Question

33.
Show that
(i) sin 420° cos 390° + COS(-300°) sin (-330°) = 1/2
(ii) cos 570° sin 510° - sin 330° cos 390º = 0
Ans:

Answer 1

Answer:

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Answer 2

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(ii)

LHS =cos (570)sin (510) + sin (- 330)cos (- 390)

= cos (570) sin (510) + [ – sin (330) ]cos (390) [ because sin( – x ) = – sin x and cos( – x ) = cos x ]

= cos (570)sin(510) – sin (330)

= cos (90 * 6 + 30) sin (90 * 5 + 60) – sin (90 * 3 + 60) cos (90 * 4 + 30)

= – cos (30) cos (60) – [ – cos (60) ] cos (30)

= – cos (30) cos (60) + cos (30) sin (60)

= 0

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