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(109.) Let a = n A and b = A.

Then sin. {(n+1). A} = sin. nA × cos. A+cos. n A

[× sin. A,

cos. nA × sin. A,

and sin. (n-1). A sin. n Ax cos. A
.. sin. (n+1). A+sin. (n − 1). A= 2 sin. n Ax cos. A*,
and sin. (n+1). A=2 sin. nAx cos. A sin. (n - 1).A.
Also sin. (n + 1). A=sin. (n - 1). A+ 2 sin. A× cos. nA,
and cos. (n+1). A=2 cos. n Ax cos. A-cos. (n-1). A
or cos. (n-1). A-2 sin. n Ax cos. A.
(110.) If in the equations (97) and (100)
a+b=A and a-b= B, then a=+ (A+B) and b = ÷ (A−B)
and sin. (a+b)=sin. ax cos. b+cos. a x sin. b
sin. a x cos. b-cos. a x sin. b,
.•. sin. (a+b) + sin. (a—b) = 2 sin. a× cos. b
and sin. (a+b)—sin. (a - b) = 2 cos. a × sin. b,


sin. (ab)

... sin. A+ sin. B= 2 . sin. ÷ (A+B) × cos. † (A−B) and sin. A-sin. B=2. cos. + (A + B) × sin. ÷ (A – B)

Hence also sin. A+ sin. B=

and sin. A sin. B =

cos. (A-B)
cos. (A + B)
sin. (A-B)

× sin. (A+B),

× ́sin. (A+B).


Hence also sin2. A-sin2. B=sin. (A + B) x sin. (A – B).

* Hence if a series of arcs be taken in arithmetic progression, radius is to twice the cosine of the common difference as the sine of any one arc taken as a mean is to the sum of the sines of any two equidistant


Also radius is to twice the cosine of the common difference as the cosine of any one arc taken as a mean is to the sum of the cosines of any two equidistant extremes.

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And in a similar manner,

cos". B-coso. A=sin. (A+B) x sin. (A-B).

(111.) To find the sin. (A+B+C). Sin. (A+B+C)=sin.(A+B) x cos. C+cos. (A+B)

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[sin. C

=sin. Ax cos. B x cos. C+sin. B x cos. Ax cos. C + sin. Cx cos. Ax cos. B-sin. Ax sin. B x sin. C.

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Prop. XX.






(112.) Having given the tangents of two arcs; to determine the tangents of two arcs which may be equal to their sum and difference.

Let AB, BC be the given arcs, AB being the greater, then AC 'is the sum of the arcs (in Fig. 1.) and their difference (in Fig. 2.); AT the tangent of their sum (1) or difference (2), BP, BO the tangents of the respective arcs AB, BC.

Draw OD perpendicular to SA cutting SB or SB produced in x, .. the triangle OBx is similar to SxD and .. to SBP; hence OB : Bx :: SB : BP, *. OB x BP=SB x Bx. Also by the similar triangles TAS, ODS,

AT : OD :: SA: SD

OD : OP :: SD: Sx,
by the sim. triangles Sx D, ODP,
:. AT : OP :: SA Sx

:: SA : SB- Bx (fig. 1)
:: SA? SB-SB x B x
:: SAS : SA - OBX BP.


Again (in Fig. 2.) AT: OP :: SA : Sx

:: SA: SB+Bx

::SA: SB2+SB x B x
:: SA: SA2+OB × BP.

Hence AT = SA ×


tang. (a + b) =


tang. 5a=


COR. 1.

(114.) COR. 2.

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Tang. 3a=



COR. 2. If a = b, tang. 2a =

tang. 4a=

Tang. (a+b)=

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If rad. 1, AB = a, BC = b, tang. atang. b tang. a x tang. b



SA2 + OB × BP'

5 tang. a-10 tang3. a + tang3. a
1-10 tang. a +5 tang4. a


-tang. a

6 tang. a- 20 tang3, a+6 tangs. a
1-15 tang2. a+15 tang4. a·
(115.) If (in Art. 113.) a = 45°,
tang. (45°+b)

1+tang. b

1 – tang. b'

sin. (a+b)
cos. (a+b)

sin. a
sin. b
cos. b

cos. a




3. tang. a-tang3. a


1-3. tang2. a



1. ax sin. b

cos, a x cos.b


4. tang. a-4 tang3. a
1 − 6 tang2. a + tang4. a

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2. tang. a

1 - tang2. a

sin. ax cos. b+cos. a x sin. b
cos. a x cos. b- sin. a x sin. b

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tang. a+tang. b
1-tang. a tang. b


and tang. (45 – b)

... tang. (45° + b) — tang. (45° - b)=


that tang. a =

cos. 2 α =


2 tang. b
1-tang2. b

1 - tang. b
1+tang. b'


(116.) Tang. a+cotang. a=2 cosec. 2 a.

sin. a cos. a
cos. a

sin. a

1 - tang3. a

1+tang. b
1- tang. b

and tang. (45+b) = tang. (45 − b) +2 tang. 2b. If the tangents of arcs less than 45° be known, the tangents of arcs greater than 45° may be determined by addition.

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= 2 tang. 2b,

For tang. a + cotang. a =

sin2. a+cos2. a


sin. a x cos. a

sin. 2 a

In the same way it may be shewn that


cotang. a tang. a 2 cotang. 2a.
and sec. a+tang. a = tang. (45°+§ a)
sin. 2 a
1+cos. 2 a' cotang. a=

sin. 2a


1 - tang.
1+tang. b

and sec. 2a =


=2 cosec. 2 a.

1- cos. 2 a

sec2. a 1- tang2. a

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1 + tang2. a


(117.) The base and altitude of a triangle being given; to find its area.

(Eucl. I. 41.) The area of the_triangle_ABC = AC × BD. (See next Fig.).

(118.) COR. 1. If a the altitude BD, and b= the

and a =




base AC, the area S=



two of the terms.. being known, the third may be found.

and b =

Let AB, AC and the angle BAC be given. Let fall the perpendicular BD, AB BD :: rad. sin. A,

ABX sin. A



(119.) Two sides of a triangle, and the included angle being given; to find its area.

= AB x sin. A,

:. BD =

(if R = 1),

... the area=AC × BD=↓ ABX ACx sin. A.

2 S
AB x sin. A'

and sin. A =

2 S

(120.) COR 1. Hence the areas of triangles which have one angle in each equal, are as the products of the sides containing those angles. Which is also true of parallelograms.

(121.) COR. 2. AC=

; any



2 S ACx sin. A’


(122.) Given two angles and a side of a triangle ;

to find its area.

Two angles being given, the third is also known.

Let AC be the given side.

Then sin. B: sin. C: AC : AB=

ACX sin. C


sin. B

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