We know that,
sinA+sinB=2sin\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
LHS = cot 4x (sin 5x + sin 3x)
\( \dfrac{cot4x}{sin4x}[2sin(\dfrac{5x+3x}{2})cos(\dfrac{5x+3x}{2})]\)
\(( \dfrac{cos4x}{sin4x})[2sin4xcosx]\)
=2cos4xcosx
Now also ,we know that,
sinA-sinB=2cos\( ( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})\)
R.H.S.= cotx(sin5x-sin3x)
= \( \dfrac{cosx}{sinx}[2cos(\dfrac{5x+3x}{2})sin(\dfrac{5x-3x}{2})]\)
= \(( \dfrac{cosx}{sinx})[2cos4xsinx]\)
=2cos4xcosx
Therefore , we can conclude that,
L.H.S.=R.H.S.
Hence proved.
Answered by Abhisek | 1 year agoprove that \(sin \dfrac{8π}{3} cos \dfrac{23π}{6} + cos \dfrac{13π}{3} sin \dfrac{35π}{6} = \dfrac{1}{2}\)
prove that \( 3 sin \dfrac{π}{6} sec \dfrac{π}{3} – 4 sin \dfrac{5π}{6} cot \dfrac{π}{4} = 1\)
prove that \( tan \dfrac{11π}{3} – 2 sin \dfrac{4π}{6} – \dfrac{3}{4} cosec^2 \dfrac{π}{4} + 4 cos^2 \dfrac{17π}{6} = \dfrac{(3 – 4\sqrt{3})}{2}\)
prove that cos 570° sin 510° + sin (-330°) cos (-390°) = 0
prove that tan (-125°) cot (-405°) – tan (-765°) cot (675°) = 0