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S. N. 8.

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(14) cota =

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Substituting 1-sin2a for cos2a, then 1-- cos2a for sina, in (6), and reducing, we have

(9) 1- cos a

2 sin2 a.

(10) 1+ cos a

... (11) sina =

... (12) cosa=

Dividing (11) by (12), then (12) by (11), we have

COS a

(13) tana

1
1+ cos a

2 cot a

1

2 cos2a.

1

COS a

2

1+ cos a

2

1 + cos a

1

cos a

Dividing (5) first by (10), then by (9), and transposing, we have

(15) tana

sin a 1+ cos a

sin a

(16) cota 1

Taking the reciprocal of (16), then of (15), we have

1

(17) tana=

(18) cota

Let the formulas of this article be expressed in words.

COS a

cos a

sin a

1 + cos a

sin a

96. Consequences of (a), (b), (c), (d).

Taking the sum and difference of (a) and (c), (d) and (b), we have

(1) sin (a+b) + sin (a - b) 2 sin a cos b.
(2) sin (a + b) — sin (a - b)=2 cos a sin b.
(3) cos (a+b)+cos (a

2 cos a cos b.

- b) (4) cos (a - b) · cos (a + b)

2 sin a sin b.

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(11)

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(12)

-b

(13)

(14)

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8,

Substituting the values of a + b, a-b, a, and b, in (1), (2), (3), and (4), we have

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sin s+sind
sin s-sind

By formula (5) of the preceding article we have

(9) sin (sd) = 2 sin (10) sin (sd) = 2 sin

(8 + d) cos 3 (s + d).
(sd) cos (8―d).
1⁄2

sin s+ sin d

cos scos d

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Dividing each of these formulas by each of the following, we have

sin s+ sin d

cos d

COS S

then

sins + sin d

sin (8+ d)

{

+ d). \

a = 1⁄2 (8
( b = 4 (8 d). S

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2 sin (s + d) cos
2 cos(8+ d) sin 1⁄2
2 cos(8+ d) cos 1
2 sin (s + d) sin

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sin (8+ d)

cos

(8+ d)

d)

cos (8
sin(s-d)

(sd).

(s — d).

- d).

(8-d).

sin(s+d) cos(sd) ___tan}(s+d)
cos(s+d) sin(s—d) tan(s—d)

(8

cos (sd)

cos

(8+ d)

=tan (8+ d).

cot (sd).

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Formula (11) gives the proportion,

sin s+ sind: sin s sind: tan(s + d) : tan 1 (s—d).

Hence, The sum of the sines of two angles is to their difference as the tangent of one-half the sum of the angles is to the tangent of one-half their difference.

Let us apply this principle in solving triangles when two sides and their included angle are given. Article 75.

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sin A+ sin B: sin A-sin B:: tan(A+B): tan(A-B).

... a+ba-b:: tan (A+B) : tan (A—B).

n.

b

97. Theorem.

The square of any side of a triangle is equal to the sum of the squares of the other sides, minus twice their product into the co-sine of their included angle.

A

1st. When the angle is acute.

(1) m b

(1)2 (2) m2 = b2 n2-2 bn.

(3) p2 = p2.

(2)+(3)=(4) m2 + p2 = b2 + n2 + p2 — 2 bn.

2

But m2 + p2 = a2 and n2+ p2 - c2, ... (4) becomes = (5) a2b2+c2 - 2 bn.

But nc cos A, which substituted in (5) gives

(6) a2b2c2-2 be cos A.

B

p

C

D

C

n

B

2d. When the angle is obtuse.

(1)

m = b + n.

(1)2=(2) m2 = b2 + n2 + 2 bn.

(3) p2 = p2.

(2)+(3) = (4) m2+p2=b2+ n2 + p2 + 2 bn.

C

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m

B

a

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But m2+p2=a2 and n2+ p2c2, ... (4) becomes

(5) a2b2+ c2+ 2 bn.

But nc cos

BAD:

.'. (6) a2= b2 + c2— 2 bc cos A.

(1) cos A

98. Problem.

To find the angles of a triangle when the sides are given.

From either formula (6) of the last article we have

b2 + c2 a2
2 bc

V (3)

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=

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c cos BAC= =c cos A.

Hence, The co-sine of any angle of a triangle is equal to the sum of the squares of the adjacent sides, minus the square of the opposite side, divided by twice the rectangle of the adjacent sides.

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Formula (1) gives the natural co-sine of A; hence, A can be found. But it is best to place the formula under such a form as to adapt it to logarithmic computation.

Adding 1 to both members of (1) we have

(2) 1+ cos A

(b + c)2 — a2 (a + b + c) (bc-a) 2 bc 2 bc But 1+ cos 4-2 cos2 A. Article 95, (10). (a+b+c) (b+c-a) p(p-2a) 2 bc

A

Let a+b+c=p, then

2 bc

Substituting these values in (2), and dividing by 2,

we have

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=

1 p(p-a)

bc

1 p ( p − a)
bc

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