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Draw

any line OR, and take M in it so that the measure of OM is unity.

Draw MP perpendicular to OM, and take P so that the measure of MP is a. Join OP. Then the angle ROP is the angle required.

MP For tan ROP

1 Thus an angle ROP has been drawn so that tan ROP=a.

a
=a.

ОМ

117. DEFINITION. One angle is called the complement of another, when the two angles added together make up a right angle.

Example 1. The complement of A is (90° - 4).
Example 2. The complement of 190° is (900 – 1900)= - 1000.
For 190° +(900 – 1900)=90°.

51 37 Example 3. The complement of is

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118. To prove that the sine of an angle A is equal to the cosine of its complement (90° - A).

[This is true whatever be the magnitude of A, as will be proved later on.]

If A be less than 90°, let ROP be A (see last figure).
Draw PM perpendicular to OR.

Then since PMO = 90°, therefore POM + OPM = 90° and therefore OPM =(90°- A).

MP
Now, sin A

= cos OPM = cos (90° – A). Q.E.D.
OP

EXAMPLES. XIX.

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(1) Draw an angle whose sine is 1.
(2) Draw an angle whose cosecant is 2.
(3) Draw an angle whose tangent is 2.
(4) Can an angle be drawn whose tangent is 431 ?
(5) Can an angle be drawn whose cosine is f ?
(6) Can an angle be drawn whose secant is 5 ?
(7) Find the complements of
(i) 300
(ii) 190°. (iii) 900 (iv) 3500

375
(v) – 250 (vi) - 320° (vii)

(viii) 4

6 (8) Prove that sin 70°=cos 20°. (9) Prove that cos 47° 16'=sin 42° 44'. (10) Prove that tan 79°=cot 110. (11) Prove that sec 36° =

=cosec 54o. (12) If A be less than 90°, prove

(i) cos A=sin (90° – A).
(ii) tan A=cot (90° – A).
(iii) sec A=cosec (90° — A).
(iv) cot A=tan (90° – A).

ON THE SOLUTION OF TRIGONOMETRICAL EQUATIONS.

119. A TRIGONOMETRICAL equation is an equation in which there is a letter, such as 0, which stands for an angle of unknown magnitude,

The solution of the equation is the process of finding an angle which, if it be substituted for 0, satisfies the equation.

Example 1. Solve cos 0 =*.

This is a Trigonometrical equation. To solve it we must find some angle such that its cosine is 1.

We know that cos 60°=.
Therefore if 60° be put for the equation is satisfied,

.: 0=60° is a solution of the equation.

Example 2. Solve sin A - cosec 6+ i=0.

The usual method of solution is to express all the Trigonometrical Ratios in terms of one of them.

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This is an equation in which 0, and therefore sin 0 is unknown. It will be convenient if we put x for sin 0, and then solve the equation for x as an ordinary algebraical equation. Thus we get

1
+3=0,

3x or,

22 + :1,

2
3x
:: x2 + +id=1+18=14

2

.: X+*=*
Whence

x= -2, or x=1.
But x stands for sin A.
Thus we get sin 0= -2, or sin 0=1.

The value - 2 is inadmissible for sin 0, for there is no angle whose sine is numerically greater than 1 (Art. 115).

.. sin 0=1 But

sin 30o=1.

.:: sin 0=sin 30°. Therefore one angle which satisfies this equation for 8 is 30°.

Example 3. To solve the equation cosec 0 - cota 0+1=0.
We have
1+cot A=cosec? A.

[Art. 108] .:. cot? A=cosec? A-1. and we get

cosec 0 - cosec? A +2=0. Let x stand for cosec 8. Then

- 22 +2=0.

.. 22 – X= +2.
...? - 3+=+*+2=2,

or

Whence

x=2, or x= -1,
.. cosec 6=2,
cosec 30°=2,
.. cosec A =cosec 30°.

but

therefore 300 is one angle which satisfies the equation for 0.

EXAMPLES. XX.

Find one angle which satisfies each of the following equations.

1 (1) sin 0 72

(2) 4 sin 0=cosec 0. (3) 2 cos 0 =sec A.

(4) 4 sin 0 - 3 cosec 0=0. (5) 4 cos 0 - 3 sec 0=0. (6) 3 tan Q=cot 0. (7) 3 sin 0 – 2 cos? 0=0. (8) 12 sin (=tan 0. (9) 2 cos 0=1/3 cot 0. (10) tan 0=3 cot 0. (11) tan 0+3 cot A=4. (12) tan 0 + cot 0=2. (13) 2 sin0+/2 cos 0=2. (14) 4 sina 0+2 sin 6=1. (15) 3 tano 0 - 4 sina 0=l. (16) 2 sin? 0+1/2 sin 0=2. (17) cos 0 - 3 cos 0+2=0. (18) cosa 6 + 2 sina 0 – sin 0=0.

* * MISCELLANEOUS EXAMPLES. XXI.

(1) Prove that 3 sin 60° — 4 sin360°=4 cos 30o – 3 cos 30o. (2) Prove that tan 30° (1 + cos 30° + cos 60°)=sin 30° + sin 60°.

(3) If 2 cosa 6 7 cos 0+3=0, show there is only one value of cos A.

(4) Find cos 0 from the equation 8 cosa 0 - 8 cos 0+1=0.

(5) Find sin 8 from the equation 8 sina 0 - 10 sin 0+3=0, and prove that one value of 0 is

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(6) Find tan 8 from the equation 12 tan20 – 13 tan 8+3=0.

(7) If 3 cosa 8+2./3.cos 0=5, show that there is only one value of cos 6, and that one value of 0 is

6 (8) Prove that the value of sino 0+ cos 0 + 2. sin0.cosA is always the same.

(9) Simplify cos4 A +2.sin? A. cos? A.

(10) Express sine A + cosé A in terms of sino A and powers of sino A. (11) Express 1+tano 0 in terms of cos 0 and its powers.

cos A +cos B sin A +sin B (12) Prove that

+

=0. sin A - sin B cos A - cos B (13) Express (sec A - tan A)? in terms of sin A.

(14) Trace the changes in the magnitude of cosec 8 as 0 increases from 0 to .

2 (15) Trace the changes in the magnitude of cot & as 0 de

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