Show that
sin 52° sin 38º- cos 52° cos 38°=0
[tex] \sin(52) \sin(38) – \cos(52) \cos(38) = 0[/tex]
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1 thought on “Show that<br /> sin 52° sin 38º- cos 52° cos 38°=0<br />[tex] \sin(52) \sin(38) – \cos(52) \cos(38) = 0[/tex]<br />​”

1. Answer:

   sin52 cos 52 – cos 52 cos 52

   1 – 1 ( sin theta + cos theta=1)

   0

   hence proved

   LHS= RHS

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