Sine and Cosine of 15 Degrees Angle

The purpose of this page is to find in closed form values of sine and cosine of the 15° angle. Our investigation will be based on the following diagram, which [R. Vakil, p. 97] attributes to Larry Hoehn:

We can place side by side the right isosceles triangle ADE and the 30-60-90 triangle BDE and then extend AD to C to form another 45-45-90 triangle ABC. For convenience, we assume that AE = 1. In ΔBCD, ∠CBD = 15°. This is the triangle whose sides we are hunting.

In ΔADE, DE = 1 and, by the Pythagorean theorem, AD = √2.

In ΔBDE, DE = 1 and ∠DBE = 30° so that BD = 2 and BE = √3.

In ΔABC, AB = 1 + √3 so that AC = BC = (1 + √3) / √2:

AC = BC = (√2 + √6) / 2.

Letting CD = x, AC = √2 + x, which gives

	x	= (√2 + √6) / (2 - √2)
		= (√6 - √2) / 2.

Since sin 15° = CD/BD = x/2, it follows that

sin 15° = (√6 - √2) / 4.

Next, from BC = (√2 + √6) / 2 and cos 15° = BC/BD:

cos 15° = (√6 + √2) / 4.

It is reassuring that the values for the sine and cosine of 15° fit into the double argument formula sin 2α = 2 sinαcosα:

	2 sin(15°) cos(15°)	= [(√6 - √2) / 4] × [(√6 + √2) / 4]
		= 2 (6 - 2) / 16
		= 1 / 2
		= sin 30°.

For the record, we can also deliver the tangent:

tan 15° = 2 - √3.

Here's another diagram that also offers a way of finding the values of trigonometric function for 15°.

Reference

D. S. Ailles, Triangles and trigonometry, Mathematics Teacher 64 (1971) 562
R. Vakil, A Mathematical Mosaic, Brendan Kelly Publishing, 2008 (Expanded Edition)

Trigonometry

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