निम्नलिखित को सिद्ध कीजिए

cos ^{2} 2 x-cos | vedclass

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Question

It is known that

$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}\right) \cos \left(\frac{A-B}{2}\right), \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}\right)

Vedclass · Accepted Answer

Vedclass · Answer

It is known that

$\cos A+\cos B=2 \cos \left(\frac{A+B}{2}ight) \cos \left(\frac{A-B}{2}ight), \cos A-\cos B=-2 \sin \left(\frac{A+B}{2}ight) \sin \left(\frac{A-B}{2}ight)$

$	herefore$ $L.H.S.$ $=\cos ^{2} 2 x-\cos ^{2} 6 x$

$=(\cos 2 x+\cos 6 x)(\cos 2 x-6 x)$

${ = \left[ {2\cos \left( {\frac{{2x + 6x}}{2}} ight)\cos \left( {\frac{{2x - 6x}}{2}} ight)} ight]}$

${\left[ { - 2\sin \left( {\frac{{2x + 6x}}{2}} ight)\sin \left( {\frac{{2x - 6x}}{2}} ight)} ight]}$

${ = [2\cos 4x\cos ( - 2x)][ - 2\sin 4x\sin ( - 2x)]}$

${ = [2\cos 4x\cos 2x][ - 2\sin 4x( - \sin 2x)]}$

${ = (2\sin 4x\cos 4x)(2\sin 2x\cos 2x)}$

${ = \sin 8x\sin 4x = R.H.S}$

निम्नलिखित को सिद्ध कीजिए

$\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$

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निम्नलिखित को सिद्ध कीजिए $\cos ^{2} 2 x-\cos ^{2} 6 x=\sin 4 x \sin 8 x$