﻿ Determine the value of \sin 15\degree

Question

Determine the value of \sin 15\degree

Collected in the board: Trigonometry

Steven Zheng posted 2 years ago

Noticed 15 is equal to the difference of 45 and 30. So sines of 15°could be transformed to the sines of the difference of 45° and 30°

Using the difference identity for sines function

\sin(\alpha -\beta ) = \sin \alpha\cos \beta -\cos \alpha \sin \beta

\sin 15\degree=\sin(45\degree -\sin 30\degree )

=\sin 45\degree \cos 30\degree -\cos 45\sin 30\degree

=\dfrac{\sqrt{2} }{2} \cdotp \dfrac{\sqrt{3} }{2} - \dfrac{\sqrt{2} }{2} \cdotp\dfrac{1}{2}

=\dfrac{\sqrt{6}-\sqrt{2} }{4}

Steven Zheng posted 2 years ago

Since 15 is the half of 30, we can use half angle identity to determine the value of sin 15°

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

Obviously, the negative value is cancelled as 15° is within quadrant I. So,

\sin 15\degree = \sqrt{\dfrac{1-\cos 30\degree }{2} }

=\sqrt{\dfrac{1-\dfrac{\sqrt{3} }{2} }{2} }

=\sqrt{\dfrac{2-\sqrt{3} }{4} }

=\dfrac{1}{4}\sqrt{8-4\sqrt{3} }

=\dfrac{1}{4}\sqrt{(\sqrt{6}-\sqrt{2} )^2}

=\dfrac{\sqrt{6}-\sqrt{2} }{4}

Another form of result of \sin 15° is \dfrac{2-\sqrt{3} }{2}

Steven Zheng posted 2 years ago

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