Trig Graphs & Equations | Sec 3 & 4 A Maths Guide

Secondary 3 & 4 Additional Mathematics · Study guide

Read periodic graphs and find every permitted solution, keeping angle units, transformed intervals and excluded values precise.

Concept → worked examples → practice32 min read · practice linked below

The angle unit and interval are part of the question

This shared Secondary 3 and 4 guide covers general angles, trigonometric graphs, equations on specified intervals and periodic models. Begin by recording whether the angles are in degrees or radians and whether each endpoint is included. A calculator's inverse-trigonometric answer is one principal angle, not usually the full solution.

One full turn is $360^\circ=2\pi$ radians, so $180^\circ=\pi$ radians. Multiply degrees by $\pi/180$ to obtain radians, and radians by $180/\pi$ to obtain degrees. For a point $(x,y)$ on a circle of radius $r>0$ , $\cos\theta=x/r$ , $\sin\theta=y/r$ and $\tan\theta=y/x$ when $x\ne0$ . This explains the signs in each quadrant and extends the ratios beyond acute angles.

Exact values and signs
Angle	Sine	Cosine	Tangent
$0^\circ=0$	0	1	0
$30^\circ=\pi/6$	$1/2$	$\sqrt3/2$	$1/\sqrt3$
$45^\circ=\pi/4$	$1/\sqrt2$	$1/\sqrt2$	1
$60^\circ=\pi/3$	$\sqrt3/2$	$1/2$	$\sqrt3$
$90^\circ=\pi/2$	1	0	Undefined

In quadrants I, II, III, IV the positive basic ratios are respectively all three, sine, tangent and cosine. Sine is odd: $\sin(-x)=-\sin x$ ; cosine is even: $\cos(-x)=\cos x$ . Also $\sin(180^\circ-x)=\sin x$ and $\cos(180^\circ-x)=-\cos x$ . The same rules use π in place of 180° in radians. Sine and cosine repeat after a full turn; tangent repeats after half a turn.

The reciprocal functions are $\sec x=1/\cos x$ , $\operatorname{cosec}x=1/\sin x$ , and $\cot x=\cos x/\sin x$ . Respect their denominators when solving. The identities guide explains these domains and the identities used below.

Inverse functions return principal values

Calculator principal ranges
Function	Input	Principal output
$\sin^{-1}u$	$-1\le u\le1$	$[-90^\circ,90^\circ]$ , or $[-\pi/2,\pi/2]$
$\cos^{-1}u$	$-1\le u\le1$	$[0^\circ,180^\circ]$ , or $[0,\pi]$
$\tan^{-1}u$	Any real u	$(-90^\circ,90^\circ)$ , or $(-\pi/2,\pi/2)$

Example 1: principal values are not unrestricted cancellation

Evaluate $\sin210^\circ$ , $\cos(-\pi/3)$ , $\sin^{-1}(\sin150^\circ)$ in degrees, and $\cos^{-1}(\cos(7\pi/6))$ in radians.

210° has reference angle 30° in quadrant III, so $\sin210^\circ=-1/2$ . Cosine is even, so $\cos(-\pi/3)=1/2$ .

Since $\sin150^\circ=1/2$ , its inverse sine is $30^\circ$ , the value within $[-90^\circ,90^\circ]$ . Since $\cos(7\pi/6)=-\sqrt3/2$ , its inverse cosine is $5\pi/6$ , within $[0,\pi]$ . Neither inverse simply returns the original angle outside its principal range.

Read amplitude, period, midline and asymptotes

For $y=a\sin(bx)+c$ or $y=a\cos(bx)+c$ , with $a\ne0$ and $b>0$ , the amplitude is $|a|$ , the midline is $y=c$ , and the range is $[c-|a|,c+|a|]$ . A negative a reflects the wave in its midline. The period is $360^\circ/b$ or $2\pi/b$ : increasing x by that amount increases the input angle by one full turn.

For $y=a\sin(x/b)+c$ and $y=a\cos(x/b)+c$ , the period is instead $360^\circ b$ or $2\pi b$ . Dividing the angle stretches the graph horizontally. If $a=0$ , the graph is the constant $y=c$ ; it has no smallest positive period.

For $y=a\tan(bx)$ , with $a\ne0$ , the range is all real numbers and there is no amplitude. The period is $180^\circ/b$ or $\pi/b$ . Vertical asymptotes occur where $bx=90^\circ+180^\circ n$ , or $bx=\pi/2+n\pi$ , for integer n. They are excluded inputs, never parts of the curve. Sketch each branch separately.

Example 2: sketch from the input angle

Describe $y=-2\cos(3x)+1$ , with x in degrees, and compare it with $y=2\sin(x/3)-1$ .

The first graph has amplitude 2, period 120°, midline $y=1$ , range $[-1,3]$ . One cycle has points $(0^\circ,-1),(30^\circ,1),(60^\circ,3),(90^\circ,1),(120^\circ,-1)$ , obtained from cosine input angles 0°, 90°, 180°, 270°, 360°. Repeat this pattern to cover any longer interval.

The second graph has amplitude 2, period 1080°, midline $y=-1$ , and range $[-3,1]$ over all real inputs. Its first maximum is at $x/3=90^\circ$ , hence $x=270^\circ$ . Multiplying the angle by 3 and dividing it by 3 have opposite effects on the period.

Example 3: tangent branches and a half-turn period

Describe $y=2\tan(2x)$ , with x in degrees, and solve $\tan(2x)=1$ for $0^\circ\le x<180^\circ$ .

The graph has period $180^\circ/2=90^\circ$ , with no amplitude. In the displayed interval, the asymptotes occur when $2x=90^\circ$ or $270^\circ$ , hence at $x=45^\circ,135^\circ$ . The zeros occur at $x=0^\circ,90^\circ$ , and also at the excluded right endpoint 180°.

To solve the equation, set $u=2x$ , so $0^\circ\le u<360^\circ$ . Tangent equals 1 at $u=45^\circ,225^\circ$ . Therefore $x=22.5^\circ,112.5^\circ$ . On the graph these are the intersections with $y=2$ . Neither is an asymptote.

The dashed vertical lines are excluded inputs. Each solid branch is drawn separately; the curve never crosses an asymptote.

Explore: a horizontal level meets a sine graph

Move the level k in $\sin x=k$ on the closed interval $0^\circ\le x\le360^\circ$ . Count solutions at k = 0 and at the turning levels ±1. The two endpoints are different x-values even though their sine values agree.

Level k:0.5

Use the slider or arrow keys. Levels outside −1 to 1 have no intersections.

A horizontal line meets the graph at every permitted solution angle.

k = 0.5. Two solutions: x = 30° and 150°.

Solve within the interval, including its boundaries

Isolate one trigonometric ratio, and reject impossible values such as $\sin x=2$ .
Find a reference angle, then use the sign and quadrants to list all angles in the specified interval.
Check the original equation and any undefined values. Include or exclude endpoints exactly as stated.

Example 4: negative angles and tangent's shorter period

Solve $\sin x=-1/2$ for $-180^\circ<x\le180^\circ$ .

The reference angle is 30°. Sine is negative below the horizontal axis. In this interval the angles are $-150^\circ$ and $-30^\circ$ . Both give sine $-1/2$ , and both are inside the stated interval. Neither endpoint solves the equation.

For comparison, solve $\tan x=-1$ on $-90^\circ<x<270^\circ$ . Its principal angle is −45°. Tangent repeats every 180°, so the permitted answers are $-45^\circ$ and $135^\circ$ . The next repeat, 315°, is outside. The angle 90° inside the interval is undefined and cannot be a solution.

Example 5: a multiple angle in radians

Solve $\cos2x=1/2$ for $0\le x<2\pi$ .

Set $u=2x$ . Its interval is $0\le u<4\pi$ , so it contains two full cycles. In that interval, cosine is one half at

u=\frac\pi3,\ \frac{5\pi}3,\ \frac{7\pi}3,\ \frac{11\pi}3.

Divide every value by 2:

x=\frac\pi6,\ \frac{5\pi}6,\ \frac{7\pi}6,\ \frac{11\pi}6.

All four lie in the original half-open interval. Solving only on $[0,2\pi)$ for the doubled angle would lose two solutions.

Example 6: shift the whole interval before solving

Solve $\sin(2x-30^\circ)=1/2$ for $0^\circ\le x\le180^\circ$ .

Let $u=2x-30^\circ$ . Then $-30^\circ\le u\le330^\circ$ . Within it, $u=30^\circ$ or $150^\circ$ . Hence $x=(u+30^\circ)/2$ , giving $x=30^\circ$ or $90^\circ$ .

Substitution gives input angles 30° and 150°, as required. Adding another full turn to either input angle would put it outside the transformed interval.

Use algebra and identities without losing solutions

A quadratic in a ratio can be solved by treating that ratio as one variable. When several ratios occur, use an identity to express them in terms of one ratio. Factor a product instead of dividing by a factor that might be zero. For fractions, state the denominator restrictions before multiplying through.

Example 7: a quadratic in cosine

Solve $2\cos^2x+\cos x-1=0$ for $0^\circ\le x<360^\circ$ .

(2\cos x-1)(\cos x+1)=0.

Thus $\cos x=1/2$ or $\cos x=-1$ . The first gives $60^\circ,300^\circ$ ; the second gives $180^\circ$ . The complete answer is $x=60^\circ,180^\circ,300^\circ$ . Substitute either ratio value into the original polynomial to verify zero.

Example 8: a zero factor must be retained

Solve $\sin2x=\cos x$ for $0^\circ\le x<360^\circ$ .

Use $\sin2x=2\sin x\cos x$ , then factor:

\cos x(2\sin x-1)=0.

The factor $\cos x=0$ gives $90^\circ,270^\circ$ . The factor $\sin x=1/2$ gives $30^\circ,150^\circ$ . Thus $x=30^\circ,90^\circ,150^\circ,270^\circ$ . Dividing immediately by $\cos x$ would discard two valid answers. At both of those angles the original equation is $0=0$ .

Example 9: a reciprocal equation with an impossible branch

Solve $2\cot^2x=1-5/\sin x$ for $0^\circ\le x<360^\circ$ .

The domain requires $\sin x\ne0$ . Use $\cot^2x=1/\sin^2x-1$ , then multiply by the nonzero $\sin^2x$ :

2-2\sin^2x=\sin^2x-5\sin x,\qquad 3\sin^2x-5\sin x-2=0.

(3\sin x+1)(\sin x-2)=0.

Reject $\sin x=2$ . For $\sin x=-1/3$ , the reference angle is $\beta=\sin^{-1}(1/3)\approx19.4712^\circ$ . Therefore $x=180^\circ+\beta$ or $360^\circ-\beta$ , giving $x\approx199.5^\circ,340.5^\circ$ .

The retained sine is nonzero. Independently, $\cot^2x=(1-1/9)/(1/9)=8$ , so both sides of the original equation equal 16.

Example 10: solve after using the R-formula

Solve $3\sin x+4\cos x=1$ for $0^\circ\le x<360^\circ$ .

Write the left side as $5\sin(x+\alpha)$ , where $\alpha=\tan^{-1}(4/3)\approx53.1301^\circ$ . Set $u=x+\alpha$ , so $\alpha\le u<360^\circ+\alpha$ . We need $\sin u=1/5$ .

Let $\beta=\sin^{-1}(1/5)\approx11.5370^\circ$ . Within the transformed interval, the valid angles are $180^\circ-\beta$ and $360^\circ+\beta$ . Subtract α to obtain

x=180^\circ-\beta-\alpha\approx115.3^\circ,\quad x=360^\circ+\beta-\alpha\approx318.4^\circ.

Keep α and β unrounded until the final answers. The principal angle β alone is outside the transformed interval, but β plus a full turn is inside it.

Build and interpret a periodic model

For a sinusoidal model, the midline is the average of the largest and smallest values; the amplitude is half their difference. Choose sine or cosine from the starting position and direction of motion. A model that starts at a maximum naturally uses positive cosine. The coefficient of time follows from the period, with its angle unit stated explicitly.

Example 11: form a model and interpret a threshold

A modelled water depth varies smoothly between 1 m and 5 m, with a 12-hour period and a maximum at $t=0$ . Form a cosine model and find when the depth is at least 4 m during $0\le t\le12$ .

The midline is $(5+1)/2=3$ and amplitude is $(5-1)/2=2$ . In radians, $2\pi/b=12$ , so $b=\pi/6$ . Hence

H(t)=3+2\cos\left(\frac{\pi t}{6}\right)\quad\text{metres}.

The threshold is $\cos(\pi t/6)\ge1/2$ . On $0\le\pi t/6\le2\pi$ , this occurs in $[0,\pi/3]$ and $[5\pi/3,2\pi]$ . Therefore $0\le t\le2$ or $10\le t\le12$ , in hours.

The endpoints are included because the question says “at least”. At t = 2 and 10 the depth is exactly 4 m; at t = 0 and 12 it is 5 m. The total time is 4 hours. This conclusion applies to the stated sinusoidal model and time interval.

Common mistakes and useful checks

Mixing angle units. A radian answer needs π-based periods and a calculator in radian mode.
Stopping at an inverse-function value. Use quadrants and the whole interval to find every solution.
Keeping the old interval for 2x or x + α. Transform both endpoints before listing angles.
Forgetting that tangent is undefined at its asymptotes. Never draw a curve across an asymptote or accept it as a solution.
Dividing out a zero factor. Factor and solve each branch, checking domains afterwards.
Changing a strict endpoint into a closed one. A solution at 360° belongs to $[0^\circ,360^\circ]$ , but not to $[0^\circ,360^\circ)$ .

Practise equations and graphs

Use the Secondary 3 trigonometric equations and graphs questions for general angles, graphs and basic equations. The Secondary 4 trigonometric equations and graphs questions add later methods and mixed revision. Review identities and R-formula when an equation needs restructuring first.