Question

Prove that: $ \cos 18{}^\circ -\sin 18{}^\circ =\sqrt{2}\sin 27{}^\circ $

Answer 1

Hint: We know that $ \sin (90{}^\circ -\theta )=\cos \theta $ .
Use the identity $ \cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B) $ .
We also know that $ \sin (-\theta )=-\sin \theta $ .
Simplify the expression and use $ \sin 45{}^\circ =\dfrac{1}{\sqrt{2}} $ .

Complete step by step answer:
We can write $ \sin 18{}^\circ =\sin (90{}^\circ -72{}^\circ )=\cos 72{}^\circ $ .
Therefore, the LHS $ \cos 18{}^\circ -\sin 18{}^\circ $ of the given relation becomes:
= $ \cos 18{}^\circ -\cos 72{}^\circ $
= $ \cos (2\times 9{}^\circ )-\cos (2\times 36{}^\circ ) $
Using the identity $ \cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B) $ , we get:
= $ -2\sin (9{}^\circ +36{}^\circ )\sin (9{}^\circ -36{}^\circ ) $
= $ -2\sin (45{}^\circ )\sin (-27{}^\circ ) $
Using $ \sin (-\theta )=-\sin \theta $ and $ \sin 45{}^\circ =\dfrac{1}{\sqrt{2}} $ , we get:
= $ -2\times \dfrac{1}{\sqrt{2}}\times (-\sin 27{}^\circ ) $
Multiplying the numerator and the denominator by $ \sqrt{2} $ , we get:
= $ 2\times \dfrac{1\times \sqrt{2}}{\sqrt{2}\times \sqrt{2}}\times \sin 27{}^\circ $
= $ 2\times \dfrac{\sqrt{2}}{2}\times \sin 27{}^\circ $
= $ \sqrt{2}\times \sin 27{}^\circ $
= RHS
Hence, proved.

Note: $ \sin 18{}^\circ =\dfrac{\sqrt{5}-1}{4} $ , $ \cos 18{}^\circ =\dfrac{\sqrt{10+2\sqrt{5}}}{4} $ and $ \sin 72{}^\circ =\dfrac{\sqrt{10+2\sqrt{5}}}{4} $ .
The values of $ \sin \theta $ and $ \cos \theta $ are positive in the range $ 0{}^\circ <\theta <90{}^\circ $ .
 $ \sin 2A+\sin 2B=2\sin (A+B)\cos (A-B) $
 $ \sin 2A-\sin 2B=2\cos (A+B)\sin (A-B) $
 $ \cos 2A+\cos 2B=2\cos (A+B)\cos (A-B) $
 $ \cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B) $