Extend the definitions of sine, cosine and tangent to any angle using the unit circle and the four quadrants, use the ASTC rule for the signs of the ratios, find the related acute angle, determine exact values of the trigonometric functions at angles around the circle, and find one ratio given another together with the quadrant, all in degrees

What this dot point is asking

In right-angled triangle trigonometry the sine, cosine and tangent of an angle were ratios of sides, and that only makes sense when the angle is acute, because the other two angles of a right-angled triangle are acute. Yet the exam happily asks for $\cos 150\degree$, $\tan 210\degree$ or every solution of $\cos\theta = -\tfrac12$ between $0\degree$ and $360\degree$. This page extends the three ratios to any angle at all using a circle in the number plane, so that an obtuse or reflex angle has a perfectly good sine, cosine and tangent. Everything here is in degrees.

The whole topic rests on one picture and two habits. The picture is the unit circle: a ray from the origin at angle $\theta$ meets a circle of radius $1$ at a point $P$, and we simply define $\cos\theta$ to be the $x$-coordinate of $P$ and $\sin\theta$ to be the $y$-coordinate. The two habits are reading the sign of a ratio from the quadrant the ray lands in (the ASTC rule), and folding any angle back to its related acute angle so the actual value comes from the familiar special angles $30\degree$, $45\degree$ and $60\degree$. Get the quadrant and the related angle and every value on this page follows.

The answer

Defining sine, cosine and tangent for any angle

Place the angle on the number plane. Start a ray along the positive $x$-axis (this is $0\degree$) and rotate it anticlockwise through $\theta$; a negative angle rotates the other way, clockwise. Now draw a circle of radius $1$ centred at the origin, the unit circle, and let the ray meet it at the point $P(x, y)$.

For an acute angle this point sits in the first quadrant, and dropping a perpendicular to the $x$-axis makes a right-angled triangle with hypotenuse $1$ (the radius). In that triangle the side opposite $\theta$ has length $y$ and the side adjacent has length $x$, so $\sin\theta = \tfrac{y}{1} = y$ and $\cos\theta = \tfrac{x}{1} = x$. The new definitions are built to keep exactly these values, but they now make sense wherever $P$ lands:

Because $P$ is always a real point with real coordinates, every angle now has a sine and a cosine, and they range between $-1$ and $1$ as $P$ swings around the circle. Tangent is the ratio $\tfrac{y}{x}$, which is the gradient of the ray, and it is undefined exactly when $x = 0$ (the ray points straight up or down).

The four quadrants and the ASTC sign rule

The axes cut the plane into four quadrants, numbered $1$ to $4$ anticlockwise starting from the top right. The quadrant of $\theta$ is just the quadrant the ray lands in: acute angles ($0\degree$ to $90\degree$) sit in quadrant 1, obtuse angles ($90\degree$ to $180\degree$) in quadrant 2, reflex angles from $180\degree$ to $270\degree$ in quadrant 3, and from $270\degree$ to $360\degree$ in quadrant 4.

The sign of each ratio depends only on the signs of $x$ and $y$, because the radius is always positive. In quadrant 1 both $x$ and $y$ are positive, so all three ratios are positive. In quadrant 2, $x < 0$ but $y > 0$, so only sine ($y$) is positive. In quadrant 3 both are negative, so sine and cosine are negative but their ratio tangent is positive. In quadrant 4, $x > 0$ but $y < 0$, so only cosine ($x$) is positive. The four "positive" letters, read anticlockwise from quadrant 1, spell A, S, T, C, remembered in New South Wales as "All Stations To Central".

A quick way to use the letter is to ask which ratio you want and whether its letter is the one in that quadrant. Want $\cos 200\degree$? That ray is in quadrant 3, whose letter is T, so cosine is not the survivor and $\cos 200\degree$ is negative. The reciprocal ratios follow their partners (secant with cosine, cosecant with sine, cotangent with tangent), but the Year 11 Advanced course only needs sine, cosine and tangent.

The related acute angle

Reflections of the unit circle in the axes carry $P$ to points whose coordinates are the same except possibly for sign. So the size of every ratio at $\theta$ equals its size at the nearest acute angle measured to the $x$-axis, and only the sign can differ. That nearest acute angle is the related acute angle (or reference angle): the acute angle between the ray and the $x$-axis.

Finding it is a one-line subtraction once you know the quadrant, because you always measure to the horizontal axis, never the vertical:

• Quadrant 1: the related angle is $\theta$ itself.
• Quadrant 2: related angle $= 180\degree - \theta$.
• Quadrant 3: related angle $= \theta - 180\degree$.
• Quadrant 4: related angle $= 360\degree - \theta$.

For example $150\degree$ is in quadrant 2, so its related acute angle is $180\degree - 150\degree = 30\degree$; the rays for $30\degree$, $150\degree$, $210\degree$ and $330\degree$ are the four reflections of one another and all make $30\degree$ with the $x$-axis. The diagram shows the construction for $150\degree$.

The single rule that ties the page together follows: the trigonometric function of $\theta$ equals the same function of its related acute angle, with a sign supplied by ASTC.

Exact values around the circle

Because the related acute angle is always one of the special angles when $\theta$ is a "nice" multiple of $30\degree$ or $45\degree$, the exact values from the two special triangles spread to the whole circle. The acute anchors are

At the four boundary angles the point $P$ sits on an axis, and the values are read straight off its coordinates: $P$ is $(1,0)$ at $0\degree$, $(0,1)$ at $90\degree$, $(-1,0)$ at $180\degree$ and $(0,-1)$ at $270\degree$. Hence

Tangent is $\tfrac{y}{x}$, so it is $0$ at $0\degree$ and $180\degree$ (where $y = 0$) and undefined at $90\degree$ and $270\degree$ (where $x = 0$). For any other multiple of $30\degree$ or $45\degree$, take the related acute angle and attach the ASTC sign, for instance $\cos 225\degree = -\cos 45\degree = -\tfrac{1}{\sqrt2}$ (quadrant 3, cosine negative) and $\sin 330\degree = -\sin 30\degree = -\tfrac12$ (quadrant 4, sine negative).

How exam questions ask about trigonometric functions of any angle

The wording tells you which tool to reach for:

• "Find the exact value of $\cos 150\degree$" (or $\tan 210\degree$, $\sin 300\degree$, ...) wants the quadrant, the related acute angle and the ASTC sign, ending in a surd, not a decimal.
• "State the sign of $\tan 200\degree$" or "in which quadrant is ..." is a pure ASTC question; name the quadrant and the surviving letter.
• "Given $\sin\theta = \tfrac35$ and $\theta$ is obtuse, find $\cos\theta$" fixes the quadrant with a word (obtuse $\to$ quadrant 2, reflex, or a range like $180\degree < \theta < 270\degree$); use $\sin^2\theta + \cos^2\theta = 1$ for the size and ASTC for the sign.
• "Solve $\cos\theta = -\tfrac12$ for $0\degree \le \theta \le 360\degree$" is the related-angle method: related acute angle from the positive value, then one angle per allowed quadrant, then trim to the range.
• "Evaluate without a calculator" or "leave your answer in surd form" both forbid the decimal and demand the related-angle working.
• "Show that ..." hands you the target value; lay the quadrant and related-angle steps out so the working lands exactly on it.

Evaluating a value stage by stage

The most reliable way to evaluate any trigonometric function of a non-acute angle is the same four-stage routine every time. Here it is for $\cos 150\degree$, the worked example above, drawn out so you can see each decision on the unit circle.

Stage 1, place the ray for the angle. Rotate anticlockwise from the positive $x$-axis through $150\degree$; the ray lands in quadrant 2.

Stage 2, read the sign from ASTC. Quadrant 2 carries the letter S, so only sine is positive; cosine is negative there. The answer will therefore carry a minus sign.

Stage 3, find the related acute angle. Measure to the $x$-axis: the related acute angle is $180\degree - 150\degree = 30\degree$, the angle between the ray and the negative $x$-axis.

Stage 4, combine the sign and the value. Cosine at the related angle is $\cos 30\degree = \dfrac{\sqrt3}{2}$, and the sign from Stage 2 is negative, so $\cos 150\degree = -\cos 30\degree = -\dfrac{\sqrt3}{2}$.

The same four stages, quadrant, sign, related angle, value, evaluate every entry on the circle and solve every equation. Equation solving is the routine run in reverse: you are handed the value and its sign, so you find the related acute angle from the positive size and then place one solution in each quadrant the sign permits.