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.

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1 Comment

  1. A Curious Square Root: Index | Mean Green Math

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