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342. The projection and erection of any directed length upon any indefinite line are obtained by multiplying the directed length by the cosine and sine respectively of the angle which the positive direction along the directed length makes with the positive direction along the indefinite line.

For consider a line to revolve round 0 from its (positive) initial direction parallel to X'X or XX' to its (positive) final direction along OH or HO (produced), and so describe an angle 0. [See fig. Art. 339.]

Then (OB), (OH)-directed from 0-are respectively base and hypothenuse.

And (BH)directed from the initial line—is the perpendicular. Thus (OB)

(BH) = cos O and

= sin 0. (OH)

(OH) .: (OB) =(OH) cos 0 and (BH)=(OH) sin 0. 343. To find the cosine and sine of the sum of two angles by projections and erections.

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Let (X0Y) be any angle A; (YOZ) any angle B; then (X0Z) = A +B.

In (OY) take any point Q : through draw QX' in the ilirection Ox, and QE making a positive right-angle with OY.

Then LX'QE is always 90° + X'QY i.e. 90° + A.

Let QE, 02-either being produced, if necessary-cut in P. Then, by Art. 340,

proj. of (OP) = proj. of (0Q) + proj. of (QP), and erec. of (OP) = erec. of (OQ) + + erec. of (QP). Taking these projections and erections upon OX,

(OP) cos (XOZ) = (02) cos (-XOY) + (QP) cos (X'QE), and (OP) sin (XOZ) = (OQ) sin (XOY) + (QP) sin (X'QE).

But (OQ), (QP) are, respectively, the projection and erection of (OP) upon (OY).

:: (0Q)=(OP) cos (YOZ) and (QP)= (OP) sin (YOZ). Substituting and dividing by (OP) we have

cos (A + B) = cos B.cos A + sin B. cos (90° + A),

sin (A + B) = cos B. sin A + sin B. sin (90° + A).
But, by Art. 326,
cos (90° + A) =- sin A and sin (90° + A) = + cos A,

cos (A + B) = cos A cos B sin A sin B, and

sin (A + B) = sin A cos B+cos A sin B.
The above proof is perfectly general.
Hence, writing - B for B,
since cos (B)= + cos B and sin (- B)=-- sin B,

cos (A B) = cos A cos B + sin A sin B, and sin (A B) = sin A cos B – cos A sin B.

344. The student should observe that the important triangular formula

csin B = b sin C,

c cos B + b cos C = A, are obtained simply by erecting and projecting upon BC. J. T.

17

§ 3.

EXTENSION TO THREE OR MORE ANGLES.

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345. To find the sine, cosine, and tangent of the sum of three angles. (i) sin (A + B + C) i.e. sin {(A + B) + C}

= sin (A + B) cos C + cos (A + B) sin C = sin A cos B cos C + cos A sin B cos C + cos A cos B sin C

– sin A sin B sin C. (ii) cos (A + B + C) i.e. cos {(A + B) + C}

= cos (A + B) cos C – sin (A + B) sin C = cos A cos B cos C

- sin A sin B cos C - sin A cos B sin C cos A sin B sin C. (iii) tan (A + B+C) i.e. tan {(A + B) + C}

tan (A + B) + tan C 1 – tan (A + B) tan C tan A + tan B

+ tan c 1- tan A tan B

(tan A + tan B) tan C
1-

1- tan A tan B
tan A + tan B + tan C – tan A tan B tan C
1-tan A tan B-tan A tan C -tan B tan c

+

346. These formulæ should be carefully noted and their symmetry examined. In the same way we may prove formula involving three angles one or more of which is subtracted. 347. The following results should be noticed :

(cos A + sin A) (cos B+ sin B) = cos A cos B + sin A cos B + cos A sin B + sin A sin B. Here first term – last term = cos (A + B), and

the middle terms = sin (A + B). Again

(1 + tan A)(1+tan B)

= l + tan A + tan B + tan A tan B. Here first term last term = den. of tan (A + B), and

the middle terms = num. of tan (A + B).

+

=

348. Again

(cos A + sin A (cos B + sin B) (cos C + sin C) (cos A cos B cos C)

+ (sin A cos B cos C + sin B cos C cos A + sin C cos A cos B)
+ (cos A sin B sin C + cos B sin C sin A + cos C sin A sin B)

+ (sin A sin B sin C). Here, arranging the terms in groups according to the number of sines in each term, we have

first
group
– third

group = cos (A + B+C),
second group - fourth group = sin (A + B + C).

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349. The result suggested by the last two articles may be shown to be true for any number of angles.

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1

350. Arrange the product

(cos A + sin A) (cos B + sin B) (cos C + sin C)... to any number of factors, in groups of terms, according to the ascending number of sines in each term. Then the expansions of

cos (A + B + C + ...) and of sin (A + B + C + ...) will consist respectively of the above groups, taken alternately but with alternate sign.

In other words, if the product of n factors (cos A + sin A) (cos B + sin B)... = S, +S,+ ... +S,+ ... + Sn (1), wherein each term of the group S, contains r sines, then cos (A + B + ...) = 8.-S,+SA S. S4

(2), sin (A + B + ...)= 8.- Sg + S S-S .........(3).

3 For assume this to be true for n angles.

Introduce an (n + 1)th angle, X say. Then, by multiplication of (1),

(cos A + sin A) (cos B + sin B)... (cos X + sin X) = ($. +S,+ + Sn) cos X + (S. +S,+ + Sn) sin X = S, cos X + (S, cos X + S, sin X) + (Secos X + S, sin X)

+(Sz cos X + S, sin X) + ... ... ... (1'),

+

=

1

=

2

where the terms are grouped according to the number of their sines.

But

cos (A + B + + X) = cos (A + B+...) cos X – sin (A + B + ...) sin X
= (S6-S+S4 - ...) cos X - (S. - Sz+Ss - ...) sin X
= S, cos X - (S, cos X+S, sin X) + (S. cos X + S, sin X) - ...(2),
by (2) and (3).

And sin (A + B + ... + X)=sin(A + B + ...) cos X + cos (A + B +...) sin X

= (S. -S,+S- ...) cos X+($. -S,+S4 - ...) sin X
= (S, cos X + S, sin X) - (Sz cos X + S, sin X) + ...... (3),
X X

' by (2) and (3).

Comparing (1'), (2), (3') we see that the proposition holds for (n + 1) angles, if it is assumed for n. Now it has been seen to be true for 2 or for 3 angles, .. it is true for any number of angles.

0

2

351. In the same way the student may show that if the product (1 + tan A) (1 + tan B) (1 + tan C)...to any number of factors is arranged in ascending groups of terms according to the number of tangents in each term, then the den. and num. of tan (A + B + C + ...) will consist respectively of the above groups, taken alternately but with alternate sign. In other words If (1+tan A)(1+tan B)... =1+T, + T', + +T, + ... + I'n

T.-T,+ T5 - ... then

.

T4

tan (A + B+C+ ...)= 1-T,+T, - ...

3

1

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+

+1,

Similarly, if (cot A + 1) (cot B + 1)... = K, + K + K,

+ K,

+ where K, denotes the group of terms containing n-r cotangents,

K-K,+ K.-... then cot (A + B + C + .)

K,-K,+K, - ...

r

0

2

4

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