I am trying to figure out the exact value of $\sin(1^\circ)$ and $\cos(1^\circ)$ so that I am able to get the exact value of every integral sine and cosine. The list of formulas that we need are:
$$\sin(\alpha+\beta)=\sin\alpha\cdot\cos\beta+\cos\alpha\cdot\sin\beta$$
$$\cos(\alpha+\beta)=\cos\alpha\cdot\cos\beta-\sin\alpha\cdot\sin\beta$$
$$\sin(2\alpha)=2\sin\alpha\cdot\cos\alpha$$
$$\cos(2\alpha)=\cos^2\alpha-\sin^2\alpha$$
$$\sin30^\circ=\frac{1}{2},~\cos30^\circ=\frac{\sqrt{3}}{2},~\sin45^\circ=\cos45^\circ=\frac{\sqrt{2}}{2}$$
My Current Progress:
We'll need another formula for "triple" angle.
$$\sin(3\alpha)=\sin(2\alpha+\alpha)=\sin2\alpha\cos\alpha+\cos2\alpha\sin\alpha$$
$$\sin(3\alpha)=2\sin\alpha\cos^2\alpha+\cos^2\alpha\sin\alpha-\sin^3\alpha$$
$$\sin(3\alpha)=3\sin\alpha\cos^2\alpha-\sin^3\alpha$$
$$\sin(3\alpha)=3\sin\alpha(1-\sin^2\alpha)-\sin^3\alpha$$
$$\sin(3\alpha)=3\sin\alpha-4\sin^3\alpha$$
$$\cos(3\alpha)=\cos(2\alpha+\alpha)=\cos2\alpha\cos\alpha-\sin2\alpha\sin\alpha$$
$$\cos(3\alpha)=\cos^3\alpha-\sin^2\alpha\cos\alpha-2\sin^2\alpha\cos\alpha$$
$$\cos(3\alpha)=\cos^3\alpha-3\sin^2\alpha\cos\alpha$$
$$\cos(3\alpha)=\cos^3\alpha-3(1-\cos^2\alpha)\cos\alpha$$
$$\cos(3\alpha)=4\cos^3\alpha-3\cos\alpha$$
Since we want to calculate a very small angle, we want to calculate a bigger but still very small angle. The smallest angle that I can calculate directly that I think of is $3^\circ$ by subtracting $72^\circ$ from $75^\circ$.
Let's find $\sin 72^\circ$ and $\cos 72^\circ$ first. We need the value of $\sin18^\circ$ or $\cos18^\circ$ first because calculating $72^\circ$ is a bit hard.
$$\sin(36^\circ)=\cos(90^\circ-36^\circ)$$
$$\sin(2\cdot18^\circ)=\cos(3\cdot18^\circ)$$
$$2\sin18^\circ\cos18^\circ=4\cos^318^\circ-3\cos18^\circ$$
$$4\cos^318^\circ-3\cos18^\circ-2\sin18^\circ\cos18^\circ=0$$
$$\cos18^\circ(4\cos^218^\circ-2\sin18^\circ-3)=0$$
$$\textrm{Since }\cos18^\circ\neq0,~4\cos^218^\circ-2\sin18^\circ-3=0.$$
$$4-4\sin^218^\circ-2\sin18^\circ-3=0$$
$$4\sin^218^\circ+2\sin18^\circ-1=0$$
$$\sin18^\circ=\frac{-2\pm\sqrt{2^2+4\cdot4}}{8}$$
$$\sin18^\circ=\frac{-1\pm\sqrt{5}}{4}$$
$$\textrm{Since }\sin^\circ>0,~\sin18^\circ=\frac{-1+\sqrt{5}}{4}.$$
Therefore, $\cos72^\circ=\sin18^\circ=\frac{-1+\sqrt{5}}{4}$ and $\sin 72^\circ=\sqrt{1-(\frac{-1+\sqrt{5}}{4})^2}=\frac{\sqrt{10+2\sqrt{5}}}{4}$.
$$\sin75^\circ=\sin(30^\circ+45^\circ)=\sin30^\circ\cos45^\circ+\cos30^\circ\sin45^\circ=\frac{\sqrt{2}+\sqrt{6}}{4}$$
$$\cos75^\circ=\cos30^\circ\cos45^\circ-\sin30^\circ\sin45^\circ=\frac{\sqrt{6}-\sqrt{2}}{4}$$
Now we can calculate the values of $\sin 3^\circ$ and $\cos 3^\circ$.
$$\sin3^\circ=\sin75^\circ\cos72^\circ-\cos75^\circ\sin72^\circ$$
$$\sin3^\circ=\frac{\sqrt{2}+\sqrt{6}}{4}\cdot\frac{-1+\sqrt{5}}{4}-\frac{\sqrt{6}-\sqrt{2}}{4}\cdot\frac{\sqrt{10+2\sqrt{5}}}{4}$$
$$\sin3^\circ=\frac{(\sqrt{2}+\sqrt{6})\cdot(-1+\sqrt{5})-(\sqrt{6}-\sqrt{2})\cdot\sqrt{10+2\sqrt{5}}}{16}$$
$$\cos3^\circ=\cos75^\circ\cos72^\circ+\sin75^\circ\sin72^\circ$$
$$\cos3^\circ=\frac{\sqrt{6}-\sqrt{2}}{4}\cdot\frac{-1+\sqrt{5}}{4}+\frac{\sqrt{2}+\sqrt{6}}{4}\cdot\frac{\sqrt{10+2\sqrt{5}}}{4}$$
$$\cos3^\circ=\frac{(\sqrt{6}-\sqrt{2})\cdot(-1+\sqrt{5})+(\sqrt{2}+\sqrt{6})\cdot\sqrt{10+2\sqrt{5}}}{16}$$
I have checked with approximate values that these two values are correct.
$$\textrm{Let }x=\sin1^\circ\textrm{ and }y=\cos1^\circ$$
$$\sin3^\circ=3x-4x^3$$
$$\cos3^\circ=4y^3-3y$$
\begin{equation}
\begin{cases}
\frac{(\sqrt{2}+\sqrt{6})\cdot(-1+\sqrt{5})-(\sqrt{6}-\sqrt{2})\cdot\sqrt{10+2\sqrt{5}}}{16}=3x-4x^3\\
\frac{(\sqrt{6}-\sqrt{2})\cdot(-1+\sqrt{5})+(\sqrt{2}+\sqrt{6})\cdot\sqrt{10+2\sqrt{5}}}{16}=4y^3-3y\\
x^2+y^2=1
\end{cases}
\end{equation}
Here is where I got stuck. I can't solve a cubic function with this giant constant term. I have three equations (sin, cos, and sum of squares) that I can use. Any help would be appreciated.
Best Answer
As pointed out in a comment, of course you can solve for $x$ in the equation $$ \frac{(\sqrt{2}+\sqrt{6})\cdot(-1+\sqrt{5})-(\sqrt{6}-\sqrt{2})\cdot\sqrt{10+2\sqrt{5}}}{16}=3x-4x^3. $$ It will just be an even much bigger and uglier expression. And it will involve square roots of negative numbers at some point while you are evaluating it; this is unavoidable. (See the answers to Can we express the value of $\sin 1^\circ$ without using the imaginary unit? and Is it possible get an algebraic expression for $\sin 1^\circ$ that does not contain roots of negative values?.)
But if you start with a larger angle than $3^\circ$ there are expressions that are not so ugly. Here is one that is very neat (albeit possibly a little cryptic at first) from The Sine of a Single Degree by Travis Kowalski: $$ \sin 1^\circ = \frac{\sqrt[90]{\sqrt{-1}} - \sqrt[90]{-\sqrt{-1}}}{2 \sqrt{-1}} $$ where all the roots are principal roots.