Trigonometric values of angles from 0° to 180° | Grade 10

In a right triangle, sine and cosine only make sense for acute angles. But a roof can slope at 120°, a door can swing open to 150°, and a ruler can be turned through any angle up to a straight line. To measure such things we need $\sin$ , $\cos$ , $\tan$ and $\cot$ for obtuse angles too. The trick is to move the triangle into a coordinate plane.

The unit semicircle

Draw the half circle above the $x$ -axis with centre $O(0, 0)$ and radius $1$ . It is called the unit semicircle: its points $(x, y)$ satisfy $x^2+y^2=1$ and $y\geq0$ .

Pick an angle $\alpha$ between $0^\circ$ and $180^\circ$ . Start at the positive $x$ -axis and turn the ray anticlockwise by $\alpha$ . The ray meets the semicircle at exactly one point $M(x_0, y_0)$ .

The point

M

for an angle

\alpha

When $\alpha$ is acute, $OHM$ is a right triangle with hypotenuse $OM=1$ . Then $\cos\alpha=\dfrac{OH}{OM}=x_0$ and $\sin\alpha=\dfrac{HM}{OM}=y_0$ . So the coordinates of $M$ are just the cosine and sine. We use this as the definition for every $\alpha$ in $[0^\circ, 180^\circ]$ .

When are tan and cot undefined?

$x_0=0$ exactly when $M=(0, 1)$ , that is $\alpha=90^\circ$ . So $\tan90^\circ$ does not exist.
$y_0=0$ exactly when $M=(\pm1, 0)$ , that is $\alpha=0^\circ$ or $\alpha=180^\circ$ . So $\cot0^\circ$ and $\cot180^\circ$ do not exist.

Signs

Since $M$ lies on the upper half circle, $\sin\alpha\geq0$ always. The sign of $\cos\alpha=x_0$ depends on the side of the $y$ -axis:

angle	$\sin\alpha$	$\cos\alpha$	$\tan\alpha$	$\cot\alpha$
acute ( $0^\circ<\alpha<90^\circ$ )	$+$	$+$	$+$	$+$
obtuse ( $90^\circ<\alpha<180^\circ$ )	$+$	$-$	$-$	$-$

Supplementary angles

Take $\alpha=60^\circ$ and $180^\circ-\alpha=120^\circ$ . Their points $M$ and $M'$ are mirror images in the $y$ -axis.

Supplementary angles give mirror-image points

Mirror images have the same height and opposite $x$ -coordinates. This gives the rule for any $\alpha$ .

For acute angles there is also the complementary rule $\sin(90^\circ-\alpha)=\cos\alpha$ and $\cos(90^\circ-\alpha)=\sin\alpha$ , which you know from right triangles.

Table of special values

Placing $M$ at $30^\circ$ , $45^\circ$ , $60^\circ$ (and the mirror points $150^\circ$ , $135^\circ$ , $120^\circ$ ) gives the values below. Learn the first half; the second half follows from the supplementary rule.

Points of the semicircle at special angles

$\alpha$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$	$120^\circ$	$135^\circ$	$150^\circ$	$180^\circ$
$\sin\alpha$	$0$	$\frac12$	$\frac{\sqrt2}{2}$	$\frac{\sqrt3}{2}$	$1$	$\frac{\sqrt3}{2}$	$\frac{\sqrt2}{2}$	$\frac12$	$0$
$\cos\alpha$	$1$	$\frac{\sqrt3}{2}$	$\frac{\sqrt2}{2}$	$\frac12$	$0$	$-\frac12$	$-\frac{\sqrt2}{2}$	$-\frac{\sqrt3}{2}$	$-1$
$\tan\alpha$	$0$	$\frac{\sqrt3}{3}$	$1$	$\sqrt3$	none	$-\sqrt3$	$-1$	$-\frac{\sqrt3}{3}$	$0$
$\cot\alpha$	none	$\sqrt3$	$1$	$\frac{\sqrt3}{3}$	$0$	$-\frac{\sqrt3}{3}$	$-1$	$-\sqrt3$	none

("none" means the value is not defined.)

Basic identities

The point $M(x_0, y_0)$ lies on the circle $x^2+y^2=1$ , so $\cos^2\alpha+\sin^2\alpha=1$ . Dividing by $\cos^2\alpha$ or $\sin^2\alpha$ gives two more.

Theorem Basic identities

$\sin^2\alpha+\cos^2\alpha=1$

$1+\tan^2\alpha=\frac{1}{\cos^2\alpha}\quad(\alpha\neq90^\circ),\qquad 1+\cot^2\alpha=\frac{1}{\sin^2\alpha}\quad(\alpha\neq0^\circ,180^\circ)$

$\tan\alpha=\frac{\sin\alpha}{\cos\alpha},\qquad \cot\alpha=\frac{\cos\alpha}{\sin\alpha},\qquad \tan\alpha\cdot\cot\alpha=1$

(the last one when both $\tan\alpha$ and $\cot\alpha$ exist, i.e. $\alpha\neq0^\circ,90^\circ,180^\circ$ ).

Finding the other values from one

Use an identity to get the square of the missing value.
Take the square root and choose the sign from the table of signs (this is where the angle's quadrant matters).
Use $\tan=\sin/\cos$ and $\cot=\cos/\sin$ .

Example Tangent given

Let $\tan\alpha=-2$ . Find $\sin\alpha$ , $\cos\alpha$ , $\cot\alpha$ .

Solution. Step 1. $\tan\alpha<0$ , so $\alpha$ is obtuse: $\cos\alpha<0$ and $\sin\alpha>0$ . Step 2. $\dfrac{1}{\cos^2\alpha}=1+\tan^2\alpha=1+4=5$ , so $\cos^2\alpha=\dfrac15$ . Step 3. $\cos\alpha<0$ , so $\cos\alpha=-\dfrac{1}{\sqrt5}=-\dfrac{\sqrt5}{5}$ . Step 4. $\sin\alpha=\tan\alpha\cdot\cos\alpha=(-2)\left(-\dfrac{\sqrt5}{5}\right)=\dfrac{2\sqrt5}{5}$ . Step 5. $\cot\alpha=\dfrac{1}{\tan\alpha}=-\dfrac12$ .

Check: $\sin^2\alpha+\cos^2\alpha=\dfrac{4}{5}+\dfrac15=1$ .

Using a scientific calculator

Keys differ slightly between models, but the ideas are the same.

First set the calculator to degree mode (the screen shows D or DEG). In radian mode the answers will be wrong.
For $\sin$ , $\cos$ , $\tan$ : press the function key, type the angle, close the bracket if needed, then press equals.
There is no $\cot$ key: use $\cot\alpha=\dfrac{1}{\tan\alpha}$ .
To find an angle from a value, press SHIFT (or 2nd) and then the function key (shown as $\sin^{-1}$ , $\cos^{-1}$ , $\tan^{-1}$ ).

Example Calculator work

(a) Find $\sin37^\circ$ to 4 decimal places. (b) Find $\alpha\in[0^\circ, 180^\circ]$ with $\cos\alpha=-0.3$ . (c) Find all $\alpha\in[0^\circ, 180^\circ]$ with $\sin\alpha=0.8$ .

Solution. (a) Entering $\sin 37$ gives $\sin37^\circ\approx0.6018$ .

(b) Entering $\cos^{-1}(-0.3)$ gives $\alpha\approx107.46^\circ$ . The inverse cosine always returns an angle in $[0^\circ, 180^\circ]$ , so this is the only answer.

(c) $\sin^{-1}(0.8)\approx53.13^\circ$ . But the inverse sine only returns acute angles. The supplementary angle has the same sine, so there is a second answer $180^\circ-53.13^\circ=126.87^\circ$ . Hence $\alpha\approx53.13^\circ$ or $\alpha\approx126.87^\circ$ .

Key terms

English	Tiếng Việt
unit semicircle	nửa đường tròn đơn vị
sine	sin (sin của một góc)
cosine	côsin
tangent	tang
cotangent	côtang
supplementary angles	hai góc bù nhau
obtuse angle	góc tù
acute angle	góc nhọn
identity	đẳng thức (hằng đẳng thức)
undefined	không xác định

Summary

For $\alpha\in[0^\circ, 180^\circ]$ and $M(x_0, y_0)$ on the unit semicircle: $\sin\alpha=y_0$ , $\cos\alpha=x_0$ , $\tan\alpha=\dfrac{y_0}{x_0}$ , $\cot\alpha=\dfrac{x_0}{y_0}$ .
$\tan90^\circ$ , $\cot0^\circ$ , $\cot180^\circ$ do not exist. $\sin\alpha\geq0$ always; for obtuse $\alpha$ , $\cos$ , $\tan$ , $\cot$ are negative.
Supplementary angles: same $\sin$ , opposite $\cos$ , $\tan$ , $\cot$ .
$\sin^2\alpha+\cos^2\alpha=1$ , $1+\tan^2\alpha=\dfrac1{\cos^2\alpha}$ , $1+\cot^2\alpha=\dfrac1{\sin^2\alpha}$ .
On a calculator use degree mode; $\sin^{-1}$ gives only acute angles, so check the supplementary angle too.

Practice

Use the supplementary rule to find $\sin150^\circ$ , $\cos150^\circ$ , $\tan150^\circ$ , $\cot150^\circ$ .
Write the coordinates of the point $M$ of the unit semicircle for $\alpha=135^\circ$ , and for $\alpha=90^\circ$ .
Which of these do not exist: $\tan90^\circ$ , $\cot0^\circ$ , $\tan180^\circ$ , $\cot180^\circ$ , $\cot90^\circ$ ? Give the value of each one that exists.
Let $\sin\alpha=\dfrac{5}{13}$ and $\alpha$ be acute. Find $\cos\alpha$ , $\tan\alpha$ , $\cot\alpha$ .
Let $\cos\alpha=-\dfrac35$ with $0^\circ\leq\alpha\leq180^\circ$ . Find $\sin\alpha$ , $\tan\alpha$ , $\cot\alpha$ .
Evaluate $\cos20^\circ+\cos160^\circ+\sin35^\circ-\sin145^\circ$ .
Let $\tan\alpha=\dfrac12$ . Find $\sin\alpha$ and $\cos\alpha$ .
With a calculator (degree mode), find to 4 decimal places $\sin72^\circ$ , $\cos123^\circ$ , $\tan148^\circ$ .
In triangle $ABC$ , show that $\sin(B+C)=\sin A$ and $\cos(B+C)=-\cos A$ . Use this to find $\sin(B+C)$ and $\cos(B+C)$ when $A=50^\circ$ in terms of $\sin 50^\circ$ and $\cos 50^\circ$ .
(Challenge) For an angle $\alpha\in[0^\circ, 180^\circ]$ it is known that $\sin\alpha+\cos\alpha=\dfrac12$ . Find $\sin\alpha$ and $\cos\alpha$ .

Show solutions

$150^\circ=180^\circ-30^\circ$ . So $\sin150^\circ=\sin30^\circ=\dfrac12$ , $\cos150^\circ=-\cos30^\circ=-\dfrac{\sqrt3}{2}$ , $\tan150^\circ=-\tan30^\circ=-\dfrac{\sqrt3}{3}$ , $\cot150^\circ=-\cot30^\circ=-\sqrt3$ .
For $\alpha=135^\circ$ : $M=(\cos135^\circ, \sin135^\circ)=\left(-\dfrac{\sqrt2}{2}, \dfrac{\sqrt2}{2}\right)$ . For $\alpha=90^\circ$ : $M=(0, 1)$ .
$\tan90^\circ$ , $\cot0^\circ$ and $\cot180^\circ$ do not exist (the denominator $x_0$ or $y_0$ is $0$ ). $\tan180^\circ=\dfrac{0}{-1}=0$ and $\cot90^\circ=\dfrac{0}{1}=0$ exist.
$\cos^2\alpha=1-\dfrac{25}{169}=\dfrac{144}{169}$ and $\alpha$ is acute, so $\cos\alpha=\dfrac{12}{13}$ . Then $\tan\alpha=\dfrac{5/13}{12/13}=\dfrac{5}{12}$ and $\cot\alpha=\dfrac{12}{5}$ .
$\sin^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}$ and $\sin\alpha\geq0$ , so $\sin\alpha=\dfrac45$ . Then $\tan\alpha=\dfrac{4/5}{-3/5}=-\dfrac43$ and $\cot\alpha=-\dfrac34$ .
$\cos160^\circ=-\cos20^\circ$ and $\sin145^\circ=\sin35^\circ$ . So the sum is $\cos20^\circ-\cos20^\circ+\sin35^\circ-\sin35^\circ=0$ .
$\tan\alpha>0$ so $\alpha$ is acute and both values are positive. $\dfrac{1}{\cos^2\alpha}=1+\dfrac14=\dfrac54$ , so $\cos^2\alpha=\dfrac45$ and $\cos\alpha=\dfrac{2}{\sqrt5}=\dfrac{2\sqrt5}{5}$ . Then $\sin\alpha=\tan\alpha\cos\alpha=\dfrac12\cdot\dfrac{2\sqrt5}{5}=\dfrac{\sqrt5}{5}$ . Check: $\dfrac15+\dfrac45=1$ .
$\sin72^\circ\approx0.9511$ , $\cos123^\circ\approx-0.5446$ , $\tan148^\circ\approx-0.6249$ . (The last two are negative because the angles are obtuse, which is a good check on the calculator mode.)
In a triangle $A+B+C=180^\circ$ , so $B+C=180^\circ-A$ . By the supplementary rule $\sin(B+C)=\sin(180^\circ-A)=\sin A$ and $\cos(B+C)=\cos(180^\circ-A)=-\cos A$ . For $A=50^\circ$ : $\sin(B+C)=\sin50^\circ$ and $\cos(B+C)=-\cos50^\circ$ (here $B+C=130^\circ$ ).
Square: $(\sin\alpha+\cos\alpha)^2=\dfrac14$ , so $1+2\sin\alpha\cos\alpha=\dfrac14$ and $\sin\alpha\cos\alpha=-\dfrac38$ . The product is negative and $\sin\alpha\geq0$ , so $\sin\alpha>0$ and $\cos\alpha<0$ . Then $(\sin\alpha-\cos\alpha)^2=1-2\sin\alpha\cos\alpha=1+\dfrac34=\dfrac74$ and $\sin\alpha-\cos\alpha>0$ , so $\sin\alpha-\cos\alpha=\dfrac{\sqrt7}{2}$ . Adding and subtracting with $\sin\alpha+\cos\alpha=\dfrac12$ : $\sin\alpha=\dfrac{1+\sqrt7}{4}\approx0.911$ and $\cos\alpha=\dfrac{1-\sqrt7}{4}\approx-0.411$ . Check: $\dfrac{(1+\sqrt7)^2+(1-\sqrt7)^2}{16}=\dfrac{16}{16}=1$ .