Here I will collect $\sin$ and $\cos$ values of as many angles as I can. --- ### Exact Trig Values of $3\degree$ or $\frac{\pi}{60}$ [[Angle Sum-or-Difference Identities]] $\sin(3\degree)=\sin(18\degree-15\degree)=\sin(18\degree) \cos (15\degree)-\cos (18\degree) \sin (15\degree)$ [[#Exact Trig Values of 18 degree or frac pi 10]] [[#Exact Trig Values of 15 degree or frac pi 12]] --- ### Exact Trig Values of $15\degree$ or $\frac{\pi}{12}$ [[Special Right Triangles#15-75-90]] [[Double-or-Half Angle Identities]] with [[#Exact Trig Values of 30 degree or frac pi 6]] $\sin(15\degree)=\sin\left(\frac{30\degree}{2}\right)=\sqrt{\frac{1-\cos(30\degree)}{2}}=\sqrt{\frac{1-\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2-\sqrt3}}{2}=\frac{\sqrt{6} -\sqrt{2}}{4}$ $\cos(15\degree)=\cos\left(\frac{30\degree}{2}\right)=\sqrt{\frac{1+\cos(30\degree)}{2}}=\sqrt{\frac{1+\frac{\sqrt{3}}{2}}{2}}=\frac{\sqrt{2+\sqrt3}}{2}=\frac{\sqrt{6} +\sqrt{2}}{4}$ --- ### Exact Trig Values of $18\degree$ or $\frac{\pi}{10}$ [[Special Right Triangles#18-72-90]] $\frac{-1+\sqrt{5}}{4}$ --- ### Exact Trig Values of $27\degree$ or $\frac{\pi}{120}$ We will find a sum and difference that include the desired value. (both depend on [[#Exact Trig Values of $36\degree$ or $\frac{ pi}{5}$]] The sum $ \begin{aligned} \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=\sin ^2 27^{\circ}+2 \sin 27^{\circ} \cos 27^{\circ}+\cos ^2 27^{\circ} \\ \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=1+\sin 2 \cdot 27^{\circ} \\ \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=1+\sin 54^{\circ} \\ \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=1+\sin \left(90^{\circ}-36^{\circ}\right) \\ \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=1+\cos 36^{\circ} \\ \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=1+\frac{\sqrt{5} +1}{4} \\ \left(\sin 27^{\circ}+\cos 27^{\circ}\right)^2&=\frac{1}{4}(5+\sqrt{ 5} ) \\ \sin 27^{\circ}+\cos 27^{\circ}&=\frac{1}{2} \sqrt{5+\sqrt{5}} \\ \end{aligned} $ The difference $ \begin{aligned} \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=\sin ^2 27^{\circ}-2 \sin 27^{\circ} \cos 27^{\circ}+\cos ^2 27^{\circ} \\ \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=1-\sin 2 \cdot 27^{\circ} \\ \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=1-\sin 54^{\circ} \\ \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=1-\sin \left(90^{\circ}-36^{\circ}\right) \\ \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=1-\cos 36^{\circ} \\ \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=1-\frac{\sqrt{5} +1}{4} \\ \left(\sin 27^{\circ}-\cos 27^{\circ}\right)^2&=\frac{1}{4}(3+\sqrt{ 5} ) \\ \sin 27^{\circ}-\cos 27^{\circ}&=\frac{1}{2} \sqrt{3+\sqrt{5}} \end{aligned} $ These can be combined to cancel the $\cos$ part, and find the $\sin$ part. $\begin{aligned} \sin 27^{\circ}+\cos 27^{\circ}&=\frac{1}{2} \sqrt{5+\sqrt{5}}\\ \sin 27^{\circ}-\cos 27^{\circ}&=\frac{1}{2} \sqrt{3+\sqrt{5}} \\ \hline2\sin{27^{\circ}}&=\frac{1}{2} \sqrt{5+\sqrt{5}}+\frac{1}{2} \sqrt{3+\sqrt{5}}\\ \sin{27^{\circ}}&=\frac{1}{4}\left( \sqrt{5+\sqrt{5}}+\sqrt{3+\sqrt{5}} \right) \end{aligned}$ Or combined to cancel the $\sin$ part, and find the $\cos$ part $\begin{aligned} \sin 27^{\circ}+\cos 27^{\circ}&=\frac{1}{2} \sqrt{5+\sqrt{5}}\\ \sin 27^{\circ}-\cos 27^{\circ}&=\frac{1}{2} \sqrt{3+\sqrt{5}} \\ \hline 2\cos{27^{\circ}}&=\frac{1}{2} \sqrt{5+\sqrt{5}}-\frac{1}{2} \sqrt{3+\sqrt{5}}\\ \cos{27^{\circ}}&=\frac{1}{4}\left( \sqrt{5+\sqrt{5}}-\sqrt{3+\sqrt{5}} \right) \end{aligned}$ --- ### Exact Trig Values of $30\degree$ or $\frac{\pi}{6}$ [[Special Right Triangles#30-60-90]] --- ### Exact Trig Values of $36\degree$ or $\frac{\pi}{5}$ With the previous knowledge: ![[Ratio of Diagonal To Side Length of Regular Pentagon is φ]] ![[special-right-triangles-in-regular-pentagon.svg]]