Question

Express sin(-36degrees) in terms of a) the sin of an acute angle and b) the cosine of an acute angle.

Answer 1

Task. Express $\sin (-36{}^{\circ})$ in terms of

a) the sin of an acute angle and

b) the cos of an acute angle.

Solution. Consider the unit circle in the plane $(x,y)$ :

Let $A$ be the point on this circle in the lower half-plane such that $\angle AOK = 36{}^\circ$ . Then $(\cos(-36), \sin(-36))$ are, by definition, the coordinates of $A$ . Hence

$$\sin (- 3 6) = - A K.$$

We should express the second coordinate of $A$ via $\sin$ and $\cos$ of acute angle.

Consider the symmetric point $B$ on the circle, laying in the upper half-plane and such that $\angle BOK = 36{}^{\circ}$ . Then $B = (\cos(36{}^{\circ}), \sin(36{}^{\circ}))$ , and in particular, $BK = \sin(36{}^{\circ})$ . It is also evident, that the triangles $OBK$ and $OAK$ are equal, so $BK = AK$ , and therefore

$$\sin (- 3 6 {}^ {\circ}) = - A K = - B K = - \sin (3 6 {}^ {\circ}).$$

Furthermore,

$$\sin 3 6 {}^ {\circ} = \sin \angle B O K = \frac {B K}{O B} = \cos K B O = \cos 5 4 {}^ {\circ}.$$

Answer. $\sin (-36{}^{\circ}) = -\sin 36{}^{\circ} = \cos 54{}^{\circ}$