We know that ,
cosA-cosB= -2sin \( ( \dfrac{A+B}{2})sin( \dfrac{A-B}{2})\)
\( cos( \dfrac{3\pi}{4}+x)-cos(\dfrac{3\pi}{4}-x)=-\sqrt{2}sinx\)
\(L.H.S= cos( \dfrac{3\pi}{4}+x)-cos(\dfrac{3\pi}{4}-x)\)
\( -2sin\{\dfrac{( \dfrac{3\pi}{4}+x)+(\dfrac{3\pi}{4}-x)}{2}\}\)
\( sin\{\dfrac{( \dfrac{3\pi}{4}+x)+(\dfrac{3\pi}{4}-x)}{2}\}\)
= \( -2sin(\dfrac{3\pi}{4})sinx\)
Since sinx repeats the same value after an interval 2π ,
we have, \( sin(\dfrac{3\pi}{4})=sin(\pi-\dfrac{\pi}{4})\)
therefore,
L.H.S= -2sin\( \dfrac{\pi}{4}sinx\)
= \( -2\times \dfrac{1}{\sqrt{2}}\times sinx\)
\( =-\sqrt{2}sinx\)
= 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