ABSTRACT
Applying Pythagoras’ theorem to the right-angled triangle shown in Fig. 25.1 gives:
a2 C b2 D c2 1
Dividing each term of equation (1) by c2 gives: a2
c2 C b
c2 D c
i.e. (a c
)2 C (b c
)2 D 1
cos 2 C sin 2 D 1 Hence cos2 qY sin2 q = 1 2 Dividing each term of equation (1) by a2 gives:
a2 C b
a2 D c
i.e. 1 C ( b
a
)2 D ( c a
)2 Hence 1Y tan2 q= sec2 q 3 Dividing each term of equation (1) by b2 gives:
2 C b
2 D c
i.e. (a b
)2 C 1 D ( c b
)2 Hence cot2 qY 1= cosec2 q 4 Equations (2), (3) and (4) are three further examples of trigonometric identities.
