Theory

Trigonometric equations and inequalities

Sine, cosine and tangent on the unit circle, the identities that relate them, and the solution methods for every type of trigonometric equation and inequality.

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The unit circle
The identities

The unit circle

1. Unit circle, sine and cosine

The unit circle is the circle with center at the origin and radius $1$ . An angle $x$ , measured in radians from the positive half of the horizontal axis in the counterclockwise direction, determines a point $P$ on it, and one sets

cos x = abscissa of P, sin x = ordinate of P .

Since $P$ lies on the circle of radius $1$ , we have $- 1 \leq sin x \leq 1$ and $- 1 \leq cos x \leq 1$ . Adding $2 π$ to the angle does not change the point $P$ , therefore sine and cosine are periodic with period $2 π$ :

sin (x + 2 π) = sin x, cos (x + 2 π) = cos x .

Because of this periodicity, a trigonometric equation that has one solution has infinitely many, and the solutions are written as families depending on an integer.

2. Tangent

tan x = \frac{sin x}{cos x}, for cos x \neq = 0, that is x \neq = \frac{π}{2} + nπ, n \in Z .

The tangent is periodic with period $π$ : adding $π$ to the angle changes the sign of both sine and cosine, therefore their ratio does not change. On the unit circle, $tan x$ is the ordinate of the point where the line through the origin and $P$ intersects the vertical line of equation $x = 1$ .

3. Special values

$x$	$0$	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{π}{2}$	$π$	$\frac{3 π}{2}$
$sin x$	$0$	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$	$1$	$0$	$- 1$
$cos x$	$1$	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$	$0$	$- 1$	$0$
$tan x$	$0$	$\frac{3}{3}$	$1$	$3$	undefined	$0$	undefined

These values, together with the symmetries of related angles, cover all the special angles of the circle. The solutions of the trigonometric equations in the exercises are almost always expressed with angles from this table.

The identities

1. Pythagorean identity

For every $x \in R$ ,

sin^{2} x + cos^{2} x = 1.

It is the Pythagorean theorem applied to the triangle with legs $∣ cos x ∣$ and $∣ sin x ∣$ and hypotenuse the radius $1$ . It is used to pass from one function to the other: given $sin x$ , the value of $cos x$ is determined up to sign, and the sign is read from the quadrant in which the angle lies. In equations it allows one to write $sin^{2} x = 1 - cos^{2} x$ , and with this substitution an equation containing both functions reduces to one containing a single function.

Example

If $sin x = \frac{3}{5}$ and $x$ lies in the second quadrant, then $cos^{2} x = 1 - \frac{9}{25} = \frac{16}{25}$ . In the second quadrant the cosine is negative, therefore $cos x = - \frac{4}{5}$ .

2. Related angles

For every $x \in R$ :

$sin (- x) = - sin x$ and $cos (- x) = cos x$ ;
$sin (π - x) = sin x$ and $cos (π - x) = - cos x$ ;
$sin (π + x) = - sin x$ and $cos (π + x) = - cos x$ ;
$sin (\frac{π}{2} - x) = cos x$ and $cos (\frac{π}{2} - x) = sin x$ .

They follow from the symmetries of the unit circle with respect to the axes and to the bisector of the first quadrant. They make it possible to reduce every angle to the first quadrant and to turn a cosine into a sine, for example in equations of the form $sin A = cos B$ . From the second line it follows that $cos \frac{5 π}{6} = cos (π - \frac{π}{6}) = - cos \frac{π}{6} = - \frac{3}{2}$ .

3. Addition and double-angle formulas

For every $a, b \in R$ :

sin (a \pm b) = sin a cos b \pm cos a sin b, cos (a \pm b) = cos a cos b \mp sin a sin b .

Setting $b = a$ in the formulas with the plus sign gives the double-angle formulas:

sin 2 a = 2 sin a cos a, cos 2 a = cos^{2} a - sin^{2} a = 1 - 2 sin^{2} a = 2 cos^{2} a - 1.

The two alternative forms of $cos 2 a$ follow from the Pythagorean identity and are used when an equation contains $cos 2 x$ together with $sin x$ or with $cos x$ : the right form turns it into a quadratic equation in a single function.

Example

With the subtraction formula for the cosine:

cos \frac{π}{12} = cos (\frac{π}{3} - \frac{π}{4}) = cos \frac{π}{3} cos \frac{π}{4} + sin \frac{π}{3} sin \frac{π}{4} = \frac{1}{2} \cdot \frac{2}{2} + \frac{3}{2} \cdot \frac{2}{2} = \frac{2 + 6}{4} .

The theory continues with 6 more sections (the graphs of sine and cosine, elementary equations, equations reducible to elementary ones…), plus all worked examples, the solutions and the PDF.

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