CHAPTER III

EQUATIONS AND SUB-MULTIPLE ANGLES

A FIRST discussion of Trigonometrical equations, for solutions from 0° to 360°, has been given in E.T., Ch. XVIII, p. 258. We shall now obtain the general solutions of such equations, and shall usually work in radians.

Example 1. Solve sin θ = .

 

FIG. 22.

There are two, and only two, solutions between 0 and 2π, (see Fig. 22), θ = or π – .

But any angle differing from these by a multiple of 2π is also a solution; therefore the general solution is

 

where n is any positive or negative integer or zero.

Example 2. Solve cos θ = .

 

FIG. 23.

The solutions between 0 and 2π are

 

(see Fig. 23);

∴ the general solution is

 

where n is any positive or negative integer or zero.

Example 3. Solve tan θ = 1.

 

FIG. 24.

The solutions between 0 and 2π are

 

(see Fig. 24) ;

∴ the general solution is

 

where n is any positive or negative integer or zero.

These examples illustrate the following general statements :

 

 

 

where n is any positive or negative integer or zero.

The solution of sin θ = sin a may also be written in the form, θ = mπ + ( – 1 )^ma, since, if m is even, this becomes 2nπ + a and, if m is odd, it becomes (2n + l)π – a.

Note. There are certain specially simple equations which can be solved at sight by thinking of the figure without recourse to the general formulae. For example the values of θ for which cos θ = 0 are evidently odd numbers of right angles; this gives the solution in the form θ = (2n + l), which is slightly better than the form 2nπ ± given by the general formula; the two forms are, of course, equivalent.

EXERCISE III. a.

Use figures to write down the solutions of Nos. 1-6.

1. sin θ = 0.

2. cos θ = 1.

3. tan θ = 0.

4. sin θ = 1.

5. cos θ = –l.

6. sin θ = – 1.

Apply the general formulae (1), (2), (3) to write down the solutions of Nos. 7-12.

7. sin θ = .

8. cos θ =.

9. tan θ = .

10. sin θ = – .

11. cos θ = – .

12. tan θ = – 1.

Solve the following :

13. cos 2θ = 1.

14. sin 3θ = 0.

15. tan 4θ = 0.

16. sin 3θ = – 1.

17. tan 3θ = –1.

18. cos 4θ = 0.

19. sin θ = cos θ.

20. sin θ + cos θ = O.

21. sec θ = 2.

22. cosec θ = cosec a.

Some useful methods of solving equations are illustrated in the following examples :

Example 4. Solve 4 sin² θ = 1.

This may be written

 

The procedure just adopted is more convenient than using equation (1) to solve sin θ = ± .

Example 5. Solve sin 7θ =sin 5θ.

From equation (1),

 

Example 6. Solve tan A = – cot 2A.

This may be written

 

where m is any integer or zero. This result is equivalent to

 

Example 7. Solve sin 2θ = 1 + cos2θ.

First Method. 2 sin θ cos θ = 2 cos² θ.

 

Second Method. This consists in applying the general process which is applicable to A cos θ + B sin θ = C. We have

 

An alternative method is to express the sine and cosine each in terms of the tangent of half the angle (see E.T., p. 263). Equations of this type should not be solved by methods involving the squaring of both sides; for since this process is not reversible the solutions obtained need not necessarily satisfy the equation, and testing becomes necessary.

Example 8. Solve sin 2θ = sin θ.

Using the same method as in Example 5, we find

 

and this result follows also by using

 

We can, however, write the equation in the form 2 sin θ cos θ = sin θ;

 

This illustrates the fact that different methods of solution sometimes give results of different forms; in the present example the aggregate of values of (2n + l) can be separated into two parts, those for which 2n + 1 is and is not a multiple of 3; the first of these taken with the values of 2nπ just make up the values of nπ and the second part is the same as the values given by 2nπ ± .

EXERCISE III. b.

Solve :

1. 2 cos² θ = 1.

2. cos 7θ = cos5θ.

3. sin 3θ = sin 7θ.

4. sin 3θ = cos 2θ.

5. sin 6θ + sin θ = 0.

6. sin 5 θ + cos 3 θ = 0.

7. cot 5θ = cot 2θ.

8. tan 3θ = cot 4θ.

9. sin 2θ = 1 –cos 2θ.

10. sec θ = sec 2θ.

11. sin θ + . cos θ = 1.

12. cos θ – sin θ = 1.

13. sin 3θ = 3 sin θ.

14. 3 sin θ + 4 cos θ = 2.

15. 4 cos θ = cosec θ.

16. 13 sin θ – 84 cos θ = 17.

17. cos θ + cos 2θ + cos 3θ = 0.

18. sin 7θ = sin θ + sin 3θ.

19. cos 3θ = cos θ cos 2θ.

20. tan θ + tan 2θ = tan 3θ.

21. 4 cos θ cos 2θ cos 3θ = 1.

22. sec θ + cosec θ = 22.

23. cos θ + sin θ = 1 + sin 2θ.

24. tan θ + sec 2θ = 1.

25. cos x + tan asin x = sec a.

26. cos 9x cos 7x = cos 5x cos 3x.

27. cot x – cosec 2x = 1.

28. cos (x – a) cos (x – β) = cosa cos β + sin²x.

29. cos³x – cos x sin x – sin³x = 1.

30. cosec 4α – cosec 4x = cot 4α – cot 4x.

31. tan (cot θ) = cot (tan θ).

32. tan^–1 = tan^–1 x.

33. Discuss the solution of sin θ + cos θ = k when (i) k = 1, (ii) k = 2, (iii) k = (l + ).

34. Find θ such that tan θ = and sec θ = – 2 simultaneously.

35. Find θ such that sin θ + sin 3θ = cos θ and sin 4θ = simultaneously.

36. Show that the aggregates of values given by

 

are identical, n being any integer or zero.

37. Are the aggregates of values of nπ + and nπ ± identical, n being any integer or zero ?

MISCELLANEOUS EQUATIONS

Example 9. Find from graphical considerations the number of real roots of x = 3π(1 – sin x), x being measured in radians.

The equation may be written, sin x = 1 – . Sketch the graphs of

 

 

FIG. 25.

The graph of y = sinx lies between the lines y = 1 and y = –1.

The graph of y = 1 – is the straight line joining (0, 1) to (6π, –1).

It is evident that the two graphs intersect at 7, and only 7, points.

∴ the equation x = 3π(1 – sin x) has 7 real roots.

Approximate values of these roots may be found by plotting the graphs carefully ; see also Chapter V, Example 4, p. 82.

Example 10. Solve sin x + sin y = sin c; cos x + cos y = cos c.

We have *

 

and

 

∴ by division,

 

since cos ≠ 0, as this would involve sin c = 0 = cos c;

 

∴ from (ii),

 

∴ if cos c ≠ 0,

 

 

∴ from (iii) and (iv),

 

These solutions may be written

 

where p, q are any integers and the upper signs are taken together, as also the lower signs.

If cos c = 0, cos must be zero; ∴ x + y = (2n + 1)π;

 

It happens that these solutions are contained in the previous general solutions, but as cos c ≠ 0 was assumed in obtaining the general solutions, it could not be anticipated that this would be the case.

EXERCISE III. c.

1. Solve graphically x² = cos x.

2. Solve graphically x = cos² x.

3. How many roots has x = 10 sin x ?

4. How many roots has 2x = 3π( 1 – cos x)?

5. Solve graphically x² = 4(1 – sinx).

6. How many roots are there of x + tan 2x = , which lie between x = 0 and x = π.

7. Find the general expression for the range of values of θ if x + = 4 cos θ is satisfied by two values of x.

8. Show that the condition for sin x(cos x + sin x) = c to have roots is (1 + ) ≥ c ≥ (l – ).

9. Prove that sec x + cosec x = c has two roots between 0 and 2π if c² < 8, and four roots if c² > 8.

10. Solve graphically by a geometrical construction

 

where a, b are given positive numbers. What limitation is there to the values of a and b to ensure that solutions exist ?

How can the cases when a or b is negative be dealt with ?

Solve the equations in Nos. 11-20 :

11. sin (x + y) = , cos (x – y) = – .

12. cos x cos y = , sin x sin y = .

13. tan x = sin 3y, sin x = tan 3y.

14. sin²x – sin x = sin x sin y = sin²y + sin y.

15. cosx+ sin y = 2 cos (x + y) = cosx – sin y.

16. sin θ + sin ϕ = ; cos θ + cos ϕ = , for values of θ, ϕ between 0° and 360°.

17. 5 sin x – 2 sin y = 1; 5 cos x – 2 cos y = 4, for values between 0° and 360°.

18. cos (x + 3y) = sin (2x + 2y); sin (3x + y) =cos (2x + 2y) given that

19. tan x = tan 2y; tan y = tan 2z; tan z = tan 2x for values between 0 and π.

20. sin x + sin 2y = sin y + sin 2z = sin z + sin 2x = 0 for values between 0 and π.

Submultiple Angles.  Given the value of cos θ, to find the values of sin θ and cos θ.

From the formulae,

 

we have

 

The ambiguous signs are due to the fact that it does not follow from cos θ = cos α that θ = α; the correct conclusion is θ = 2nπ ± a, from which

 

and

 

The ambiguity of sign can, of course, be removed if the actual value of θ is given, and this is done most easily by reference to a figure.

 

FIG. 26.

For example, if θ = 400°, cos ( = cos 200°) is negative and sin (= sin 200°)is also negative, this, in that case, the minus sign must be taken in both formulae. For the general results, see Ex. III. d, Nos. 2, 3.

Given the value of sin θ, to find the values of sin θ and cos θ.

We begin by finding sin + cos and sin – cos .

Since

 

and similarly

 

On account of the two ambiguous signs, there are four possibilities. From sin θ = sin a, the conclusion is

 

whence

 

and

 

If the actual value of 0 is given, it is possible to remove the ambiguities. For example if θ = 320°, then = 160°, so sin is positive and cos is negative; also the cosine is numerically the greater; thus, in that case,

 

and

 

whence

 

For the general results, see Ex. III. d, Nos. 9, 10.

We can also determine for what values of θ a particular formula, such as

 

will hold; for this requires

 

Now

 

which is positive for ; therefore the first result holds if the angle ( = XOP) is such that its arm OP lies within the angle shown in Fig. 27.

 

FIG. 27.

 

FIG. 28.

Similarly,

 

and so the second result holds if OP lies within the angle shown in Fig. 28. Thus, for both results to be true the arm must lie within the angle shown in Fig. 29,

i.e.

 

or

 

 

FIG. 29.

Given the value of cos θ, to find the values of cos .

There will be three values ; for if cos θ = cos a, θ = 2nπ ± a

 

and so cos equals cos, cos , or cos .

Hence from the identity cos θ = 4 cos³ – 3 cos , it follows that the values are the roots of the equation 4 cos³ – 3 cos = cos a, which is a cubic for cos .

Cubic Equations.  The general cubic equation

 

can be transformed by the substitution

 

where H = ac – b² and G = a²d – 3abc + 2b³.

The further substitution y = k cos θ gives

 

and if k is chosen so that k³ : 3Hk = 4 : – 3, this becomes

 

or

 

and

 

From this, it may be possible to find 3θ, and three possible values of cos θ, giving three values of y and hence of x.

The conditions for the possibility are that H should be negative, and numerically less than unity. Both conditions are included in

 

This is precisely the condition for the cubic equation to have three real roots, and this method of solution is applicable therefore just to that case in which the usual algebraic solution breaks down.

EXERCISE III. d.

1. Give the signs to be used in the formulae

 

when θ is

(i) 70°;

(ii) 110°;

(iii) 200°;

(iv) – 50°;

(v) 300°;

(vi) 3000°.

2. Show that

 

3. Obtain results like those in No. 2 for sin θ.

4. Determine the signs of sin θ + cos θ and sin θ – cos θ when θ is (i) 340°, (ii) 480°, (iii) 1360°.

5. If prove 2 sin θ = – (1 + sin θ) + (1 – sin θ), and obtain the formula for 2 sin θ, when θ is

 

6. Determine the signs in 2 cos θ = ± (1 + sin θ) ± (1 – sin θ), when θ is in the neighbourhood of 280°.

7. Determine the signs in

 

8. Determine the signs in

 

when 2θ lies between and .

9. By writing sin + cos in the form . sin , show that sin + cos = + (1 + sin θ), if (4n – ) π < θ < (4n + ) π, and that it = – (1 + sin θ), if (4n + ) π < θ < (4n + ) π.

10. Obtain results corresponding to those in No. 9 for sin – cos . Determine the ranges of values of θ for which the following results (Nos. 11-14) hold:

11. .

12. .

13. .

14. .

15. Draw figures and use them to obtain the possible values of sin , when (i) cos θ = cos 120°; (ii) cos θ = cos 300°.

16. Draw figures and use them to obtain the possible values of cos , when (i) sin θ = sin 60°; (ii) sin θ = sin 240°.

17. Prove that , and find sin .

18. If cos θ = and (4n – 2)π < θ < 4nπ, find sin .

19. If cos θ = – and (4n – 1) π < θ < (4n + 1) π, find cos .

20. Express tan in terms of tan θ. Determine the ambiguous sign for the cases when θ is between

 

21. Prove tan x + cot x = 2 cosec 2x, and use it to express tan in terms of sin θ.

22. Prove and show how to determine the sign.

23. If sin θ has the given value sin α, find the possible values of .

24. Draw figures and use them to obtain the possible values of cos , when (i) cos θ = cos 60°, (ii) cos θ = cos 210°.

25.* Prove that .

26. Solve (i) x³ – 12x + 8 = 0 ; (ii) x³ – 12x = 4.

27. Solve (i) x³ – 27x – 27 = 0; (ii) x³ – 27x + 40 = 0.

28. Solve x³ + 3x² – 9x – 3 = 0.

29. By putting x = k cos θ, reduce the equation x⁵ – 5x³ + 5x + 1 = 0 to the form cos 5θ = c and hence solve it.

Inverse Functions. The equation y = sin x, regarded as an equation for x in terms of a given number y (between – 1 and +1) has an unlimited number of solutions.

 

FIG. 30.

If x = α is one solution, the others are

 

The equation x = sin^–1y will be used, at present, to signify that x is equal to one of these values.

Similarly cos^–1k will denote any one of the angles whose cosine is k, and tan^–1k any one of the angles whose tangent is k.

Thus sin^–1k, cos^–1k, tan^–1k are mapy-valued functions of k.

Relations between Inverse Functions. Ordinary trigonometrical identities can often, with advantage, be expressed in terms of the inverse functions. For example, from

 

by putting tan θ = m, tan θ′ = m′ we get

 

which may be written

 

or

 

This form happens to be more convenient for certain purposes; it only implies that when any value of tan^–1m′ is subtracted from any value of tan^–1m the result is one of the values of

 

Similarly from the expansion of tan(θ + θ′) we deduce the result

 

Similarly from

 

by putting sin a = x, sin β = y, we get

 

and from cos (α + β) = cos α cos β – sin α sin β, we get

 

Equations like (7) and (8) are really alternative statements of the same fact, and the reader should try to pass from one to the other without going through the process of substitution.

EXERCISE III. e.

1. (i) Prove that cos^–1x = ±sin^–1 (l – x²).

    (ii) If cos^–1x = tan^–1p = cosec^–1q, express p and q in terms of x.

2. If tan^–1 x = sin^–1p = cos^–1q = cot^–1r, express p, q, r in terms of x.

3. How can you construct geometrically cosec^–1 1? Use the figure to express this angle in the forms, cos^–1p, tan^–1q.

4. Prove that sin (cos^–1x) = ± (1 – x²).

5. Express in terms of x (i) cos (sin^–1x); (ii) tan (sin^–1x).

6. Prove that the general value of 2 cos^–1x equals

 

Give two simple values of 2 cos^–1x + cos^–1(2x² – 1).

7. Express 2 sin^–1x in the form sin^–1y.

8. Prove that .

9. Find the simplest value of tan–1 + tan^–1 .

10. Find the simplest value of

 

11. Prove that .

12. Simplify cos (2 sin^–1x).

13. Find a value of x, such that .

14. Express 3 sin^–1x in the form sin^–1y.

15. Find a simple value of cosec^–1 5 + cot^–13.

16. Find the general value of tan^–1 (cot x)+ cot^–1 (tan x).

17. Evaluate cos 2 {tan^–1x + tan^–1 y}.

18. .

19. Prove

 

20. Express sin^–1 p = cos^–1 q as an algebraic relation between p and q.

21. If tan^–1 p + tan^–1x + tan^–1r = , prove qr + rp + pq = l.

22. If cos^–1 a + cos^–1b + cos^–1 c = π, prove that

 

Find the simplest solutions of the following equations :

23. cot^–1 2 = cot^–1 x + cot^–1 7.

24. tan^–1 x + tan^–1 (1 – x) = tan^–1 .

25. .

26. 6 cos^–1(2x² – 1) = π.

27. .

MISCELLANEOUS EXAMPLES.

EXERCISE III. f.

Solve the equations : (Nos. 1-10).

1. tan x + tan 2x = 0.

2. sin 3x + cos 2x = 0.

3. sin x + sin 3x = 2 cos x.

4. sin² x+ cos 3x cos x = l.

5. tan x + tan (x + α) + tan (x + β) = tan x tan (x + α) tan (x + β).

6. tan x cot (x + α) = tan β cot (β + α).

7. cos x cos c + sin a sin b = cos (x – a)cos (x – b).

8. .

9. 70 cos θ – 24 sin θ = 37.

10. sin θ + sin 2θ + sin 3θ + sin 4θ = 0.

11. If sec α sec θ + tan α tan θ = sec β, find tan θ.

12. Find for what values of θ, 2 sin θ – tan θ > 0.

13. Solve x + y = 2a, cot x + cot y = 2 cot a.

14. Solve a² cos θ + b² cos ϕ = c² cos a; a² sin θ + b² sin ϕ = c² sin a.

15. Find all values of x, y, such that

 

16. Solve

 

17. Show that the roots of

 

are θ = nπ or nπ + tan^–1 (tan α + tan β + tan γ + tan α tan β tan γ).

18. Prove that the roots of tan³ x tan = 1 also satisfy

 

19. Investigate how many values of sin θ (between –1 and +1) satisfy sin² θ – 2c sin θ + 5c – 6 = 0 for various values of c.

20. Discuss the solution of cos² x – 2m cos x + 4m² + 2m – 1 = 0 for various values of m.

21. Show that if bh cos θ + ak sin θ = ab has roots for cos θ, they always determine values of θ.

22. Express sin in terms of sin θ. when θ is in the neighbourhood of 420°. For what precise neighbourhood is the result valid ?

23. Prove that tan is one of the values of , and find the other values.

24. Prove . (See footnote, p. 46.)

25. If p is an integer and – 1 < q < 1, find the number of possible values of sin x, such that (i) sin 2px = q, (ii) sin (2p + 1)x = q.

26. Solve x⁵ – 5k²x³ + 5k⁴x = 2k⁵ cos a, for x in terms of a and k.

27. Simplify .

28. Prove that .

29. Use the result of No. 28 to express in the form tan^–1 + tan^–1 . Also express tan^–1 and tan–1 each in the form where m and n are positive integers.

30. Prove that .

31. Prove that .

32. Prove that .

33. Find x if one value of cos^–1 x + cos^–1 2x equals .

34. Find a value of x between 0 and , such that

 

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* This solution illustrates some points of importance; other methods would be shorter for this particular example; e.g. the answers might be written down by geometrical considerations.

* [x] denotes the greatest integer that is not greater than x; thus [] = 3, [5] = 5, [ –4] = –4, and [ – ] = – 3.