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OP,

OP = - cos A.

16. To show that sin (180° + A) = – sin A,

cos (180° + A) = - cos A, Let AOP, = A,

And let the revolving line take a position such that P2P, is a straight line.

Then, evidently, ZAOP, AN = 180° + A. Then, as in last Article

P:N, = - PAN,

ON, = – ON. - PV HenceSin (180° + A) = sin AOP, = PN =-PN = - sin A. Cos (180° + A) = cos AOP, = 0N3 = - ON, -

And similarly, Tan (180° + A) = tan A, cot (180° + A) = cot A. Sec (180° + A) = – sec A, cosec (180° + A) = – cosec A. 17. To show that sin (- A) = – sin A, and

cos ( - A) = cos A, Let Z AOP = A,

And let the revolving line describe an angle AOP! = - A.

Then evidently, if PNP'
be drawn perpendicular to
OA, we have (Euc. I., 26)
P'N = - PN.
Hence-

PN.
Sin (- A) = sin AOP! - PN -

Op =- OP = - sin A,

1. ON ON Cos (- A) = cos AOP' =

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OPOP = cos A.

And similarly

Tan ( - A) = - tan A, cot (- A) = – cot A,
Sec (- A) = sec A, cosec (- A) = - cosec A.

Although the results of Arts. 12, 15, 16, 17 have been proved from diagrams where A is less than a right angle, the

student will have no difficulty, if he has understood the proofs, in deducing the same results for any les of any magnitude whatever. 18. To show that tan 1 - cos A

sin A

-A Let ZAOC = A; Bisect it by the straight line OB, so that 2 AOB = and draw CD perpendicular to OA, meeting OB in E. A

on ED Then tan = tan EOD = OD ......................(1).

OD ED Now (Euc. VI., 2), 00 = BCand .-00+0D = CD or

OD ED ED CD CD(OC - OD) - CD(OC – OD) OD – OC + OD = OC - OD = CD OC - OD OC - OC cos A - 1 - cos A. Q.E.D. CD - OC sin

A

sin A

Cor 1. Hence, squaring-
Tan? A - (1 – cos A) _ (1 – cos A): _ 1 - cos A
* 2 sin’ A 1 - cos? A – 1 + cos A

lo tane A

.:. Art. 64, page 215,

2 _ cos A

AT 1 + tan 5

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cos? – sino ................(2.) = COSA - (1 - cose A) = 2 coso A - 1.............(3.)

= 2 (1 – sino 4) – 1=1 – 2 sino (4) 119. To find the trigonometrical ratios of 15°, 75°, 120°,

135°, 150°.
(1.) Ratios of 120°.

Sin 120° = sin (180° - 1209) = sin 60° =

Cos 120o = - cos (180° - 1209) = - cos 60o =

Tan 120o = - tan (180o – 120°) = – tan 60° = - 13,&c, (2.) Ratios of 150°.

Sin 150° = sin (180o – 150°) = sin 30o = =
Cos 150° = – cos (180° - 150°) = - cos 30o = -

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Tan 150° = - tan (180° - 150°) = - tan 30 =

&c.

(3.) Ratios of 15°.

A 1 - cos A By last Art., tan * 2 sin A

**; put A = 30', or

1 - 13

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then tan 15° - 1 - cos 30° 1

= sin 30=-1

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2 From this result we easily get, Art. 8,

cos 15

2 /Z' (4.) Ratios of 75o. We have, sin 75o = sin (90° - 159) = cos 15o =

Sin 15o = 13 - 1

2 V2 , &c.

13 + 1 212'

cos 75o = sin (90° – 15%) = sin 15o = .

212'

tan 75o = tan (90° – 15°) = cot 15o =

= 2 + V3, &c. (5.) Ratios of 135o. We have, sin 135o = sin (1800 – 135°) = sin 45o =

cos 135o = - cos (180° – 135°) = - cos 45°

N2

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- tan 45o

tan 135o = - tan (180° - 1359 =

= – 1, &c.

Ex. III. 1. Define a negative angle, and show that tan (- A) = - tan A, when I lies between – 90° and – 180°.

2. Trace the changes of sign of sin A .cos A through the four quadrants.

3. Trace the changes of sign of cos A + sin A, and of cos A - sin A, as A changes from – 45° to 315o.

4. Assuming generally that cos 2 A = cosa A - sino A, trace the changes of sign of cos 2 A as A changes from – 45° to 315o.

5. Write down the sines of 210°, 165°, 240°, - 120°.

6. Show that sin (90° + A) = cos A, and cos (90° + A) = – sin A, for any value of A from 0° to 180°.

7. Assuming generally that 2 cos m = 1 + cos A, and 2 sino 4 = 1 - cos A, show that /2 cos m = - VI + cos A and JZ sin = - VI - cos A, when A lies between 360° and 540°.

8. Given cos A = 1 - 2 sin , show that sin A = 2 sin cos

9. Hence show that 2 cos = - v1 + sin A - VI - sin A, when lies between 135° and 225°.

Solve the following equations :10. Cos’ A + cos A = 1's. 11. Tan 0 + 5 cot a = 6. 12. Sin A + sec A = Ta + . 13. 2 coso A = 3 sin A. 14. Sin (A + B) = cos (A - B) = 3 15. Tan’ A = 2 sinA. 16. Sin (3 A + 75°) = cos (2 A – 15°). 17. Sec 0 + cos 0 = 7Tz: tan 0. 18. Tan 0 + cot 0 = 4.

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