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CHAPTER VII.

ON THE RELATIONS BETWEEN THE TRIGONOMETRICAL

RATIOS OF THE SAME ANGLE.

102. The following relations are evident from the definitions : 1

1 cosec

1 sin ?

cot A cos o

tan A

sec

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We have sin 0 =

perpendicular
hypotenuse

base
and
cos A

hypotenuse
sin perpendicular

= tan 0. base

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104. We may prove similarly cot e

cos a

sin 0
cos
sin 0.

Or thus, coto

1 tan 0

=

105. Euclid I. 47 gives us that in any right-angled triangle the square on the hypotenuse the sum of the squares on the perpendicular and on the base,

or, (hypotenuse)' = (perpendicular)* + (base)”.

(i)
Divide each side of this identity by

(hypotenuse), and we get
hypotenuse /perpendicular base
hypotenuse, hypotenuse

1 = sino A + cosa 0.

2

3
+

=

(hypotenuse

)

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that is,

=

(ii) Divide each side of the same identity by

(base)', and we get hypotenuse

perpendicular) base

base

base)

2

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base\ 2

=

+

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(iii) Divide each side of the same identity by

(perpendicular), and we get hypotenuse perpendicular perpendicular) perpendicular,

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2

2

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106. Thus the three results

(i) cos 0 + sin 0 = 1 (ii) 1 + tan' 0 = sec 0

(iii) 1+ cot' 0 = coseco 0 are each a statement in Trigonometrical language of Euc. I. 47.

107. We give the above proof in a different form.
To prove that cos 0 + sin 0 = 1.
Let ROE be any angle 0.

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In OE take any point P, and draw PM perpendicular to OR. Then with respect to 0, MP is the perpendicular, OP is the hypotenuse, and OM is the base;

MP2

OM .. sino 0

cos? 0 = OP2

OP2 We have to prove that sino 0 + cos 0 = 1,

MP2 OM that is, that

= 1, OP2 OP

MP + OM OP2 i.e. that

OP2 OP2)

i.e. that MP + OM=OP?. But this is true by Euclid I. 47.

Therefore coso A + sin? 0 =1.
Similarly we may prove that

1 + tan’ 0 = sec , and that

1 + coto 0 = cosec.

108. The following is a LIST OF FORMULÆ with which the student must make himself familiar :

1 cosec 0

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sin ?

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109. In proving Trigonometrical identities it is often convenient to express the other Trigonometrical Ratios in terms of the sine and cosine.

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Example. Prove that tan A +cot A= =sec A. cosec A.

sin A Since tan A

cot A=

sin A' 1

1 sec A=

and cosec A COS A

sin A' we have to prove that

sin A
cos A

1 1
+

sin A COS A'sin A' or that sin? A +cos? A

1 cos A, sin A

COS A. sin A' and this is true, because sin? A + cos2 A=1.

2

COS A

110. Sometimes it is more convenient to express all the other Trigonometrical Ratios in terms of the sine only, or in terms of the cosine only.

Example. Prove that sin4 6+2 sin8. cosa 0=1 - cost 0.

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Hence, putting 1 – cosa 0, and (1 - cos? 6)2 for sin? 6 and sin4 0 respectively, we have to prove that

(1 - cos? 0)2 +2.(1 - cosa ). cos? A=1 - cost 0, or that

1 - 2 cos0 + cos4 6+2 cos0 - 2 cos4 A=1-cost ,

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This example may be proved directly, by reversing the steps of the above proof; thus

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.: (1 – 2 cosa 0 + cos4 0) + 2 cos' 0 - 2 cos4 6=1 - cost 0,

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.: (1 - cosa 0)2 + 2 cos? 0 (1 - cosa 0)=1 - cos4 0,
.: (sinề0)2 + 2 cos? 0. sin0=1-cos'0. Q.E.D.

NOTE. (1 – cos 6) is called the versed sine of 0; it is abbreviated thus

versin 0.

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