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6.5.1 Proving Trigonometric Identities Using Addition Formula and Double Angle Formulae (Examples)


Example 2:
Prove each of the following trigonometric identities.
(a) 1 + cos 2 x sin 2 x = cot x (b) cot A sec 2 A = cot A + tan 2 A (c) s i n x 1 − c o s x = cot x 2

Solution:
(a)
L H S = 1 + cos 2 x sin 2 x = 1 + ( 2 cos 2 x − 1 ) 2 sin x cos x = 2 cos 2 x 2 sin x cos x = cos x sin x = cot x = R H S (proven)



(b)
R H S = cot A + tan 2 A = cos A sin A + sin 2 A cos 2 A = cos A cos 2 A + sin A sin 2 A sin A cos 2 A = cos A ( cos 2 A − sin 2 A ) + sin A ( 2 sin A cos A ) sin A cos 2 A = cos 3 A − cos A sin 2 A + 2 sin 2 A cos A sin A cos 2 A = cos 3 A + cos A sin 2 A sin A cos 2 A = cos A ( cos 2 A + sin 2 A ) sin A cos 2 A = cos A sin A cos 2 A ← sin 2 A + cos 2 A = 1 = ( cos A sin A ) ( 1 cos 2 A ) = cot A sec 2 A


(c)
L H S = s i n x 1 − c o s x = 2 s i n x 2 cos x 2 1 − ( 1 − 2 s i n 2 x 2 ) ← sin x = 2 s i n x 2 cos x 2 , cos x = 1 − 2 sin 2 x 2 = 2 s i n x 2 cos x 2 2 s i n 2 x 2 = cos x 2 s i n x 2 = cot x 2 = R H S (proven)


Example 3:
(a) Given that sinP= 3 5  and sinQ= 5 13 ,  such that P is an acute angle and Q is an obtuse angle, without using tables or a calculator, find the value of cos (P + Q).

(b) Given that sinA=− 3 5  and sinB= 12 13 ,  such that A and B are angles in the third and fourth quadrants respectively, without using tables or a calculator, find the value of sin (A – B).

Solution:
(a)
sin P = 3 5 , cos P = 4 5 sin Q = 5 13 , cos Q = − 12 13 cos ( P + Q ) = cos A cos B − sin A sin B = ( 4 5 ) ( − 12 13 ) − ( 3 5 ) ( 5 13 ) = − 48 65 − 15 65 = − 63 65



(b)


sin A = − 3 5 , cos A = − 4 5 sin B = − 5 13 , cos B = 12 13 s i n ( A − B ) = s i n A cos B − c o s A sin B = ( − 3 5 ) ( 12 13 ) − ( − 4 5 ) ( − 5 13 ) = − 36 65 − 20 65 = − 56 65

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