side, one side and two angles must be known. This is the situation described by the AAS congruence test, so only one triangle will be possible. The sine rule should be learned in verbal form because the triangle being solved could have any names, or could be unnamed.

33

USING THE SINE RULE TO FIND A SIDE: In the AAS congruence situation:

unknown side sine of opposite angle =

known side sine of opposite angle.

WORKEDEXERCISE: Find x in the given triangle. SOLUTION: x sin 120◦ = 7 sin 45◦ × sin 120◦ x = 7 sin 120◦ sin 45◦ = 7 × √ 3 2 × √ 2 45º 120º 7 x = 7 2 √ 6

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CHAPTER4:Trigonometry 4H The Sine Rule and the Area Formula 141

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Using the Area Formula — The SAS Congruence Situation: The area formula requires the SAS situation where two sides and the included angle are known.

34 U

SING THE AREA FORMULA: In the SAS congruence situation:

area = (half the product of two sides) × (sine of the included angle) .

WORKEDEXERCISE: Find the area of the given triangle. SOLUTION: Area = 1 2 ×3 × 4 × sin 135◦ = 6 ×√1 2× √ 2 √ 2 135º 3 4 = 3√2 square units.

Using the Sine Rule to Find an Angle — The Ambiguous ASS Situation: It is well known that the SAS congruence test requires that the angle be included between the two sides. When two sides and a non-included angle are known, the situation

In document Maths Textbook (Page 142-170)