Question

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

Answer 1

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 $0{}^\circ <\theta <90{}^\circ$ .
$\sin 2A+\sin 2B=2\sin (A+B)\cos (A-B)$
$\sin 2A-\sin 2B=2\cos (A+B)\sin (A-B)$
$\cos 2A+\cos 2B=2\cos (A+B)\cos (A-B)$
$\cos 2A-\cos 2B=-2\sin (A+B)\sin (A-B)$