We know that,
cosA+cosB=2cos\( ( \dfrac{A+B}{2})cos( \dfrac{A-B}{2})\)
cosA-cosB=-2sin \( ( \dfrac{A+B}{2})sin( \dfrac{A-B}{2}) \)
L.H.S. =cos22x-cos26x
=(cos2x+cos6x)(cos2x-6x)
\([ 2cos(\dfrac{2x+6x}{2})cos( \dfrac{2x-6x}{2})]\)
\([ - 2sin(\dfrac{2x+6x}{2})sin( \dfrac{2x-6x}{2})]\)
We get
= [2 cos 4x cos (-2x)] [-2 sin 4x sin (-2x)]
It can be written as
= [2 cos 4x cos 2x] [–2 sin 4x (–sin 2x)]
So we get
= (2 sin 4x cos 4x) (2 sin 2x cos 2x)
= sin 8x sin 4x
= RHS
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