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  • Show that cos 38^o cos 52^o- sin 38^o sin 52^o = 0

2.  Show that :

   (ii) \cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

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\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}= 0

Taking Left Hand Side (LHS)
=\cos 38^{o}\cos 52^{o}-\sin 38^{o}\sin 52^{o}
=\cos 38^{o}\cos (90^o-38^{o})-\sin 38^{o}\sin (90^o-38^{o})
=\cos 38^{o}\sin38^{o}-\sin 38^{o}\cos 38^{o} [it is known that \sin(90^0-\theta) =\cos \theta and \cos(90^0-\theta) =\sin \theta ]
= 0

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manish

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