=

24. Use of the Preceding Formulæ. I. To express all the other functions in terms of the sine. Since sin? 0 + cos20

.. cos 0

#V1-sin20. sin A

sin 0 tan0 =

3 cos A

V1 – sino 0 1 V1 - sino coto

+ tan A

sin 0 1

1 cos A

V1 – sin20

1 coseco

sin II. To express all the other functions in terms of the tangent.

sin 0 Since tan A =

sec

=

Cos 0 =

sec

sin 0 = tan 0 cos A

tan 0

tan
==
sec A

V1+tan20
1

1
+

V1+tan'a
1
coto

= + V1 + tan”0.
tan

1 V1+tan20 cosec 0 :

+ sin

tan 8

sec A

Similarly, any one of the functions of an angle may be expressed in terms of any other function of that angle. The sign of the radical will in all cases depend upon the quadrant in which the angle 0 lies.

25. Graphic Method of finding All the Functions in Terms of One of them.

To express all the other functions in terms of the cosecant.

Construct a right triangle ABC, having the side BC 1. Then

А

cosec

C

cosec A'

AB AB
cosec A

- AB.
BC 1
.: AC=+ V cosec’A 1.

BC 1
Now

sin A

AB
AC Vcosec- A – 1

==
AB

cosec A BC

1 tan A

==

AC Vcosec’A -1 and similarly the other functions may be expressed in terms of cosec A.

COS A

=

= CB.

m

mv2

2

26. To find the Trigonometric Functions of 45°. — Let ABC be an isosceles right triangle in which

B
CA= CB.
Then CAB= CBA = 45°.
Let

AC=m=
Then
AB’ = AC + CBP = m2 + m2 = 2 m2.

A
.:. AB = m V2.
BC

1
.. sin 45°
AB

V2
AC

1
cos 45o =
AB

V2
BC
tan 45o =

1. cot 45o=1.
AC
sec 45°= =V2. cosec 45°= V2.

m

m

m V2

m

m V2

in

m

27. To find the Trigonometric Functions of 60° and 30°. — Let AB be an equilateral triangle. Draw AD perpendicular to BC. Then AD bisects the angle BAC and the side BC. Therefore BAD= 30°, and ABD = 60°.

B

2 m

B

;

3

3 1. Given sin 0 find the other

5 trigonometric functions.

Let BAC be the angle, and BC be perpendicular to AC. Represent BC by 3, AB by 5, and consequently AC

А by 25 -9=4.

AC 4 Then

AB

4

cos A

or

5. Given tan 6 =V3; find sin and cos 0.

$V3,

12

; find cos 0. 13

6. Given sin A :

7. Given cosec 0 = 5; find seco and tan 0.

find cos 0, tan 0, and cot 0.

V7 3V7 V7 4 7 3

6 find tan .

V c? — 62

28. Reduction of Trigonometric Functions to the 1st Quadrant. - All mathematical tables give the trigonometric functions of angles between 0° and 90° only, but in practice we constantly have to deal with angles greater than 90°. The object of the following six Articles is to show that the trigonometric functions of any angle, positive or negative, can be expressed in terms of the trigonometric functions of an angle less than 90°, so that, if a given angle is greater than 90°, we can find an angle in the 1st quadrant whose trigonometric function has the same absolute value.

29. Functions of Complemental Angles. — Let AA', BB'

be two diameters of a circle at right angles, and let OP and OP' be the positions of the radius for

B any angle AOP = A, and its complement AOP'= 90° — A (Art. 12).

Draw PM and P'M' at right angles to OA.

A

M M

A Angle OP'M'= BOP'= AOP= A. Also OP=OP.

B' Hence the triangles OPM and OP'M' are equal in all respects.

P'M' OM .. P'M'= OM.

OP OP .. sin (90° — A)

= cos AOP

OM' PM Also, OM'= PM.

OP

OP .. cos (90° — A) = sin AOP : sin A. Similarly, tan (90°– A)=tan AOP'=

P'M' OM

= cot A.

OP OP The other relations are obtained by inverting the above.

30. Functions of Supplemental Angles. — Let OP and OP' be the positions of the radius for any angle AOP = A, P and its supplement AOP'= 180o – A

A!

A (Art. 12).

M' Since OP = OP', and POA = POA', the triangles POM and P'OM' are geometrically equal.

B

P'M PM .. sin (180° — A)=sin AOP'=

OP' OP

OM' - OM cos (180° — A)= cos AOP'=

cos A, ОР! OP

B

M

= sin A,

=

=