Sine and cosine

The Euler e-function is a series that converges for every complex number. Put a pure imaginary number $ix$ into that series. The even powers are real and the odd powers are imaginary. On the plane, are those two real series the horizontal and vertical segments of the point reached by turning through a real angle $x$?

Turn through a real angle

This picture keeps the angle real, and it keeps the radius equal to $1$. The angle is measured in radians. Drag it. The point stays on the unit circle. Its horizontal segment and its vertical segment are the two coordinates. Drag through a full turn, $2\pi$, and the point comes back. Drag through $0$ and the turn runs clockwise: the vertical segment changes sign, and the horizontal segment does not.

Read the same two segments as heights against the angle. This picture keeps the angle real, and it keeps the radius equal to $1$. Drag the angle from $0$ to $\pi$. A horizontal segment carries the vertical segment across onto the sine. The cosine height matches the horizontal segment on the circle. At a right angle the cosine sits on the axis and the sine height is $1$.

For every complex $x$,

$\sin \left(\frac{\pi }{2}+x\right)=\cos x,\cos \left(\frac{\pi }{2}+x\right)=-\sin x.$

The radius is frozen at $1$, and the angle on this picture is real. Both of those are extra. The two coordinates still have to be named for a complex number, where there is no segment to measure.

The exponential splits

For every complex $x$,

$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots ={\sum }_{k=0}^{\infty }(-1{)}^k\frac{x^{2k+1}}{(2k+1)!},$

$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots ={\sum }_{k=0}^{\infty }(-1{)}^k\frac{x^{2k}}{(2k)!}.$

Both series converge for every complex $x$. When $x$ is real, they are the two segments on the circle.

Substitute $ix$ into the series for the Euler e-function. Because $i^2=-1$, the real part is the cosine series and the imaginary part is the sine series. For every complex $x$,

$e^{ix}=\cos x+i\sin x,e^{-ix}=\cos x-i\sin x.$

Solving that pair for the two series gives, again for every complex $x$,

$\sin x=\frac{e^{ix}-e^{-ix}}{2i},\cos x=\frac{e^{ix}+e^{-ix}}{2}.$

The identities that follow are this pair, read again.

Addition is one exponential

The Euler e-function multiplies when the inputs add: $e^{u+v}=e^u{e}^v$. Set $u=ix$ and $v=iy$, and compare real and imaginary parts. For all complex $x$ and $y$,

$\sin (x\pm y)=\sin x\cos y\pm \cos x\sin y,$

$\cos (x\pm y)=\cos x\cos y\mp \sin x\sin y.$

This picture keeps both angles real and acute, so the first step points outward and the second step turns counterclockwise. Drag either angle. The first step has length $\cos y$ along the angle $x$. The second step has length $\sin y$ and stands perpendicular to the first. Their end is the point at angle $x+y$. The identity is the one just written for complex $x$ and $y$.

Twice the angle is one square

Square the exponential. For every complex $x$,

$(\cos x+i\sin x{)}^2=\cos 2x+i\sin 2x,$

and the imaginary and real parts are

$\sin 2x=2\sin x\cos x,\cos 2x={\cos }^{2}x-{\sin }^{2}x.$

This picture keeps the angle real and at most a right angle, so twice the angle stays within a half turn. Drag it. The second arc copies the first. The squared point sits at the end of the two arcs, and its two segments are $\sin 2x$ and $\cos 2x$.

A right angle is an extra freeze

Keep the angle real and freeze it so that $0<x<\pi /2$, with one right angle in the triangle. Then both segments are positive. If the hypotenuse has length $c$, the opposite side $a$ and the adjacent side $b$ satisfy

$\sin x=\frac{a}{c},\cos x=\frac{b}{c}.$

A right triangle already uses $a=c\sin \alpha$ and $b=c\cos \alpha$ when the right angle is frozen and the hypotenuse is not $1$. On the first picture the radius is $1$, so the opposite side is the sine itself. The same ratio times $10000000$ is how that side was tabulated before the hypotenuse was taken to be $1$.

On that unit circle the two segments and the radius form a right triangle, so

${\cos }^{2}x+{\sin }^{2}x=1$

for real $x$. The same equality holds for every complex $x$. The picture only shows the real case.

The acute angles fix the values below. A degree sign marks a degree. Everywhere else the unit is the radian.

Past a right angle, the sines at $2\pi /3$, $3\pi /4$, $5\pi /6$, and $\pi$ stay $\sqrt{3}/2$, $\sqrt{2}/2$, $1/2$, and $0$. The cosines change sign: $-\frac{1}{2}$, $-\sqrt{2}/2$, $-\sqrt{3}/2$, and $-1$. In degrees those angles are $120^{\circ }$, $135^{\circ }$, $150^{\circ }$, and $180^{\circ }$.

The same angles fix the tangent and the cotangent. Where both entries are numbers, each is the reciprocal of the other. A dash is not a real value.

The zeros line up

On the real line the sine meets the axis at the integer multiples of $\pi$, and the cosine meets it halfway between them. For every integer $k$,

$\sin x=0\text{ at }x=k\pi ,\cos x=0\text{ at }x=k\pi +\frac{\pi }{2}.$

A complex zero of either function is still one of these real numbers.

The sine built from those zeros is, for every complex $x$,

$\sin \pi x=\pi x{\prod }_{k=1}^{\infty }\left(1-\frac{x^2}{k^2}\right).$

At each integer the product has a factor $0$, and that factor appears once. Drag the number of factors. The dashed curve is the partial product. Between $1$ and $2$ it moves toward the sine, and it meets the axis at each integer already included.

One step rises by the cosine

For every complex $x$,

$\frac{\mathrm{d}\sin x}{\mathrm{d}x}=\cos x,\frac{\mathrm{d}\cos x}{\mathrm{d}x}=-\sin x.$

This picture keeps $x$ real. Drag the point on the sine. The horizontal step has length $1$. The vertical step equals the cosine at that point.

Sine and cosine are the imaginary and real parts of the exponential at a pure imaginary input. A full turn brings a real angle back to the same point, and a right angle with positive segments is the extra freeze in which those segments are side ratios. The quotient of the two segments is the natural next function.