10B - Trigonometric ratios beyond acute angles

Angles of any magnitude

How can $\sin120^\circ$ have a meaning when a right-angled triangle cannot contain an angle of $120^\circ$?

Start a radius on the positive horizontal axis and turn it anticlockwise. At $120^\circ$, its endpoint is above and to the left of the centre. The vertical coordinate is still positive; the horizontal coordinate is now negative. Signed coordinates extend the familiar right-triangle ratios beyond acute angles.

[DIAGRAM: asset_name: Coordinates extend trigonometric ratios to every quadrant; asset_slug: unit-circle-signs; caption: Cosine is the horizontal coordinate and sine is the vertical coordinate on a unit circle; description: A unit circle shows a radius at 120 degrees, with endpoint minus one half, square root of three over two. Dashed horizontal and vertical projections connect the point to the axes. Each quadrant lists the signs of cosine, sine and tangent; tangent is undefined at the top and bottom of the circle.]

For an acute angle the coordinates are the adjacent and opposite lengths with hypotenuse $1$, so these definitions agree with the triangle ratios you already know. Sine and cosine are defined for every real angle and lie between $-1$ and $1$. Tangent is a quotient and need not lie in that range.

Number the quadrants anticlockwise, starting at the top right. Signs follow the coordinates, without a separate sign rule to guess:

Points on the axes are not inside a quadrant. For example, at $90^\circ=\pi/2$ the endpoint is $(0,1)$, so cosine is $0$, sine is $1$ and tangent is undefined. At $180^\circ=\pi$, the endpoint is $(-1,0)$, giving cosine $-1$, sine $0$ and tangent $0$.

Reflect the point below the horizontal axis next. Both coordinates are negative, but their quotient becomes positive.

A full turn returns to the same endpoint. Adding $360^\circ$, or $2\pi$ radians, therefore leaves all three ratios unchanged wherever tangent is defined.

Exact trigonometric values

Two special right triangles give exact values without a calculator.

An isosceles right triangle with legs $1$ and $1$ has hypotenuse $\sqrt2$ by Pythagoras. Both acute angles are $45^\circ$, so $\sin45^\circ=\cos45^\circ=1/\sqrt2=\sqrt2/2$, and $\tan45^\circ=1$.

Next split an equilateral triangle of side $2$ into two equal right triangles. Each has hypotenuse $2$, shorter leg $1$ and height $\sqrt{2^2-1^2}=\sqrt3$. The angle opposite the shorter leg is $30^\circ$; the other is $60^\circ$. Reading opposite, adjacent and hypotenuse gives the remaining values.