Exact values and links between the functions

Geometry, grade 10 (4 units) Lesson 10 of 20Learning

Exact values and links between the functions

The angle 45°

In an isosceles right-angled triangle both acute angles are $\displaystyle 45^{\circ}$ . If the legs are $\displaystyle 1$ long, the hypotenuse by Pythagoras’ theorem is $\displaystyle {\sqrt{1^{2}+1^{2}}}={\sqrt{2}}$ .

$\displaystyle {\sin45^{\circ}}={\cos45^{\circ}}={\frac{1}{\sqrt{2}}}={\frac{\sqrt{2}}{2}}$ – multiply the numerator and the denominator by $\displaystyle \sqrt{2}$ , and the denominator has no root;
$\displaystyle {\tan45^{\circ}}={\frac{1}{1}}={1}$ .

The angles 30° and 60°

Half an equilateral triangle is a right-angled triangle with angles of $\displaystyle 30^{\circ}$ and $\displaystyle 60^{\circ}$ . If the hypotenuse is $\displaystyle 2$ long, the side opposite $\displaystyle 30^{\circ}$ is half the hypotenuse, $\displaystyle 1$ , and the other leg $\displaystyle {\sqrt{2^{2}-1^{2}}}={\sqrt{3}}$ .

$\displaystyle {\sin30^{\circ}}={\frac{1}{2}}$ , $\displaystyle {\cos30^{\circ}}={\frac{\sqrt{3}}{2}}$ , $\displaystyle {\tan30^{\circ}}={\frac{1}{\sqrt{3}}}={\frac{\sqrt{3}}{3}}$ ;
$\displaystyle {\sin60^{\circ}}={\frac{\sqrt{3}}{2}}$ , $\displaystyle {\cos60^{\circ}}={\frac{1}{2}}$ , $\displaystyle {\tan60^{\circ}}={\frac{\sqrt{3}}{1}}={\sqrt{3}}$ .

The table of exact values

	$\displaystyle 30^{\circ}$	$\displaystyle 45^{\circ}$	$\displaystyle 60^{\circ}$
$\displaystyle \sin$	$\displaystyle \frac{1}{2}$	$\displaystyle \frac{\sqrt{2}}{2}$	$\displaystyle \frac{\sqrt{3}}{2}$
$\displaystyle \cos$	$\displaystyle \frac{\sqrt{3}}{2}$	$\displaystyle \frac{\sqrt{2}}{2}$	$\displaystyle \frac{1}{2}$
$\displaystyle \tan$	$\displaystyle \frac{\sqrt{3}}{3}$	$\displaystyle 1$	$\displaystyle \sqrt{3}$

In a calculation that goes on, write these values, not their approximations: $\displaystyle \frac{\sqrt{3}}{2}$ , not $\displaystyle 0.87$ .

Exercise 1

Calculate exactly, with the values in the table.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – check and a new line · * – times · ^ – power, ↓ – end of the power

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Two links between the functions

In a right-angled triangle with legs $\displaystyle a$ , $\displaystyle b$ and hypotenuse $\displaystyle c$ :

divide $\displaystyle {\sin\alpha}={\frac{a}{c}}$ by $\displaystyle {\cos\alpha}={\frac{b}{c}}$ : the $\displaystyle c$ cancels, and $\displaystyle \frac{a}{b}$ is left – the tangent:

\displaystyle \tan\alpha

\displaystyle =

\displaystyle \frac{\sin\alpha}{\cos\alpha}

square and add: $\displaystyle {\sin^{2}\alpha+\cos^{2}\alpha}={\frac{a^{2}+b^{2}}{c^{2}}}$ , and by Pythagoras’ theorem $\displaystyle {a^{2}+b^{2}}={c^{2}}$ :

\displaystyle \sin^{2}\alpha+\cos^{2}\alpha

\displaystyle =

\displaystyle 1

$\displaystyle \sin^{2}\alpha$ is $\displaystyle (\sin\alpha)^{2}$ .

The two acute angles of a triangle

The two acute angles of a right-angled triangle add up to $\displaystyle 90^{\circ}$ : if one is $\displaystyle \alpha$ , the other is $\displaystyle 90^{\circ}-\alpha$ . The side opposite one is the side adjacent to the other, so:

\displaystyle \sin(90^{\circ}-\alpha)

\displaystyle =

\displaystyle \cos\alpha

\displaystyle \cos(90^{\circ}-\alpha)

\displaystyle =

\displaystyle \sin\alpha

For example $\displaystyle {\sin60^{\circ}}={\cos30^{\circ}}$ , and $\displaystyle {\cos70^{\circ}}={\sin20^{\circ}}$ .

Exercise 2

Calculate with the links between the functions.
Check with a calculator.

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 3

$\displaystyle \alpha$ is an acute angle, and $\displaystyle {\sin\alpha}={\frac{3}{5}}$ .
Calculate exactly.
(A hint: $\displaystyle {\cos^{2}\alpha}={1-\sin^{2}\alpha}$ .)

\displaystyle \cos\alpha

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \tan\alpha

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Exercise 4

$\displaystyle \beta$ is an acute angle, and $\displaystyle {\cos\beta}={\frac{1}{3}}$ .
Calculate exactly.

\displaystyle \sin\beta

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

\displaystyle \tan\beta

/ – fraction bar · Tab or the arrows – between numerator and denominator, then a note above or below · Enter – a new line · * – times · ^ – power, ↓ – end of the power

Formulas: geometry, grade 10 (4 units)

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Analytic geometry

The distance between the points $\displaystyle (x_1,y_1)$ and $\displaystyle (x_2,y_2)$
$\displaystyle d$ $\displaystyle =$ $\displaystyle \sqrt{(x_2-x_1)^{2}+(y_2-y_1)^{2}}$
The midpoint of the segment
$\displaystyle M\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)$
The slope of a line through two points
$\displaystyle m$ $\displaystyle =$ $\displaystyle \frac{y_2-y_1}{x_2-x_1}$ $\displaystyle x_1\neq x_2$
A line’s equation in explicit form – $\displaystyle m$ the slope, $\displaystyle b$ where it meets the $\displaystyle y$ -axis
$\displaystyle y=mx+b$
A line’s equation in implicit form
$\displaystyle Ax+By+C=0$
The line of slope $\displaystyle m$ through the point $\displaystyle (x_1,y_1)$
$\displaystyle y-y_1$ $\displaystyle =$ $\displaystyle m(x-x_1)$
Parallel lines
$\displaystyle m_1$ $\displaystyle =$ $\displaystyle m_2$
Perpendicular lines
$\displaystyle m_1\cdot m_2$ $\displaystyle =$ $\displaystyle -1$
The slope – $\displaystyle \alpha$ the angle between the line and the positive direction of the $\displaystyle x$ -axis
$\displaystyle m$ $\displaystyle =$ $\displaystyle \tan\alpha$

Similar triangles and midsegments

The similarity ratio – corresponding sides
$\displaystyle \frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}=k$
The ratio of the perimeters
$\displaystyle \frac{P_1}{P_2}$ $\displaystyle =$ $\displaystyle k$
The ratio of the areas
$\displaystyle \frac{S_1}{S_2}$ $\displaystyle =$ $\displaystyle k^{2}$
The ratio of corresponding altitudes, medians and angle bisectors
$\displaystyle \frac{h_1}{h_2}=\frac{m_1}{m_2}=\frac{d_1}{d_2}=k$
A triangle’s midsegment – $\displaystyle DE$ parallel to $\displaystyle BC$
$\displaystyle DE$ $\displaystyle =$ $\displaystyle \frac{1}{2}BC$
A trapezoid’s midsegment – $\displaystyle a$ , $\displaystyle b$ the bases
$\displaystyle EF$ $\displaystyle =$ $\displaystyle \frac{a+b}{2}$

Trigonometry

Sine – $\displaystyle a$ the side opposite the angle, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{c}$
Cosine – $\displaystyle b$ the side adjacent to the angle
$\displaystyle \cos\alpha$ $\displaystyle =$ $\displaystyle \frac{b}{c}$
Tangent – the opposite side over the adjacent side
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{a}{b}$
The tangent from the sine and the cosine
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$
The identity from Pythagoras
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
Angles whose sum is $\displaystyle 90^{\circ}$
$\displaystyle \sin(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos(90^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
The sine of an obtuse angle
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
A triangle’s area – $\displaystyle \gamma$ the angle between the sides $\displaystyle a$ and $\displaystyle b$
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}ab\sin\gamma$

Exact values

$\displaystyle \sin30^{\circ}=\cos60^{\circ}=\frac{1}{2}$
$\displaystyle \sin45^{\circ}=\cos45^{\circ}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin60^{\circ}=\cos30^{\circ}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan30^{\circ}$ $\displaystyle =$ $\displaystyle \frac{\sqrt{3}}{3}$
$\displaystyle \tan45^{\circ}$ $\displaystyle =$ $\displaystyle 1$
$\displaystyle \tan60^{\circ}$ $\displaystyle =$ $\displaystyle \sqrt{3}$

Special lines in a triangle

Where the medians meet, $\displaystyle G$ divides the median $\displaystyle AD$ in the ratio $\displaystyle 2:1$
$\displaystyle AG$ $\displaystyle =$ $\displaystyle 2GD$
Where the medians meet, in coordinates
$\displaystyle G\left(\frac{x_1+x_2+x_3}{3},\ \frac{y_1+y_2+y_3}{3}\right)$
The median $\displaystyle AD$ splits the triangle into two triangles of equal area
$\displaystyle S_{ABD}$ $\displaystyle =$ $\displaystyle S_{ACD}$
The angle bisector theorem – $\displaystyle AD$ bisects the angle $\displaystyle A$
$\displaystyle \frac{BD}{DC}$ $\displaystyle =$ $\displaystyle \frac{AB}{AC}$