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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

1 cosec A

coto

tan A

sec A

sin '

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We have sin A

perpendicular
hypotenuse

base
and

hypotenuse sin perpendicular

= tan . base

cos

cos e

COS

104. We may prove similarly cot 0

1 Or thus, cot 0 =

tan 0

sin cos A sin 8°

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 hypotenuse that is,

1 = sino 0 + cos* 0.

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=

+

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(ii) Divide each side of the same identity by

(base)', and we get
hypotenuse perpendicular base\?
base

base

base)

2

+

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(üi) Divide each side of the same identity by

(perpendicular)", and we get hypotenuse

) * = (perpendicular) base perpendicular perpendicular perpendicular)

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2

+

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(i) cos 0 + sin 0 = 1 (ii) 1 + tano 0 = sec 0

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

+

107. We give the above proof in a different form.
To prove that cose 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 ;

MP

OM .. sino 0

cosA OP2 :

OP2

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

MP2 OM that is, that

=1, OP2 OP2

MP2 + OM OP2 i.e. that

OP2

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

Therefore cos 0 + sin’ 0 =1.
Similarly we may prove

that

1 + tano 0 = sec 0, and that

1 + cota A = cosec? 0.

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

1 cosec

sin o'

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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.

COS A'

sin A'

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

sin A
tan A

COS A cot A=

sin A' 1

1 sec A=

and cosec A=

COS A we have to prove that

sin A
cos A

1 1
+
COS A

sin A COS A'sin A'
or that
sino A + cos2 A

1 COS A. sin A

cos A. sin A' and this is true, because sinA + cosA=1.

2

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 8+2 sin? 8. cos? 0=1 - cos4 0.

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

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

or that

1-2 cosa 0 + cos4 6+2 cos0 - 2 cos4 0=1 - cos4 0,

or that

1-cos40=1 - cos4 ,

which is true.

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

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.: (1-2 cos? 0 + c054 ) + 2 cos' 0 - 2 cos4 0=1 - cost 0,

.: (1 – cosa 0)2 + 2 cos 0 (1 - cos20)=1 - cost 0,

.. (sin? 0)2 + 2 cos? 8. sinA=1 - cos 8. Q.E.D.

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

versin 0.

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