Equations with a linear argument

Trigonometry, grade 10 (5 units) Lesson 12 of 21Learning

Equations with a linear argument

The way:

\displaystyle {u}={ax+b}

In an equation like $\displaystyle {2\sin\left(2x-\frac{\pi}{3}\right)+1}={0}$ the sine is of a linear expression in $\displaystyle x$ – a linear argument. It is solved in three steps:

isolate the sine: $\displaystyle {\sin\left(2x-\frac{\pi}{3}\right)}={-\frac{1}{2}}$ ;
write the general solution for the whole argument: $\displaystyle {\sin\left(-\frac{\pi}{6}\right)}={-\frac{1}{2}}$ , so $\displaystyle {2x-\frac{\pi}{3}}={-\frac{\pi}{6}+2\pi k}$ or $\displaystyle {2x-\frac{\pi}{3}}={\pi+\frac{\pi}{6}+2\pi k}={\frac{7\pi}{6}+2\pi k}$ ;
isolate $\displaystyle x$ – and divide $\displaystyle 2\pi k$ too: $\displaystyle {2x}={\frac{\pi}{6}+2\pi k}$ or $\displaystyle {2x}={\frac{3\pi}{2}+2\pi k}$ , that is

\displaystyle x=\frac{\pi}{12}+\pi k,\quad x=\frac{3\pi}{4}+\pi k

The distance between solutions in the same family is $\displaystyle \pi$ – the period of $\displaystyle \sin\left(2x-\frac{\pi}{3}\right)$ .

Two more examples

$\displaystyle {\cos\frac{x}{2}}={\frac{\sqrt{3}}{2}}$ : $\displaystyle {\frac{x}{2}}={\pm\frac{\pi}{6}+2\pi k}$ , so $\displaystyle {x}={\pm\frac{\pi}{3}+4\pi k}$ . Multiply by 2 – $\displaystyle 2\pi k$ too.
$\displaystyle {\tan3x}={1}$ : $\displaystyle {3x}={\frac{\pi}{4}+\pi k}$ , so $\displaystyle {x}={\frac{\pi}{12}+\frac{\pi k}{3}}$ .

Careful: dividing $\displaystyle \frac{\pi}{4}$ and forgetting $\displaystyle \pi k$ . Then the “solution” $\displaystyle {x}={\frac{\pi}{12}+\pi k}$ loses two solutions of every three.

The number of solutions in an interval

The equation $\displaystyle {\sin2x}={\frac{1}{2}}$ : the period of $\displaystyle \sin2x$ is $\displaystyle \pi$ , so in the interval $\displaystyle 0\leq x\leq2\pi$ it has two full periods – and twice as many solutions as $\displaystyle {\sin x}={\frac{1}{2}}$ .

Mark on the board where $\displaystyle {f(x)}={\sin2x}$ and $\displaystyle {g(x)}={\frac{1}{2}}$ meet in the interval $\displaystyle 0\leq x\leq2\pi$ .

\displaystyle f(x)=\sin2x

\displaystyle g(x)=\frac{1}{2}

Solve the equation in the interval $\displaystyle 0\leq x\leq2\pi$ , and check that the number of solutions is the number of points you marked.

Exercise 1

Solve. Write the general solution.

Exercise 2

Solve in the given interval.

$\displaystyle 0\leq x\leq4\pi$

$\displaystyle 0\leq x\leq\pi$

Exercise 3

Without solving: how many solutions does the equation $\displaystyle {\sin5x}={0.3}$ have in the interval $\displaystyle 0\leq x\leq2\pi$ ? Explain with the period.

/ – 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 10 (5 units)

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The unit circle (radians)

The unit circle
$\displaystyle x^2+y^2=1$
The point of the number $\displaystyle t$ on the unit circle
$\displaystyle P(t)=(\cos t,\ \sin t)$
All the numbers with the same point as $\displaystyle t$ ( $\displaystyle k$ an integer)
$\displaystyle t+2\pi k$
Radians and degrees
$\displaystyle \pi$ $\displaystyle =$ $\displaystyle 180^{\circ}$
The length of an arc of $\displaystyle \alpha$ radians ( $\displaystyle R$ – the radius)
$\displaystyle \ell$ $\displaystyle =$ $\displaystyle \alpha R$
The area of a sector of $\displaystyle \alpha$ radians ( $\displaystyle R$ – the radius)
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}\alpha R^2$

Special values

$\displaystyle \sin\frac{\pi}{6}=\cos\frac{\pi}{3}=\frac{1}{2}$
$\displaystyle \sin\frac{\pi}{4}=\cos\frac{\pi}{4}=\frac{\sqrt{2}}{2}$
$\displaystyle \sin\frac{\pi}{3}=\cos\frac{\pi}{6}=\frac{\sqrt{3}}{2}$
$\displaystyle \tan\frac{\pi}{6}=\frac{\sqrt{3}}{3},\quad\tan\frac{\pi}{4}=1,\quad\tan\frac{\pi}{3}=\sqrt{3}$

Identities

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

Graphs of trigonometric functions

A general sine (and a cosine likewise)
$\displaystyle y=A\sin(ax+b)+B$
The period of $\displaystyle \sin(ax+b)$ and of $\displaystyle \cos(ax+b)$
$\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{2\pi}{|a|}$
The range of $\displaystyle A\sin(ax+b)+B$
$\displaystyle B-|A|\leq y\leq B+|A|$
The period of $\displaystyle \tan(ax+b)$
$\displaystyle T$ $\displaystyle =$ $\displaystyle \frac{\pi}{|a|}$

Trigonometric equations ($k$ an integer)

The solutions of $\displaystyle {\sin x}={\sin\alpha}$
$\displaystyle x=\alpha+2\pi k,\quad x=\pi-\alpha+2\pi k$
The solutions of $\displaystyle {\cos x}={\cos\alpha}$
$\displaystyle x=\pm\alpha+2\pi k$
The solutions of $\displaystyle {\tan x}={\tan\alpha}$
$\displaystyle x=\alpha+\pi k$

Trigonometry in a triangle (degrees)

In a right-angled triangle: $\displaystyle a$ the leg opposite $\displaystyle \alpha$ , $\displaystyle b$ the leg beside it, $\displaystyle c$ the hypotenuse
$\displaystyle \sin\alpha=\frac{a}{c},\quad\cos\alpha=\frac{b}{c},\quad\tan\alpha=\frac{a}{b}$
$\displaystyle \sin(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle \sin\alpha$
$\displaystyle \cos(180^{\circ}-\alpha)$ $\displaystyle =$ $\displaystyle -\cos\alpha$
The area of a triangle ( $\displaystyle \alpha$ – the angle between the sides $\displaystyle b$ and $\displaystyle c$ )
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{1}{2}bc\sin\alpha$
The area of a triangle ( $\displaystyle \beta$ and $\displaystyle \gamma$ – the angles at the side $\displaystyle a$ )
$\displaystyle S$ $\displaystyle =$ $\displaystyle \frac{a^2\sin\beta\sin\gamma}{2\sin(\beta+\gamma)}$
The law of sines ( $\displaystyle R$ – the radius of the circle that circumscribes the triangle)
$\displaystyle \frac{a}{\sin\alpha}=\frac{b}{\sin\beta}=\frac{c}{\sin\gamma}=2R$
The law of cosines ( $\displaystyle \gamma$ – the angle between the sides $\displaystyle a$ and $\displaystyle b$ )
$\displaystyle c^2$ $\displaystyle =$ $\displaystyle a^2+b^2-2ab\cos\gamma$