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Hence, The sine of the sum of two angles is equal to the sine of the first into the co-sine of the second, plus the cosine of the first into the sine of the second.

From the diagram we find the following relations:

(1) CP

CM NL.

(2) CP = cos OCB cos (a + b).

(3) CM= cos MCL × CL = cos a cos b.
(4) NL
sin NBL × LB

sin a sin b.

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C

B

A

L

P M

Substituting the values of CP, CM, and NL, found in (2), (3), and (4), in (1), we have

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

Hence, The co-sine of the sum of two angles is equal to the product of their co-sines minus the product of their sines.

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90. Problems.

1. Prove that formulas (a) and (b) become homogeneous by introducing R.

2. Prove that formulas (a) and (b) are true when (a+b) is in the second quadrant.

6. Develop sin (45°+30°) by formula (a).

7. Develop cos 105° by formula (b).

3. Prove that formulas (a) and (b) are true when (a+b) is in the third quadrant.

4. Prove that formulas (a) and (b) are true when (a+b) is in the fourth quadrant.

5. Deduce formula (b) from formula (a) by substituting 90° a for a, and b for b, and reducing by articles 85-86.

91. Problem.

To find the sine and co-sine of the difference of two angles.

Let a = the angle OCA, and b the angle BCA.

Draw BL perpendicular to CA, LP and BM perpendicular to CO, and BN parallel to CO.

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The triangles NLB and PCL are similar, since their sides are respectively perpendicular; hence, the angle NLB, opposite the side NB, equals the angle PCL opposite the homologous side PL. But the angle PCL = a; hence, the angle NLB = a. Then we shall have

L

sin (a — b).

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P M

sin a cos b.

cos a sin b.

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Substituting the values of MB, PL, and NL, found in (4), (5), and (6), in (3), we have

(c) sin (a - b) = sin a cos_b

cos a sin b.

Hence, The sine of the difference of two angles is equal to the sine of the first into the co-sine of the second, minus the co-sine of the first into the sine of the second.

From the diagram we find the following relations:

(1) CM CP + NB.

(2) CM =cos OCB cos (a - b).

(3) CP = cos PCL × CL = cos a cos b.
(4) NB = sin NLB × LB

sin a sin b.

Substituting in (1) the values of CM, CP, and NB found in (2), (3), and (4), we have

(d) cos (a - b) = cos a cos b+ sin a sin b.

Hence, The co-sine of the difference of two angles is equal to the product of their co-sines, plus the product of their sines.

92. Problems.

1. Prove that formulas (c) and (d) become homogeneous by introducing R.

2. Deduce formulas (c) and (d) from (a) and (b), respectively, by substituting -b for b, and reducing by

article 85..

3. Prove that formulas (c) and (d) are true when (a - b) is in the second quadrant.

4. Prove that formulas (c) and (d) are true when (ab) is in the third quadrant.

5. Prove that formulas (c) and (d) are true when (ab) is in the fourth quadrant.

93. Problem.

To find the tangent and co-tangent of the sum or difference of two angles.

Dividing (a) by (b), we have

sin (a + b) sin a cos b + cos a sin b
cos (a + b) cos a cos b- sin a sin b

=

Dividing both terms of the fraction in the second member by cos a cos b, reducing, and recollecting that

the sine of an are divided by its co-sine is equal to its tangent, we have

(e) tan (a + b)

Hence, The tangent of the sum of two angles is equal to the sum of their tangents, divided by 1 minus the product of their tangents.

(f) cot (a + b)

Dividing (b) by (a), and reducing, we have cot a cot b 1

cot a cot b

(g) tan (a

tan atan b

1 tan a tan b

- b)

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Hence, The co-tangent of the sum of two angles is equal to the product of their co-tangents, minus 1, divided by the sum of their co-tangents.

Dividing (c) by (d), and reducing, we have

tan a tan b

1 tan a tan b

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Hence, The tangent of the difference of two angles is equal to the tangent of the first minus the tangent of the second, divided by 1 plus the product of their tangents.

Dividing (d) by (c), and reducing, we have
cot a cot b+1
cot b
cot a

(h) cot (a b)

94. Problems.

Hence, The co-tangent of the difference of two angles is equal to the product of their co-tangents, plus 1, divided by the cotangent of the second, minus the co-tangent of the first.

1. Prove that (e), (f), (g), (h) become homogeneous by introducing R.

2. Deduce (g) from (e) by substituting -b for b. 3. Deduce (h) from (f) by substituting -b for b. 4. Deduce (f) from (e) by taking the reciprocal of each member, substituting for tan a, for tan b,

1

1

cot a

cot b

and reducing.

5. Deduce, in like manner, (h) from (g).

4

6. Find the value of sin (a+b+c) by substituting bc for bin (a).

7. Find the value of cos (a+b+c) by substituting bc for bin (b),

8. Find the value of tan (a+b+c) by substituting bc for bin (e).

9. Find the value of cot (a+b+c) by substituting bc for b in (ƒ).

95. Functions of Double and Half Angles.

Making b=a in (a),

(b), (e), and (ƒ), we have

(1) sin 2 a

2 sin a cos a.

(2) cos 2 a

Substi

S

(3) tan 2a =

(4) cot 2a =

cos2 a - sin2 a.

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tuting a for a in (1), (2), (3), (4), we have

Dividi(5) sin a= 2 sina cosa.

member 6)

cos α = = cos 2a-sin2 a.

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