A curious square root (Part 2)

There are two natural ways of computing $\sin 15^o$ using trig identities.

Method #1.

$\sin 15^o = \sin(45^o - 30^o)$

$\sin 15^o = \sin 45^o \cos 30^o - \cos 45^o \sin 30^o$

$\sin 15^o = \displaystyle \frac{\sqrt{2}}{2} \cdot \frac{1}{2} - \frac{\sqrt{2}}{2} \cdot \frac{1}{2}$

$\sin 15^o = \displaystyle \frac{\sqrt{6} - \sqrt{2}}{4}$

The same expression would be obtained if we had started with $15^o = 60^o - 45^o$ .

Method #2. Since $15^o$ is in the first quadrant,

$\sin 15^o = \sin \displaystyle \left( \frac{1}{2} \cdot 30^o \right)$

$\sin 15^o = \displaystyle \sqrt{ \frac{1 - \cos 30^o}{2} }$

$\sin 15^o = \displaystyle \sqrt{ \frac{1 - \displaystyle \frac{\sqrt{3}}{2}}{2} }$

$\sin 15^o = \displaystyle \sqrt{ \frac{2 -\sqrt{3}}{4} }$

$\sin 15^o = \displaystyle \frac{ \sqrt{2 -\sqrt{3}}}{2}$

Therefore,

$\displaystyle \frac{\sqrt{6} - \sqrt{2}}{4} = \displaystyle \frac{ \sqrt{2 -\sqrt{3}}}{2}$ ,

which may be verified by squaring both sides.

One thought on “A curious square root (Part 2)”

Published by John Quintanilla