Trigonometric Functions of General Angles

Q: How do I find r?

Use $r = \sqrt{x^2 + y^2}$, the distance from the origin to the point. It is always positive.

Q: What are the trig ratios for a general angle?

$\sin\theta = \dfrac{y}{r}$, $\cos\theta = \dfrac{x}{r}$, $\tan\theta = \dfrac{y}{x}$, using the point on the terminal side.

Q: Where do the signs come from?

From $x$ and $y$, since $r$ is always positive. The point's quadrant sets which ratios are negative.

When an angle's terminal side passes through a point $(x, y)$ , you can find its trig values directly. Compute the distance $r$ to the point, then use the coordinate ratios.

Sine is $\dfrac{y}{r}$ , cosine is $\dfrac{x}{r}$ , and tangent is $\dfrac{y}{x}$ . This extends the right-triangle definitions to any angle, including negative and obtuse ones.

Find r, then the ratios

The distance from the origin to $(x, y)$ is $r = \sqrt{x^2 + y^2}$ , always positive. It plays the role of the hypotenuse.

Then $\sin\theta = \dfrac{y}{r}$ , $\cos\theta = \dfrac{x}{r}$ , and $\tan\theta = \dfrac{y}{x}$ . The signs of $x$ and $y$ carry the quadrant information.

Signs by quadrant

Because $r$ is always positive, the sign of each trig value comes from $x$ and $y$ . In the fourth quadrant, $y < 0$ , so sine is negative.

Keeping the point's coordinate signs gives the correct sign automatically — no separate rule to memorize.

In the diagram below the terminal side lands in Quadrant II, where $x$ is negative. So $\cos\theta = \dfrac{x}{r}$ comes out negative, while $\sin\theta = \dfrac{y}{r}$ stays positive because $y > 0$ .

Worked examples

Example 1: sine from a point

The terminal side of $\theta$ passes through $(8, -15)$ . Find $\sin\theta$ .

Compute r

r = \sqrt{8^2 + (-15)^2} = 17

Sine is y over r

\dfrac{-15}{17}

Answer: $-\dfrac{15}{17}$

Example 2: cosine from the same point

For the point $(8, -15)$ with $r = 17$ , find $\cos\theta$ .

Cosine is x over r

\dfrac{8}{17}

Answer: $\dfrac{8}{17}$

Example 3: tangent in quadrant II

The terminal side of $\theta$ passes through $(-6, 8)$ . Find $\tan\theta$ .

Tangent is y over x

\dfrac{8}{-6}

Reduce; x is negative here, so tangent is negative

-\dfrac{4}{3}

Answer: $-\dfrac{4}{3}$

Try one yourself

Common questions

How do I find r?

Use $r = \sqrt{x^2 + y^2}$ , the distance from the origin to the point. It is always positive.

What are the trig ratios for a general angle?

$\sin\theta = \dfrac{y}{r}$ , $\cos\theta = \dfrac{x}{r}$ , $\tan\theta = \dfrac{y}{x}$ , using the point on the terminal side.

Where do the signs come from?

From $x$ and $y$ , since $r$ is always positive. The point's quadrant sets which ratios are negative.

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Trigonometric Functions of General Angles

Find r, then the ratios

Signs by quadrant

Worked examples

Example 1: sine from a point

Example 2: cosine from the same point

Example 3: tangent in quadrant II

Try one yourself

Common questions

Keep learning

Angles & Angle Measure

Graphing Sine & Cosine Functions

Trigonometric Identities