Sin6 theta+cos 6 theta=1-3sin×sintheta×cos×costheta
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Sia ? 6 years, 3 months ago
{tex}\sin ^ { 6 } \theta + \cos ^ { 6 } \theta = 1 - 3 \sin ^ { 2 } \theta \cos ^ { 2 } \theta{/tex}
{tex}\text { L.H.S. } = \sin ^ { 6 } \theta + \cos ^ { 6 } \theta{/tex}
{tex}= \left( \sin ^ { 2 } \theta \right) ^ { 3 } + \left( \cos ^ { 2 } \theta \right) ^ { 3 }{/tex}
{tex}= \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta \right) \left( \sin ^ { 4 } \theta + \cos ^ { 4 } \theta - \sin ^ { 2 } \theta \cos ^ { 2 } \theta \right){/tex}{tex}\left[ \because a ^ { 3 } + b ^ { 3 } = ( a + b ) \left( a ^ { 2 } + b ^ { 2 } - a b \right) \right]{/tex}
{tex}= 1 \left( \sin ^ { 4 } \theta + \cos ^ { 4 } \theta + 2 \sin ^ { 2 } \theta \cdot \cos ^ { 2 } \theta - 2 \sin ^ { 2 } \theta \cos ^ { 2 } \theta - \sin ^ { 2 } \theta \cos ^ { 2 } \theta \right]{/tex}{tex}\text { since, } \sin ^ { 2 } A + \cos ^ { 2 } A = 1{/tex}
{tex}= \left( \sin ^ { 2 } \theta + \cos ^ { 2 } \theta \right) ^ { 2 } - 3 \sin ^ { 2 } \theta \cos ^ { 2 } \theta{/tex}
{tex}= 1 - 3 \sin ^ { 2 } \theta \cos ^ { 2 } \theta{/tex}
= R.H.S. proved.
1Thank You