Trigonometry

Sine and cosine are usually met as ratios in a right triangle: opposite over hypotenuse, adjacent over hypotenuse. That definition works and then stops working. A right triangle has no angle of $120°$, so $\cos (2\pi /3)$ has no meaning under it, and neither does a negative angle, nor one past a full turn.

The unit circle removes the ceiling. Walk $\theta$ units counterclockwise around the circle $x^2+y^2=1$ starting from $(1,0)$; the point you reach has coordinates $(\cos \theta ,\sin \theta )$. Every real number names a point, so every real number has a sine and a cosine.

Almost everything else then follows from the picture rather than from memory. The values cannot leave $[-1,1]$, because a point on the circle has coordinates no larger than its radius. The functions repeat every $2\pi$, because the walk returns to where it began. Tangent is undefined at $\pi /2$, because the first coordinate is zero there and the quotient $\sin /\cos$ divides by it.

The Pythagorean identity is the clearest case: ${\cos }^2\theta +{\sin }^2\theta =1$ is not a theorem to learn but the circle's own equation with the coordinates renamed. It cannot fail without the point leaving the circle.

What the circle does not supply is exact values, and those come from two triangles, the half-square and the half-equilateral, which between them fix every entry in the table calculus uses throughout.

Why the circle definition is the general one. The triangle definition reads a ratio of side lengths, and side lengths are positive, so it can only produce values for acute angles with positive sines and cosines. The circle definition reads coordinates, which carry signs, so obtuse and reflex angles work without amendment: at $2\pi /3$ the point sits in the second quadrant, giving $\cos =-\frac{1}{2}$ and $\sin =+\frac{\sqrt{3}}{2}$.

The two agree wherever both apply. For an acute $\theta$ the point $(\cos \theta ,\sin \theta )$ is the corner of a right triangle with hypotenuse 1, so the coordinates are the ratios.

Why radians and not degrees. The angle $\theta$ in the definition is an arc length along a circle of radius 1. A pure number, not a count of arbitrary subdivisions. That matters beyond tidiness: the limit $\underset{\theta \to 0}{lim}\frac{\sin \theta }{\theta }=1$ holds in radians and fails in degrees, where the same limit is $\frac{\pi }{180}\approx 0.01745$. Every derivative formula depends on it, so $\frac{d}{dx}\sin x=\cos x$ is false for $x$ in degrees.

A degree measure would force a constant into every subsequent formula, which is why analysis fixes radians and treats degrees as a conversion.

Why ${\sin }^2\theta$ is written that way. ${\sin }^2\theta$ abbreviates $(\sin \theta {)}^2$. The value squared. It does not mean $\sin ({\theta }^2)$. The notation is a compression that avoids parentheses in the identities, and misreading it inverts the order of operations: at $\theta =2$, $(\sin 2{)}^2\approx 0.827$ while $\sin (4)\approx -0.757$.

Why the range is exactly $[-1,1]$. The coordinates of a point at distance 1 from the origin satisfy $|x|\le 1$ and $|y|\le 1$, with equality at the axis crossings. So $\sin \theta =2$ is not merely hard to solve. It has no solution, because no point on the unit circle has second coordinate 2.

Why tangent has period $\pi$, not $2\pi$. Advancing by $\pi$ sends $(\cos \theta ,\sin \theta )$ to $(-\cos \theta ,-\sin \theta )$, the diametrically opposite point. Both coordinates change sign, so their quotient is unchanged. Sine and cosine need a full revolution to return; their ratio needs only half.

Why the identity list is short. The Pythagorean identity restates $x^2+y^2=1$. The double-angle formulas are angle addition with $\beta =\alpha$. The reciprocal identities $1+{\tan }^2={\sec }^2$ follow by dividing the Pythagorean identity by ${\cos }^2\theta$. Only the angle-addition formulas themselves require separate proof, so the material to remember is one equation and one pair of formulas, everything else is derived in a line.

The 45-45-90 triangle. Cut a unit square along its diagonal. The two legs are $1$ and $1$, and the hypotenuse $h$ satisfies $1^2+1^2=h^2$, so $h=\sqrt{2}$, checked: $(\sqrt{2}{)}^2=2.0000000000$.

The two acute angles are equal and sum to $90°$, so each is $45°=\pi /4$. Scaling the triangle to hypotenuse 1 divides every side by $\sqrt{2}$:

Confirmed: $\sin (\pi /4)=0.7071067812$.

The 30-60-90 triangle. Take an equilateral triangle of side 2 and drop a perpendicular from one vertex. It bisects both the opposite side and the angle, leaving a right triangle with hypotenuse 2, short leg 1, and long leg $\ell$ where $1^2+{\ell }^2=2^2$, so $\ell =\sqrt{3}$, checked: $1+3=4=2^2$.

The angles are $30°$ and $60°$, and dividing by the hypotenuse 2 gives

Confirmed: $\sin (\pi /6)=0.5000000000$ and $\sin (\pi /3)=0.8660254038=\sqrt{3}/2$.

Every entry in the special-angle table comes from these two pictures. Nothing else is memorised.

The Pythagorean identity, in one line. The definition places $(\cos \theta ,\sin \theta )$ on the circle $x^2+y^2=1$. Substituting the coordinates:

Both triangles follow from that one construction. Confirmed at $17°$, $133°$ and $271°$, each giving $1.000000000000000$, and it holds for every real $\theta$, since every $\theta$ names a point on the circle.

Dividing it gives two more. Dividing through by ${\cos }^2\theta$ (where nonzero):

and dividing by ${\sin }^2\theta$ gives ${\cot }^2\theta +1={\csc }^2\theta$. Neither is a new fact.

Double angles from angle addition. Taking $\beta =\alpha$ in $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$:

and in the cosine formula, $\cos 2\alpha ={\cos }^2\alpha -{\sin }^2\alpha$.

Confirmed at $\alpha =23°$: $\sin 46°=0.719339800339$ and $2\sin 23°\cos 23°=0.719339800339$; $\cos 46°=0.694658370459$ and ${\cos }^{2}23°-{\sin }^{2}23°=0.694658370459$. At $\alpha =77°$ and $\alpha =150°$ the agreement is identical to twelve places.

A new exact value from the old ones. $75°=30°+45°$, so

Confirmed: $\sin 75°=0.965925826289$ and $(\sqrt{6}+\sqrt{2})/4=0.965925826289$. Angle addition extends the table to every angle expressible as a sum or difference of special ones.

1. An exact value outside the first quadrant: $\cos {\displaystyle \frac{2\pi }{3}}$.

The angle $2\pi /3=120°$ places the point in the second quadrant, where the first coordinate is negative and the second positive. Its reference angle, the acute angle to the horizontal axis, is $\pi -2\pi /3=\pi /3$.

From the 30-60-90 triangle, $\cos (\pi /3)=\frac{1}{2}$ and $\sin (\pi /3)=\frac{\sqrt{3}}{2}$. Applying the quadrant's signs:

Check: $\cos (2\pi /3)=-0.5000000000$ and $\sin (2\pi /3)=+0.8660254038=\sqrt{3}/2$.

No sign rule was memorised. The quadrant supplied it.

2. One function from another: given $\cos t=\frac{3}{5}$ with , find $\sin t$ and $\tan t$.

From the Pythagorean identity,

The stated condition selects $\sin t=\frac{4}{5}$. Then

Check: ${\cos }^2+{\sin }^2=1.000000000000000$ and $4/3=1.3333333333$.

The sign condition is not decoration: without it both values are possible, and the identity alone cannot choose.

3. A new exact value by angle addition: $\sin 75°$.

Write $75°=30°+45°$ and apply $\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta$:

Check: $\sin 75°=0.965925826289$ and $(\sqrt{6}+\sqrt{2})/4=0.965925826289$.

The companion computation gives $\cos 75°=\cos 30°\cos 45°-\sin 30°\sin 45°=\frac{\sqrt{6}-\sqrt{2}}{4}\approx 0.258819045103$, with the minus sign that the cosine formula carries.

4. A double angle: $\sin 46°$ from $\alpha =23°$.

Check: both sides give. Likewise $\cos 46°={\cos }^{2}23°-{\sin }^{2}23°=0.694658370459$, matching directly.

These are angle addition with $\beta =\alpha$, not separate formulas.

5. A periodic equation: solve $2\sin \theta =1$ on $[0,2\pi )$, then in general.

Divide first: $\sin \theta =\frac{1}{2}$. The second coordinate equals $\frac{1}{2}$ at two points of the circle, one in the first quadrant and one in the second, mirror images across the vertical axis:

Check: $2\sin (\pi /6)=1.0000000000$ and $2\sin (5\pi /6)=1.0000000000$.

The general solution adds whole revolutions to each:

Check at $k=1$ and $k=2$: every value returns $2\sin \theta =1.0000000000$.

Reporting only $\pi /6$ would answer a different question. Periodicity makes the solution set infinite, and the interval, when one is given, is what makes the answer finite.