Trigonometric equations solved with identities

Trigonometry, grade 10 (5 units) Lesson 13 of 21Learning

Trigonometric equations solved with identities

Identity and equation

An identity holds for every $\displaystyle x$ where the expressions are defined: $\displaystyle {\sin^2x+\cos^2x}={1}$ , $\displaystyle {\tan x}={\frac{\sin x}{\cos x}}$ , $\displaystyle {\sin(\pi-x)}={\sin x}$ .
An equation holds only for some numbers, and you solve it to find them.

This lesson uses the identities to turn an equation into one or more basic equations: $\displaystyle {\sin u}={a}$ , $\displaystyle {\cos u}={b}$ , $\displaystyle {\tan u}={c}$ .

Sine and cosine: dividing by the cosine

The equation $\displaystyle {\sin x}={\cos x}$ : if $\displaystyle {\cos x}={0}$ , then $\displaystyle {\sin x}={\pm1}$ , and the equality does not hold. So dividing by $\displaystyle \cos x$ is allowed:

$\displaystyle {\frac{\sin x}{\cos x}}={1}$ , that is $\displaystyle {\tan x}={1}$ ;
$\displaystyle {x}={\frac{\pi}{4}+\pi k}$ .

In this way every equation of the form $\displaystyle {a\sin x}={b\cos x}$ becomes an equation in the tangent.

A quadratic equation in sine

The example below solves $\displaystyle {2\cos^2x+3\sin x-3}={0}$ . It has both sine and cosine, so first go over to one function.

$\displaystyle {\cos^2x}={1-\sin^2x}$ : $\displaystyle {\ 2(1-\sin^2x)+3\sin x-3}={0}$ , that is $\displaystyle {2\sin^2x-3\sin x+1}={0}$ .
Substitute $\displaystyle {t}={\sin x}$ : $\displaystyle {2t^2-3t+1}={0}$ , so $\displaystyle {t}={1}$ or $\displaystyle {t}={\frac{1}{2}}$ .
$\displaystyle {\sin x}={1}$ : $\displaystyle {x}={\frac{\pi}{2}+2\pi k}$ .
$\displaystyle {\sin x}={\frac{1}{2}}$ : $\displaystyle {x}={\frac{\pi}{6}+2\pi k}$ or $\displaystyle {x}={\frac{5\pi}{6}+2\pi k}$ .

Every value of $\displaystyle t$ is checked: a $\displaystyle t$ outside $\displaystyle -1\leq t\leq1$ gives no solution.

Factoring, not dividing

The equation $\displaystyle {\sin x\cos x}={\sin x}$ : dividing by $\displaystyle \sin x$ gives $\displaystyle {\cos x}={1}$ , and loses all the numbers where $\displaystyle {\sin x}={0}$ . The right way – move everything to one side and factor:

$\displaystyle {\sin x\cos x-\sin x}={0}$ , that is $\displaystyle {\sin x(\cos x-1)}={0}$ ;
$\displaystyle {\sin x}={0}$ – $\displaystyle {x}={\pi k}$ ; or $\displaystyle {\cos x}={1}$ – $\displaystyle {x}={2\pi k}$ , which are already among the solutions of $\displaystyle {\sin x}={0}$ .

The solution: $\displaystyle {x}={\pi k}$ .

The same sine, the same cosine

From the general solution:

$\displaystyle {\sin f}={\sin g}$ when $\displaystyle {f}={g+2\pi k}$ or $\displaystyle {f}={\pi-g+2\pi k}$ ;
$\displaystyle {\cos f}={\cos g}$ when $\displaystyle {f}={g+2\pi k}$ or $\displaystyle {f}={-g+2\pi k}$ .

For example, $\displaystyle {\sin3x}={\sin x}$ :

$\displaystyle {3x}={x+2\pi k}$ – $\displaystyle {x}={\pi k}$ ;
$\displaystyle {3x}={\pi-x+2\pi k}$ – $\displaystyle {4x}={\pi+2\pi k}$ , $\displaystyle {x}={\frac{\pi}{4}+\frac{\pi k}{2}}$ .

A sine against a cosine becomes two sines through $\displaystyle {\cos u}={\sin\left(\frac{\pi}{2}-u\right)}$ .

Exercise 1

Solve. Write the general solution.

Exercise 2

Solve by factoring.

$\displaystyle 0\leq x\leq2\pi$

Exercise 3

Solve. Write the general solution.

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 $\displaystyle b$ )
$\displaystyle c^2$ $\displaystyle =$ $\displaystyle a^2+b^2-2ab\cos\gamma$