An exact value for sin 15 ∘ \sin 15^{\circ} sin 1 5 ∘
Question
Without using tables, show for θ = 15 ∘ \theta=15^\circ θ = 1 5 ∘ sin θ = 6 − 2 4 \sin\theta=\dfrac{\sqrt{6}-\sqrt{2}}{4} sin θ = 4 6 − 2 cos θ \cos\theta cos θ tan θ \tan\theta tan θ
In the triangle A B C ABC A B C B B B E E E B C BC B C B B B C C C B E = B A BE=BA B E = B A D D D A C AC A C A A A C C C A D = D E = E C AD=DE=EC A D = D E = E C D C DC D C A B AB A B
Suggestion
For what angles do we know the exact values of sin θ \sin\theta sin θ
What trigonometric formulae might be useful here? How about expanding sin ( a − b ) \sin(a-b) sin ( a − b ) a a a b b b
Solution
We will use the identity
sin ( x − y ) = sin x cos y − sin y cos x \sin(x-y) = \sin x \cos y - \sin y \cos x sin ( x − y ) = sin x cos y − sin y cos x
. We have that
sin 15 ∘ = sin ( 45 − 30 ) ∘ = sin 45 ∘ cos 30 ∘ − cos 45 ∘ sin 30 ∘ = 1 2 3 2 − 1 2 1 2 = 2 3 2 × 2 − 2 2 × 2 = 6 − 2 4 . \begin{align*} \sin 15^\circ=\sin(45-30)^\circ &=\sin45^\circ\cos30^\circ-\cos45^\circ\sin30^\circ \\ &=\frac{1}{\sqrt{2}}\frac{\sqrt{3}}{2}-\frac{1}{\sqrt{2}}\frac{1}{2} \\ &=\frac{\sqrt{2}\sqrt{3}}{2\times 2}-\frac{\sqrt{2}}{2\times 2} \\ &=\frac{\sqrt{6}-\sqrt{2}}{4}. \end{align*} sin 1 5 ∘ = sin ( 45 − 30 ) ∘ = sin 4 5 ∘ cos 3 0 ∘ − cos 4 5 ∘ sin 3 0 ∘ = 2 1 2 3 − 2 1 2 1 = 2 × 2 2 3 − 2 × 2 2 = 4 6 − 2 . We now have
cos 2 15 ∘ = 1 − sin 2 15 ∘ = 1 − 6 + 2 − 2 6 2 16 = 8 + 2 12 16 = 6 + 2 + 2 2 6 16 = ( 6 + 2 ) 2 4 2 , \begin{align*} \cos^2 15^\circ=1-\sin^2 15^\circ &=1-\frac{6+2-2\sqrt{6}\sqrt{2}}{16} \\ &=\frac{8+2\sqrt{12}}{16}\\ &=\frac{6+2+2\sqrt{2}\sqrt{6}}{16} \\ &=\frac{(\sqrt{6}+\sqrt{2})^2}{4^2}, \end{align*} cos 2 1 5 ∘ = 1 − sin 2 1 5 ∘ = 1 − 16 6 + 2 − 2 6 2 = 16 8 + 2 12 = 16 6 + 2 + 2 2 6 = 4 2 ( 6 + 2 ) 2 , and so, since cos θ \cos \theta cos θ 0 ∘ 0^\circ 0 ∘ 90 ∘ 90^\circ 9 0 ∘ cos 15 ∘ = 6 + 2 4 . \cos 15^\circ=\frac{\sqrt{6}+\sqrt{2}}{4}. cos 1 5 ∘ = 4 6 + 2 .
Finally, we have
tan 15 ∘ = sin 15 ∘ cos 15 ∘ = 6 − 2 6 + 2 = ( 6 − 2 ) 2 ( 6 + 2 ) ( 6 − 2 ) = 6 + 2 − 2 2 6 6 − 2 = 8 − 2 12 4 = 2 − 3 . \begin{align*} \tan 15^\circ=\frac{\sin15^\circ}{\cos15^\circ} &=\frac{\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}} \\ &=\frac{(\sqrt{6}-\sqrt{2})^2}{(\sqrt{6}+\sqrt{2})(\sqrt{6}-\sqrt{2})} \\ &=\frac{6+2-2\sqrt{2}\sqrt{6}}{6-2} \\ &=\frac{8-2\sqrt{12}}{4} \\ &=2-\sqrt{3}. \end{align*} tan 1 5 ∘ = cos 1 5 ∘ sin 1 5 ∘ = 6 + 2 6 − 2 = ( 6 + 2 ) ( 6 − 2 ) ( 6 − 2 ) 2 = 6 − 2 6 + 2 − 2 2 6 = 4 8 − 2 12 = 2 − 3 . First let’s draw what’s happening, filling in the angles by putting in a a a A C B ACB A C B b b b C A E CAE C A E
The angle-sum in triangle A B C ABC A B C a + b = 45 ∘ a+b = 45^\circ a + b = 4 5 ∘ D D D a = 2 b a = 2b a = 2 b b = 15 ∘ , a = 30 ∘ b = 15^\circ, a = 30^\circ b = 1 5 ∘ , a = 3 0 ∘
We need the ratio of D C DC D C A B AB A B C E = 1 CE = 1 C E = 1 C D = 2 cos 30 ∘ CD = 2 \cos 30^\circ C D = 2 cos 3 0 ∘ E A = 2 cos 15 ∘ EA = 2\cos 15^\circ E A = 2 cos 1 5 ∘
Now, by Pythagoras, A B = 2 cos 15 ∘ 2 AB = \dfrac{2\cos 15^\circ}{\sqrt{2}} A B = 2 2 cos 1 5 ∘
Thus
D C : A B = 2 cos 30 ∘ : 2 cos 15 ∘ 2 = 3 : 2 6 + 2 4 = 4 3 : 12 + 2 = 2 3 : 3 + 1. \begin{align*} DC:AB &= 2\cos 30^\circ: \dfrac{2\cos 15^\circ}{\sqrt{2}}\\ &= \sqrt{3} : \sqrt{2}\frac{\sqrt{6}+\sqrt{2}}{4} \\ &= 4\sqrt{3} : \sqrt{12}+2 \\ &=2\sqrt{3} : \sqrt{3}+1. \end{align*} D C : A B = 2 cos 3 0 ∘ : 2 2 cos 1 5 ∘ = 3 : 2 4 6 + 2 = 4 3 : 12 + 2 = 2 3 : 3 + 1.
