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41. To shew that if A be less than 45°,

cos A = } {V (1 + sin 2 A) + V (1 – sin 2 A)}
then
sin A = {{V

(1 + sin 2 A) - V (1 – sin 2 A)}. By Art. 31, if A < 45°, cos A is > sin A, and they are both positive; therefore cos A + sin A and cos A - sin A are both positive quantities when A is < 45° ;

:: cos A + sin A = + (1 + sin 2 A),

and cos A – sin A = + (1 – sin 2 A). .. by addition and subtraction,

2 cos A= (1 + sin 2 A) + (1 - sin 2 A),
2 sin A = V(1 + sin 2 A) - V(1 – sin 2 A);
cos A = }{V (1 + sin 2 A) + ✓(1 – sin 2 A)},
\sin A = } {V (1 + sin 2 A) - ✓(1 – sin 2 A)}.

42. The equations deduced in Art. 40 can be applied in any case to determine the sine and cosine of A from the sine

of 2 A.

For example, if A be an angle > 3 x 45° but <4 x 45', (i. e. any angle comprehended under the form 180° B, where B is < 45o), cos A is negative, and greater in magnitude than sin A, which is a positive quantity. In this case therefore,

cos A + sin A ✓(1 + sin 2 A),

cos A – sin A = - ✓(1 – sin 2 A);
... cos A -}{V(1 + sin 2 A) + V (1 – sin 2 A)},
sin A =

}{V(1 – sin 2 A) - V (1+ sin 2 A)}. If at first sight this value of sin A appear to be negative, it is to be remembered that sin 2 A is a negative quantity, (2 A being between 270° and 360") and therefore 1 - sin 2 A is greater than 1+ sin 2 A,—and therefore the value of sin A is here a positive quantity, as it ought to be.

43. Given the. tangents of two angles, to find the tangents of their sum and their difference.

sin (A + B) Tan (A + B)

cos (A + B)

=

sin A.cosB + cos. A sin B

cos A. cos B-sin. A sin B' and dividing the numerator and the denominator by cos A.cos B,

sin A sin B

+ tan (A + B)

sin A sin B 1

cos A cos B

COS A

cos B

tan A + tan B 1-tan A . tan B

Similarly, tan (A B)

tan A

tan B
1 + tan A. tan Bo

2 tan A
Cor. 1. If B = A, tan 2 A =

1 - (tan A)*
Cor. 2. If B = 45', since tan 45o = 1, (Art. 32.)

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sin A or =

sin A + cos A

..(4);

Cor. 4.

1+tan A tan A-1 Tan (A+45°) + tan (A – 45°) =

+

1-tan A tan A+1 4 tan A 1 - (tan A) = 2 tan 2 A, by Cor. 1..............(5).

If A be < 45° ;

Since tan (A – 45°) = tan – (45o – A) - tan (45o – A),

(Art. 23.) the last expression becomes, tan (45° + A) – tan (45o – A) = 2 tan 2 A.........(6).

tan A + tan B sin (A + B) 44. To shew that

tan A tan B

sin (A B)
sin A sin B

+
tan A + tan B
tan A – tan B sin A sin B

COS A

cos B

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sin (A + B)

sin (A B) 45. Given tan A, tan B, tan C, to find tan (A + B + C). tan (A + B + C) = tan {(A + B) + C} tan (A + B) + tan C

(Art. 43.),
1 – tan (A + B). tan C
tan A + tan B

+ tan c
tan A . tan B
tan A + tan B

tan C
1 – tan A. tan B

=

1

1

.

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

E

Cor.

If A + B + C = (2n + 1). 180', n being an integer,

tan (A + B + C) = 0;

.. tan A + tan B + tan C – tan A . tan B. tan C = 0,

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tan A + tan B + tan C = tan A. tan B. tan C.

And since, if n = 0, A + B + C = 180°, this relation between the tangents of A, B, C, is that which obtains between the tangents of the angles of any plane triangle.

In the same manner the tangent of the sum of four or more angles might be found in terms of the tangents of the simple angles.

46. To find the values of sin 2 A, and cos 2 A, in terms of tan A.

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47. The following values of sin 2 A, cos 2 A, tan 2 A, are of frequent occurrence, and necessary to be remembered; those of them which have not been proved already are easily deduced after the method used in the last Article.

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48. In the same way, the following values of sin 2 A, and of cos 2 A, can be found in terms of cot A, cosec A, and versin A.

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sin®2 A = 2.(1 – versin A). ✓ {2 vers A - (vers A)?}.

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