Sine is just a shadow.

Send a point once around a circle of radius 1. Forget the point — watch only its height. That height, plotted against the angle turned, is the whole of $\sin$.

scroll ↓

On the unit circle the point sits at $(\cos\theta,\ \sin\theta)$. Sine is the height, cosine the shadow on the floor. No triangles required — the circle defines them for every angle, including obtuse and negative ones.

Drop the point onto the horizontal axis and you get $\cos\theta$. As the angle opens past $90^\circ$ that shadow slides left of the origin and cosine goes negative — which is all the "CAST diagram" is ever telling you.

Extend the radius until it strikes the vertical line $x = 1$. The height it reaches there is $\tan\theta$ — because the little triangle and the big one are similar, so $\tan\theta = \sin\theta / \cos\theta$. Tangent is the radius's slope, read off as a length.

Look at the triangle the point makes: legs $\cos\theta$ and $\sin\theta$, hypotenuse $1$. It isn't a new fact to learn — it is Pythagoras, and it holds for every angle because the radius never changes.

Now the payoff. Lay a horizontal line across the wave — say $y = \tfrac{1}{2}$ — and every crossing is a solution of $\sin x = \tfrac{1}{2}$. The graph turns "solve this equation" into "count the crossings".

Your calculator hands back $30^\circ$ and stops. The wave shows you the other one: it is symmetric about $90^\circ$, so $150^\circ$ sits the same distance the other side. Miss it and you lose the marks — the graph never lets you forget.

That's the circle behind the waves.

Exact values, the $\sin$, $\cos$ and $\tan$ graphs, both identities and twelve graded exam questions with full worked solutions are waiting.

Read the full chapter › Jump to the questions ›