Solutions in a given interval

Trigonometry, grade 11 (5 units) Lesson 11 of 18Learning

Solutions in a given interval

Putting in values of

\displaystyle k

When the solutions in an interval are asked for, write the general solution first, then put in $\displaystyle {k}={0, 1, 2, \ldots}$ (and $\displaystyle {k}={-1, -2, \ldots}$ too) and keep only what lies in the interval.

$\displaystyle {\sin2x}={\frac{1}{2}}$ , in the interval $\displaystyle 0\le x\le2\pi$ .

$\displaystyle {x}={\frac{\pi}{12}+\pi k}$ or $\displaystyle {x}={\frac{5\pi}{12}+\pi k}$ .
$\displaystyle {k}={0}$ : $\displaystyle \frac{\pi}{12}$ , $\displaystyle \frac{5\pi}{12}$ . $\displaystyle {k}={1}$ : $\displaystyle \frac{13\pi}{12}$ , $\displaystyle \frac{17\pi}{12}$ . $\displaystyle {k}={2}$ is already out of the interval, and so is $\displaystyle {k}={-1}$ .
Four solutions: $\displaystyle \frac{\pi}{12}$ , $\displaystyle \frac{5\pi}{12}$ , $\displaystyle \frac{13\pi}{12}$ , $\displaystyle \frac{17\pi}{12}$ .

A check: the period of $\displaystyle \sin2x$ is $\displaystyle \pi$ long, and each period holds two solutions – an interval $\displaystyle 2\pi$ long holds four.

Where

\displaystyle \tan x

is defined

An equation with $\displaystyle \tan x$ is defined only where $\displaystyle \cos x\neq0$ . A solution with $\displaystyle {\cos x}={0}$ is ruled out.

$\displaystyle {\tan x}={\sin2x}$ , in the interval $\displaystyle 0\le x\le\pi$ .

$\displaystyle {\frac{\sin x}{\cos x}}={2\sin x\cos x}$ , so $\displaystyle {\sin x}={2\sin x\cos^2x}$ (multiplying by $\displaystyle \cos x\neq0$ ).
$\displaystyle {\sin x(1-2\cos^2x)}={0}$ , that is $\displaystyle {\sin x}={0}$ or $\displaystyle {\cos2x}={0}$ .
In the interval: $\displaystyle {x}={0}$ , $\displaystyle {x}={\pi}$ , $\displaystyle {x}={\frac{\pi}{4}}$ , $\displaystyle {x}={\frac{3\pi}{4}}$ . None of them has $\displaystyle {\cos x}={0}$ , so all are solutions.

Squaring

Squaring both sides may add solutions: $\displaystyle {a}={-b}$ also gives $\displaystyle {a^2}={b^2}$ . So after squaring every solution must be checked in the original equation – or another way chosen.

$\displaystyle {\sin x+\cos x}={1}$ : squaring gives $\displaystyle {1+\sin2x}={1}$ , that is $\displaystyle {\sin2x}={0}$ , and $\displaystyle {x}={\frac{\pi k}{2}}$ . But $\displaystyle {x}={\pi}$ gives $\displaystyle {0+(-1)}={-1}$ – it is no solution. The one-sine way needs no check: $\displaystyle {\sqrt{2}\sin\left(x+\frac{\pi}{4}\right)}={1}$ .

Exercise 1

Solve in the interval $\displaystyle 0\le x\le2\pi$ .

Exercise 2

Solve in the interval $\displaystyle 0\le x\le2\pi$ .

Exercise 3

Solve in the interval $\displaystyle 0\le x\le2\pi$ .

Exercise 4

Solve in degrees, in the interval $\displaystyle 0^\circ\le x\le360^\circ$ .

Exercise 5

Solve in the interval $\displaystyle 0\le x\le\frac{3\pi}{4}$ .

Exercise 6

How many solutions does the equation $\displaystyle {\cos3x}={\frac{1}{2}}$ have in the interval $\displaystyle 0\le x\le2\pi$ ? Explain without finding them.

/ – 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

Formulas: trigonometry, grade 11 (5 units)

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Basic identities

The Pythagorean identity
$\displaystyle \sin^{2}\alpha+\cos^{2}\alpha$ $\displaystyle =$ $\displaystyle 1$
The tangent
$\displaystyle \tan\alpha$ $\displaystyle =$ $\displaystyle \frac{\sin\alpha}{\cos\alpha}$ $\displaystyle \cos\alpha\neq0$
$\displaystyle 1+\tan^{2}\alpha$ $\displaystyle =$ $\displaystyle \frac{1}{\cos^{2}\alpha}$
$\displaystyle \sin(-\alpha)$ $\displaystyle =$ $\displaystyle -\sin\alpha$
$\displaystyle \cos(-\alpha)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \sin(\pi-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
$\displaystyle \cos(\pi-\alpha)$ $\displaystyle =$ $\displaystyle -\cos\alpha$
$\displaystyle \sin\left(\frac{\pi}{2}-\alpha\right)$ $\displaystyle =$ $\displaystyle \cos\alpha$
$\displaystyle \cos\left(\frac{\pi}{2}-\alpha\right)$ $\displaystyle =$ $\displaystyle \sin\alpha$

The sum and the difference of angles

$\displaystyle \cos(\alpha\pm\beta)$ $\displaystyle =$ $\displaystyle \cos\alpha\cdot\cos\beta\mp\sin\alpha\cdot\sin\beta$
$\displaystyle \sin(\alpha\pm\beta)$ $\displaystyle =$ $\displaystyle \sin\alpha\cdot\cos\beta\pm\cos\alpha\cdot\sin\beta$
$\displaystyle \tan(\alpha\pm\beta)$ $\displaystyle =$ $\displaystyle \frac{\tan\alpha\pm\tan\beta}{1\mp\tan\alpha\cdot\tan\beta}$

The double angle

$\displaystyle \sin2\alpha$ $\displaystyle =$ $\displaystyle 2\sin\alpha\cdot\cos\alpha$
$\displaystyle \cos2\alpha=\cos^{2}\alpha-\sin^{2}\alpha=2\cos^{2}\alpha-1=1-2\sin^{2}\alpha$
$\displaystyle \tan2\alpha$ $\displaystyle =$ $\displaystyle \frac{2\tan\alpha}{1-\tan^{2}\alpha}$
$\displaystyle \cos^{2}\alpha$ $\displaystyle =$ $\displaystyle \frac{1+\cos2\alpha}{2}$
$\displaystyle \sin^{2}\alpha$ $\displaystyle =$ $\displaystyle \frac{1-\cos2\alpha}{2}$

Sums and differences of sines and of cosines

$\displaystyle \sin\alpha+\sin\beta$ $\displaystyle =$ $\displaystyle 2\sin\frac{\alpha+\beta}{2}\cdot\cos\frac{\alpha-\beta}{2}$
$\displaystyle \sin\alpha-\sin\beta$ $\displaystyle =$ $\displaystyle 2\sin\frac{\alpha-\beta}{2}\cdot\cos\frac{\alpha+\beta}{2}$
$\displaystyle \cos\alpha+\cos\beta$ $\displaystyle =$ $\displaystyle 2\cos\frac{\alpha+\beta}{2}\cdot\cos\frac{\alpha-\beta}{2}$
$\displaystyle \cos\alpha-\cos\beta$ $\displaystyle =$ $\displaystyle -2\sin\frac{\alpha+\beta}{2}\cdot\sin\frac{\alpha-\beta}{2}$

Equations – the general solution ($k$ a whole number)

$\displaystyle \sin x=\sin\alpha\iff x=\alpha+2\pi k\ \text{or}\ x=\pi-\alpha+2\pi k$
$\displaystyle \cos x=\cos\alpha\iff x=\pm\alpha+2\pi k$
$\displaystyle \tan x=\tan\alpha\iff x=\alpha+\pi k$

A triangle

The law of sines ( $\displaystyle R$ – the radius of the circumscribed circle)
$\displaystyle \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$
The law of cosines ( $\displaystyle \gamma$ – the angle between $\displaystyle a$ and $\displaystyle b$ )
$\displaystyle c^{2}$ $\displaystyle =$ $\displaystyle a^{2}+b^{2}-2a\cdot b\cdot\cos\gamma$
The area of a triangle ( $\displaystyle \alpha$ – the angle between $\displaystyle b$ and $\displaystyle c$ )
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}b\cdot c\cdot\sin\alpha$
The area of a triangle ( $\displaystyle \beta$ , $\displaystyle \gamma$ – the angles next to $\displaystyle a$ )
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a^{2}\cdot\sin\beta\cdot\sin\gamma}{2\cdot\sin(\beta+\gamma)}$

A circle ($\alpha$ in radians, $R$ – the radius)

The area of a disc
$\displaystyle S$ $\displaystyle =$ $\displaystyle \pi R^{2}$
The circumference of a circle
$\displaystyle P$ $\displaystyle =$ $\displaystyle 2\pi R$
The length of an arc
$\displaystyle l$ $\displaystyle =$ $\displaystyle \alpha\cdot R$
The area of a sector
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}\alpha\cdot R^{2}$

Solids ($S$ – the base area, $h$ – the height, $l$ – the slant height)

The volume of a prism and a cylinder
$\displaystyle V$ $\displaystyle =$ $\displaystyle S\cdot h$
The volume of a pyramid and a cone
$\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{S\cdot h}{3}$
The lateral surface of a cone
$\displaystyle M$ $\displaystyle =$ $\displaystyle \pi R\cdot l$
The volume of a sphere
$\displaystyle V$ $\displaystyle =$ $\displaystyle \frac{4}{3}\pi R^{3}$
The surface area of a sphere
$\displaystyle M$ $\displaystyle =$ $\displaystyle 4\pi R^{2}$