Sum

Prove that:

\[\cos\frac{\pi}{12} - \sin\frac{\pi}{12} = \frac{1}{\sqrt{2}}\]

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#### Solution

\[LHS = \cos \frac{\pi}{12} - \sin \frac{\pi}{12}\]

\[ = \cos \left( \frac{\pi}{2} - \frac{5\pi}{12} \right) - \sin \frac{\pi}{12}\]

\[ = \sin \left( \frac{5\pi}{12} \right) - \sin \frac{\pi}{12}\]

\[ = 2\sin \left( \frac{\frac{5\pi}{12} - \frac{\pi}{12}}{2} \right) \cos \left( \frac{\frac{5\pi}{12} + \frac{\pi}{12}}{2} \right) \left\{ \because \sin A - \sin B = 2\sin \left( \frac{A - B}{2} \right) \cos \left( \frac{A + B}{2} \right) \right\}\]

\[ = 2\sin \left( \frac{\pi}{6} \right) \cos \left( \frac{\pi}{4} \right)\]

\[ = 2 \times \frac{1}{2} \times \frac{1}{\sqrt{2}}\]

\[ = \frac{1}{\sqrt{2}}\]

Hence, LHS = RHS.

Concept: Transformation Formulae

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