﻿ Simplify the square roots expression \sqrt{2+\sqrt{2+\sqrt{3} } } - \sqrt{3} \sqrt{2-\sqrt{2+\sqrt{3} } }

#### Question

Simplify the square roots expression

\sqrt{2+\sqrt{2+\sqrt{3} } } - \sqrt{3} \sqrt{2-\sqrt{2+\sqrt{3} } }

Collected in the board: Trigonometry

Steven Zheng posted 4 months ago

The special values in

\sqrt{2+\sqrt{2+\sqrt{3} } } - \sqrt{3} \sqrt{2-\sqrt{2+\sqrt{3} } }
(1)

reminds us of the trigonometric values of special angles.

From our previous posts, we have derived the exact values for \sin 15° and \cos 15°,

\sin 15° = \dfrac{\sqrt{2-\sqrt{3} } }{2}
(2)
\cos 15° = \dfrac{\sqrt{2+\sqrt{3} } }{2}
(3)

We are going to use the value of the special angle to simplify the square root expression.

Rearranging (3) gives

\sqrt{2+\sqrt{3} } = 2\cos 15°
(4)

Substitute (4) to (1), we get

\sqrt{2+2\cos 15° } - \sqrt{3} \sqrt{2-2\cos 15°}
=\sqrt{2} \sqrt{1+\cos 15°} - \sqrt{6} \sqrt{1-\cos 15°}
(5)

Then, using half angle identities

\sin \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1-\cos \alpha }{2} }
\cos \dfrac{\alpha }{2}=\pm\sqrt{\dfrac{1+\cos \alpha }{2} }

gives

\sqrt{1-\cos 15°} = \sqrt{2} \sin 7.5°
\sqrt{1+\cos 15°} = \sqrt{2} \cos 7.5°

Substitute them to (5), we get

\begin{aligned} 2 \cos 7.5° - 2\sqrt{3} \sin 7.5° &=4(\dfrac{1}{2} \cos 7.5° - \dfrac{\sqrt{3} }{2} \sin 7.5°) \\&=4(\sin 30° \cos 7.5° - \cos 30° \sin 7.5°) \\&= 4\sin(30°-7.5° )\\&=4\sin 22.5° \end{aligned}

In summary, the square root express

\sqrt{2+\sqrt{2+\sqrt{3} } } - \sqrt{3} \sqrt{2-\sqrt{2+\sqrt{3} } }

is simplified as

4\sin 22.5°

Steven Zheng posted 4 months ago

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