Trig Functions for 0° to 180°

In Grade 8 you learned about sine, cosine, tangent, and cotangent of acute angles. Now we extend these definitions to any angle $\alpha$ where $0° \leq \alpha \leq 180°$ .

The Unit Semicircle

Consider the upper half of the unit circle (radius 1, center at the origin). For any angle $\alpha$ with $0° \leq \alpha \leq 180°$ , there is a unique point $M$ on this semicircle such that $\angle MOA = \alpha$ , where $O = (0,0)$ and $A = (1,0)$ .

📐Definition — Sine and Cosine (0° to 180°)

The cosine and sine of angle $\alpha$ (where $0° \leq \alpha \leq 180°$ ) are defined as the $x$ - and $y$ -coordinates of the point $M$ on the unit semicircle corresponding to $\alpha$ :

$\cos \alpha = x_M, \qquad \sin \alpha = y_M$

Special values: $\sin 0° = 0$ , $\cos 0° = 1$ , $\sin 90° = 1$ , $\cos 90° = 0$ , $\sin 180° = 0$ , $\cos 180° = -1$ .

Since the point $M$ lies on the upper semicircle, its $y$ -coordinate satisfies $0 \leq y \leq 1$ , so $0 \leq \sin\alpha \leq 1$ for all $0° \leq \alpha \leq 180°$ . Its $x$ -coordinate satisfies $-1 \leq x \leq 1$ , so $-1 \leq \cos\alpha \leq 1$ .

Key observation: The cosine of an obtuse angle is negative.

Drag the slider to see how $\sin\alpha$ and $\cos\alpha$ change as $\alpha$ goes from $0°$ to $180°$ , and how the supplement point $M'$ mirrors $M$ :

α = 50.0°

α = 50°sin α = 0.7660cos α = 0.6428tg α = 1.1918ctg α = 0.8391180°−α = 130°sin(180°−α) = 0.7660

Supplement Formulas

For angles $\alpha$ and $180° - \alpha$ , the corresponding points on the unit semicircle are reflections of each other across the $y$ -axis. Therefore their $y$ -coordinates are equal but their $x$ -coordinates are opposite:

$\boxed{\sin(180° - \alpha) = \sin\alpha, \qquad \cos(180° - \alpha) = -\cos\alpha}$

✎Example — Finding Trig Values

Find $\sin 120°$ , $\cos 120°$ .

Solution. Since $120° = 180° - 60°$ :

$\sin 120° = \sin(180° - 60°) = \sin 60° = \frac{\sqrt{3}}{2}$

$\cos 120° = \cos(180° - 60°) = -\cos 60° = -\frac{1}{2}$

Pythagorean Identity

Since $M = (\cos\alpha,\, \sin\alpha)$ lies on the unit circle $x^2 + y^2 = 1$ :

$\boxed{\sin^2\alpha + \cos^2\alpha = 1}$

This identity holds for all $0° \leq \alpha \leq 180°$ .

Tangent and Cotangent

📐Definition — Tangent and Cotangent

The tangent of angle $\alpha$ (where $0° \leq \alpha \leq 180°$ , $\alpha \neq 90°$ ) is:

$\operatorname{tg}\alpha = \frac{\sin\alpha}{\cos\alpha}$

The cotangent of angle $\alpha$ (where $0° < \alpha < 180°$ ) is:

$\operatorname{ctg}\alpha = \frac{\cos\alpha}{\sin\alpha}$

Note: $\operatorname{tg} 90°$ is undefined (since $\cos 90° = 0$ ). Similarly $\operatorname{ctg} 0°$ and $\operatorname{ctg} 180°$ are undefined.

The supplement formulas extend to tangent and cotangent:

$\operatorname{tg}(180° - \alpha) = -\operatorname{tg}\alpha, \qquad \operatorname{ctg}(180° - \alpha) = -\operatorname{ctg}\alpha$

✎Example — Proving the Tangent Supplement Formula

Prove that $\operatorname{tg}(180° - \alpha) = -\operatorname{tg}\alpha$ .

Proof. $\operatorname{tg}(180° - \alpha) = \frac{\sin(180° - \alpha)}{\cos(180° - \alpha)} = \frac{\sin\alpha}{-\cos\alpha} = -\operatorname{tg}\alpha \qquad \blacktriangleleft$

Comparing Trig Values

For $0° \leq \alpha < \beta \leq 90°$ : $\sin\alpha < \sin\beta$ and $\cos\alpha > \cos\beta$ .

For $90° \leq \alpha < \beta \leq 180°$ : $\sin\alpha > \sin\beta$ and $\cos\alpha < \cos\beta$ .

For any $0° \leq \alpha \leq \beta \leq 180°$ : $\cos\alpha \geq \cos\beta$ (cosine is decreasing on $[0°, 180°]$ ).