Question

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Answer 1

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Given
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$= > \sqrt{2} \sin( \frac{\pi}{4} + \theta ) \\ \\ = > \sqrt{2}( \sin( \frac{\pi}{4} ) \cos( \theta) + \cos( \frac{\pi}{4} ) \sin( \theta ) ) \\ \\ = > \sqrt{2} ( \frac{1}{ \sqrt{2} } \cos( \theta ) + \frac{1}{ \sqrt{2} } \sin( \theta ) ) \\ \\ = > \frac{ \sqrt{2} }{ \sqrt{2} } ( \cos( \theta ) + \sin( \theta ) ) \\ \\ = > \cos( \theta ) + \sin( \theta )$
Hence proved..
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