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is the form of the general quadratic factor of x2+1.

When n is even and =2m there are m such factors.

When n is odd and 2m+1 there are m such factors; the remaining factor is x+1, as is clear from the figure.

EXAMPLES. IX.

(1) Find the roots of the equation x2+1=0, and write down the quadratic factors of 4+1.

(2) Write down the quadratic factors of x6 +1.

(3) Write down the general quadratic factor of x2o +1=0. (4) Find all the values of 1. (5) Find the factors x13+1. (6) Find a general expression for all the values of -1.

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where scr stands for the sum of the products of the sines taken r together each multiplied by the product of the remaining n-r cosines.

(2) With the notation of Ex. 1, prove that

cos (a1+a2+az n terms)=cn- Cn−2$2 + Cn−484 — etc.

...

(3) Write down the expansion of

sin (a+B+y+d+e) and of cos(a+B+y+d+e).

(4) Prove that in the series of expressions formed by giving

to r in

(

COS

0+2rπ
q

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in succession, the product of any two equidistant from the beginning and the end is constant.

15

(5) One value of (√3+ √-1) is 27 (√-1+1).

(6) From the identity

+

+

(x-b)(x-c)(x-c) (x − a) (x − a) (x —b) ̧
— 1,
(a-b) (a-c) (b-c) (b-a) (c-a) (c-b)

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sumptions for a, b and c that

sin (0-8) sin (0-y)
sin (a-ẞ) sin (a-y)

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deduce by assuming x-cos 20+i sin 20, and corresponding as

two similar expressions=0.

(7) Prove that the n nth roots of unity form a series in G. P.

CHAPTER III.

RESULTS OF DE MOIVRE'S THEOREM.

18. We proceed to deduce many important results from De Moivre's Theorem.

We shall generally in this chapter write i for √-1.

19. By Art. 12, when n is an integer we have
cos n✪ + i sin no = (cos 0 + i sin 0)" ;

Expand the right-hand side of this identity by the binomial theorem, remembering that -1 and that i=+1. Equate the real part of the result to cos no. This gives us

cos no cos"0

n (n − 1)

=

cos"-20. sin'0

12

+

n (n − 1) (n − 2) (n − 3)

|4

cos"-40. sin10 - etc.

Equate the imaginary part to i sin no. This gives us

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20. In the above n is a positive integer, and the last terms in the series for cos no and for sin ne will be different according as n is even or odd.

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(1) sin 40-4 cos30. sin 0-4 cos 0. sin3 0.

(2) cos 40=cos1 0-6 cos20. sin20+ sin1 0.

(3) The last term in the expansion of cos 100 is - sin1o 0. (4) The last term in the expansion of sin 120 is

- 12 cos. sin11 0.

(5) When n is even the last term in the expansion of cos ne is

n

(-1) sin" 0.

(6) When n is odd the last term in the expansion of cos ne is

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EXPONENTIAL VALUES OF SINE AND COSINE.

21. By De Moivre's Theorem, when n is any commensurable number, and x any angle,

(cos nx + i sin nx) is a value of (cos x + i sin x)”. For x put the unit of angular measurement; then (cos n + i sin n) is a value of (cos 1 + i sin 1)”. Let k stand for (cos 1 + i sin 1), then

(cos n + i sin n) is a value of k",

where k is independent of n.

Whatever other values (cos 1+ i sin 1)" may have, in what follows we shall only use the value (cos n + i sin n).

22. This important result is a symbolical statement of the fact that expressions of the form cos n + i sin n are combined by the laws of indices.

23. Let the unit of angle be a radian.

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[E. 59.] [Art. 21.]

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1

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= 2 {log + (log) + etc.}. [Art. 3.]

13

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where R is finite for all values of (since sin 0 is always less

than 0, and.. log.k is finite).

Let be infinitely diminished.

Then, since is the

sin 0

circular measure of the angle, the limit of is 1. [E. 290.]

Also the limit of the right-hand side is log.k.

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Therefore, when is the circular measure of the angle,

cos + √ 1 sin 0 = e√—1o ̧

i0

=e

-io

24. Since cos + i sin =e and cos i sin = e ;

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are exponential values of the cosine and sine respectively, when the angle is expressed in circular measure. These results may be applied to prove any general formula in Elementary Trigonometry.

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prove the

Use the exponential values of the sine and cosine to following: (1) cos2a+sin2a=1. (2) cos 2a = cos2 a - sin2 a.

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a

cos 0=cos ( − 8).

- sin2 ß=cos2 ß – sin2 a.

(7) sin 30-3 sin 0 - 4 sin3 0.

(5) cos (a+B). cos (a — ß)=cos2 a
(6) cos 30=4 cos3 0 – 3 cos 0.
(8) 2 cos na. cos a=cos (n+1) a+cos (n − 1) a.
(9) 2 sin na (-1) 22 sin2 a

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