A,

3

3

N
1

3

PN

3

OP

16. To show that sin (180° + A)

sin A, cos (180° + A)

cos A, Let AOP

And let the revolving line take a position such that P2P, is a straight line. Then, evidently, ZAOP, A N3

A 180° + A. Then, as in last Article-P,N,

- PLN,
ON
ON 1

P.
Hence-
Sin (180° + A)
= sin AOP
P,N

sin A. OP OP

ON ON
Cos (180° + A) = cos AOP;

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

A)

sin A, and

A) = cos A,
Let 2 AOP A,

And let the revolving line describe an angle AOP

- A.
Then evidently, if PNP'

A
be drawn perpendicular to
OA, we have (Euc. I., 26)
PN PN.
Hence

PIN PN
Sin (- A)
sin AOP

OP OP

ON ON
Cos (- A)

ОРІОР
And similarly-

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

cosec A.

COS

N

sin A,

= cos AOP

1C

A

COS 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 B

same results for angles of any magnitude whatever.

A 18. To show that tan

2 1

sin A
D
-A Let ZAOC

A;

A Bisect it by the straight line OB, so that ZAOB =

2' and draw CD perpendicular to OA, meeting OB in E. A

ED Then tan tan EOD

(1). 2

OD
OD ED

OD

ED Now (Euc. VI., 2),

and .. OC EC

OC + OD CD' ED CD CD(OC - OD) CD(OC - OD) OD OC + OD OC - OD

CD?
OC - OD OC OC cos A 1

Q.E.D.
CD
OC sin A

sin A

or

COS A

°

19. To find the trigonometrical ratios of 15°, 75°, 120°, 135', 150° . (1.) Ratios of 120°. Sin 120° = sin (180° sin (180° - 120°) = sin 60°

2

Cos 120°

cos (180° - 120°)

cos 60°

1 2'

Tan 120° tan (180° - 120°) tan 60° (2.) Ratio of 150°

1
Sin 150° = sin (180°
sin (180° - 150°)

= sin 30°

2'

=

COS A

A = 15,

=

cos 30°

cos 15° 2 2:

= cos 15°

=

(3.) Ratios of 15°

A 1 By last Art., tan

; put A = 30°, or 2 sin A

2 13

1 1

2 then tan 15°

2 – 13. sin 30°

2 From this result we easily get, Art. 8, 13 - 1

13 + 1 Sin 15°

&c.

2 V2 (4.) Ratios of 75

13 + 1 We have, sin 75o = sin (90° - 159)

2/2

1

V3 - 1 sin (90° – 159) = sin 15°

22

1 tan 75o = tan (90° 15°) = cot 15°

2 13 = 2 + V3, &c. (5.) Ratios of 135o.

1 We have, sin 135o = sin (180° - 135°) = sin 45o =

V2' cos 135o =

cos (180° – 135°)

1

72
tan 135°
tan (180° 1359

tan 45° 1, &c.

cos 750

cos 45o

=

Ex. III. 1. Define a negative angle, and show that tan (- A)

tan A, when A 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

sin A, as A changes from 45° to 315o,

cos A

=l

4. Assuming generally that cos 2 A cos' 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°.

Α.
7. Assuming generally that 2 cosa 1 + cos A, and

2 А

A 2 sino 2 cos A, show that J2 cos

1 + cos A

2 A and 2 sin vī 2

cos A, when A lies between 360° and 540°.

A 8. Given cos A = 1 2 sin? show that sin A

2'
А. Α.
2 sin
2
2

A
9. Hence show that 2 cos VI + sin A

2

A
-VI - sin A, when lies between 135o and 225°,

2

COS

Solve the following equations :10. Cos’ A + jcos A 16. 11. Tan 0 + 5 cot o 6.

2 12. Sin A + sec A

13

+ 1. 13. 2 cos? A 3 sin A. 14. Sin (A + B) = cos (A - B) 15. Tan? A = 2 sino A. 16. Sin (3 A + 75°) = cos (2 A