Trigonometry

• One calculator value is never an answer. It is the seed. The reference angle plus the quadrant signs plus the period is the answer.
• If you revisit only one thing: a calculator gives you the reference angle, never the answer. Everything else in this chapter is a consequence of taking that seriously.

• 10.1 — State and use all six trigonometric functions for an angle of any magnitude, give each reciprocal definition in sine/cosine form, and say exactly where each function is undefined.
• 10.1 — Use the unit circle to write down the sign of \(\sin\), \(\cos\) and \(\tan\) in any quadrant, and to turn a reference angle into every angle with the same trigonometric value.
• 10.2 — Read amplitude, centre line, maximum, minimum and period straight off \(y=a\sin bx+c\), \(y=a\cos bx+c\) and \(y=a\tan bx+c\), in degrees or radians.
• 10.3 — Draw a complete labelled cycle of a sine or cosine graph using exact quarter-cycle coordinates.
• 10.3 — Draw a tangent graph with every asymptote's \(x\)-coordinate labelled and every branch drawn separately.
• 10.4 — Quote the three supplied identities, rearrange them fluently, and choose the one that reduces an equation to a single trigonometric function.
• 10.5 — Solve an equation involving any of the six functions over a stated degree or radian domain, finding all the solutions and no extra ones.
• 10.5 — Handle a complete angle such as \(2\theta+30^\circ\) by expanding its domain first, solving there, and only then returning to \(\theta\).
• 10.5 — Reject a substituted root that lies outside a function's range, such as \(\sec x=-\tfrac13\), and explain why.
• 10.6 — Prove a relationship between the six functions by transforming one side only, and state the common domain on which the result holds.

Aim for six out of six with no notes. Anything less than five is a signal to reread the section named beside the question rather than to try the mixed challenge below.

Vertical Asymptote

A vertical asymptote of a tangent graph is a vertical line at an x-value where the function is undefined, and towards which the curve rises or falls without limit while never reaching or crossing it. For y = a tan bx + c the asymptotes occur where the angle bx equals ninety degrees plus any multiple of one hundred and eighty degrees, which in radians is pi over two plus any multiple of pi, so they are spaced exactly one period apart. Each asymptote separates the graph into an independent branch, and a correct sketch never joins one branch to the next.

Reference Angle

The reference angle of an angle theta is the acute angle between the terminal radius of theta and the x-axis. Every angle with the same trigonometric value up to sign shares the same reference angle, so the reference angle supplies the size of a solution while the quadrant supplies its sign. In a domain of zero to three hundred and sixty degrees a single reference angle generates two solutions for each of sine, cosine and tangent, positioned in the two quadrants where that function has the required sign.

Quarter-Cycle Points

Quarter-cycle points are the five equally spaced x-values that determine one complete cycle of a sine or cosine graph: the start of the cycle, the three points a quarter, a half and three quarters of the way through it, and the end. They are spaced one quarter of a period apart, and at them the curve is always at a maximum, a minimum or a centre-line crossing, so plotting them and joining with a smooth curve reproduces the cycle exactly without any further calculation.

Complete Angle

A complete angle is the entire expression that a trigonometric function acts upon, such as two theta plus thirty degrees in the equation sine of two theta plus thirty degrees equals a half. It must be solved for as a single quantity. Because the complete angle changes faster than the variable itself, its domain is wider than the stated domain: substituting the endpoints of the given range into the expression gives the expanded domain in which every solution must be sought, and only after all of those are found is each one converted back to the variable. Solving for the variable first and then trying to add periods is the standard route to a missing solution.

Pythagorean Identities

The three Pythagorean identities supplied in the Additional Mathematics 4037 formula list are sin squared A plus cos squared A equals one, sec squared A equals one plus tan squared A, and cosec squared A equals one plus cot squared A. The second is obtained by dividing the first through by cos squared A and the third by dividing it through by sin squared A, so all three carry the same content. An identity is true for every angle at which both sides are defined, which is why the second requires cos A to be non-zero and the third requires sin A to be non-zero. Their practical use is to replace a squared trigonometric term so that an equation contains only one trigonometric function.

Amplitude and Period

For a sine or cosine graph of the form y = a sin bx + c or y = a cos bx + c, the amplitude is the modulus of a, which is half the vertical distance between the maximum and the minimum, and the period is the horizontal length of one complete cycle, equal to 360 degrees divided by the modulus of b or two pi divided by the modulus of b. The centre line is y = c, the maximum is c plus the modulus of a and the minimum is c minus the modulus of a. A tangent graph has no amplitude because it is unbounded, and its period is 180 degrees divided by the modulus of b, or pi divided by the modulus of b.

Reciprocal Trigonometric Functions

Secant, cosecant and cotangent are the three reciprocal trigonometric functions, defined as sec theta = 1/cos theta, cosec theta = 1/sin theta and cot theta = 1/tan theta = cos theta/sin theta. Each is undefined wherever its denominator is zero, and each carries the same sign as the function it is the reciprocal of. They are reciprocals, not inverse functions: sec theta returns a ratio, whereas the inverse function cos to the power minus one returns an angle.

Complete Solution Set

A complete solution set is every angle in the stated domain that satisfies the equation, with no angle omitted and none included that does not satisfy it or that lies outside the domain. Because trigonometric functions are periodic and many-to-one, a single calculator value is only the reference angle; the complete set is built by combining that reference angle with the quadrants in which the function has the required sign, and then adding or subtracting whole periods until the domain is exhausted. Completeness is the whole requirement, so a correct method that stops at one angle has not answered the question.

Trigonometric Proof

A trigonometric proof establishes that two expressions are equal for every angle at which both are defined. It is carried out by choosing one side, usually the more complicated one, and transforming it through valid algebra and the supplied identities until it becomes the other side. The result being proved must never be used as a starting point, and no step may divide or cancel by an expression that could be zero unless that expression is known to be non-zero on the domain. The proof concludes by stating the common domain, that is the set of angles at which both original sides are defined.

• "\(\sin^{-1}(0.5)=30^\circ\), so \(\theta=30^\circ\)." Why it fails The inverse function is defined to return exactly one value, because a function cannot return several. That single value is the reference angle, not the solution set. Sine is many-to-one, so it takes the value \(0.5\) at infinitely many angles. Fix Treat the calculator output as \(\alpha\). Then apply quadrant signs and periodicity: \(\theta=30^\circ\) and \(180^\circ-30^\circ=150^\circ\) in \(0^\circ\le\theta<360^\circ\).
• "I found the angle in quadrant I, so I am finished." Why it fails Each of \(\sin\), \(\cos\) and \(\tan\) takes every attainable value in exactly two quadrants per revolution. Stopping after one leaves half the solution set unwritten, and the question asks for the set, not for its first entry. Fix Before writing any angle, name the two quadrants from the sign of the value. Then count: how many periods does the domain span? Two solutions per period is the expected total.
• "The domain said \(0\le x\le2\pi\), so I solved it in degrees and converted at the end." Why it fails Converting at the end is legitimate arithmetic, but the calculator was in the wrong mode throughout, so every intermediate value was wrong — and any exact answer such as \(\dfrac{5\pi}{6}\) has been destroyed and replaced by a decimal. Fix Read the domain first and set the mode before touching the equation. A domain containing \(\pi\) is a radian domain and wants radian answers, exact where possible.
• "\(y=\tan 3x\) has period \(\dfrac{360^\circ}{3}=120^\circ\)." Why it fails \(360^\circ\) is the period of sine and cosine. Tangent repeats twice as often, because rotating by \(180^\circ\) negates both \(\sin\theta\) and \(\cos\theta\) and leaves their ratio unchanged. Using the wrong base halves the number of asymptotes and branches. Fix Base \(180^\circ\) for tangent, base \(360^\circ\) for sine and cosine. Here the period is \(\dfrac{180^\circ}{3}=60^\circ\).
• "\(y=4\tan 2x+1\) has amplitude \(4\)." Why it fails Amplitude is defined as half the distance between the maximum and the minimum. A tangent graph has neither, so the definition has nothing to work with. The statement is not merely wrong, it is meaningless. Fix Write "no amplitude — the graph is unbounded". If asked what \(a\) does, say it is a vertical stretch factor.
• "\(y=3\sin 2x-1\) has maximum \(3\)." Why it fails This reads the amplitude as the maximum and ignores \(c\) entirely. The wave oscillates about \(y=-1\), not about \(y=0\), so its maximum is \(-1+3=2\). Fix Always write the centre line down first, then maximum \(=c+|a|\) and minimum \(=c-|a|\). Sanity check: the midpoint of your maximum and minimum must equal \(c\).
• "Period of \(y=\cos 4x\) is \(2\pi\times4=8\pi\)." Why it fails \(b\) produces a horizontal stretch of factor \(\dfrac1b\), which is a compression when \(b>1\). Multiplying asserts the opposite: that squashing the graph makes each cycle longer. Fix Divide: \(\dfrac{2\pi}{4}=\dfrac{\pi}{2}\). Then sanity-check the direction: bigger \(b\), shorter cycle, more cycles on the page.
• "I drew the tangent curve; the dashed lines are just decoration." Why it fails The asymptotes are part of the graph's description, not an ornament: they are where the function is undefined. The syllabus requires each asymptote's \(x\)-coordinate to be labelled, so an unlabelled sketch is incomplete however accurate the curve. Fix Solve \(bx=90^\circ+180^\circ k\) first, draw and label the dashed lines, and only then draw the branches between them.
• A single sweeping curve drawn straight through an asymptote. Why it fails It claims the function is defined and continuous at a point where it is neither. It also usually implies the curve passes through a finite value there, when in fact it runs to \(+\infty\) on one side and returns from \(-\infty\) on the other. Fix Lift the pen at every dashed line. Each branch starts and ends by running alongside an asymptote without touching it.
• Quarter-cycle points plotted by eye, so the second half of the cycle is narrower than the first. Why it fails The four quarters of a period are equal by definition. Unequal spacing changes the period part-way through the sketch, which is a different function. Fix Compute \(\dfrac{\text{period}}{4}\) once, mark that interval repeatedly along the axis with a ruler, and plot only at those marks.
• "\(\sec\theta\) and \(\cos^{-1}\theta\) are two ways of writing the same thing." Why it fails They have different inputs and different outputs. \(\sec\theta\) takes an angle and returns a ratio; \(\cos^{-1}\theta\) takes a ratio and returns an angle. The only connection is a notational accident: \(\cos^2\theta\) means \(\left(\cos\theta\right)^2\), but \(\cos^{-1}\theta\) does not mean \(\left(\cos\theta\right)^{-1}\). Fix Whenever you want \(\left(\cos\theta\right)^{-1}\), write \(\sec\theta\). Reserve the \({}^{-1}\) notation for inverse functions and never mix the two in one line of working.
• "\(\cot 180^\circ=\dfrac{\cos 180^\circ}{\sin 180^\circ}=\dfrac{-1}{0}=-\infty\), so I will use it." Why it fails Division by zero is undefined, not equal to infinity. \(\cot 180^\circ\) simply does not exist, so any working that evaluates or manipulates it is invalid from that line onwards. Fix Before using a reciprocal function, check its denominator on the domain. If a candidate angle makes it zero, that angle is excluded from the solution set and must be stated as such.
• "\(2t^2+t-1=0\) gives \(t=\tfrac12\), so \(\theta=26.6^\circ\)." Why it fails Two things have gone at once: the second root \(t=-1\) has been abandoned, and the surviving root has produced only its quadrant I angle. A quadratic with two usable roots over a \(360^\circ\) domain normally yields four angles; this answer has one of them. Fix Write both trigonometric values on their own line — "\(\tan\theta=\tfrac12\) or \(\tan\theta=-1\)" — before finding any angle at all. Then solve each separately and count the total.
• "\(\sec x=-\tfrac13\), so \(x=\cos^{-1}(-3)\)… the calculator gave an error, so I left it blank." Why it fails The root is genuinely impossible, because \(|\sec x|\ge1\) wherever \(\sec x\) is defined. Leaving it blank throws away the one line that disposes of that half of the question: recognising the root as impossible and rejecting it with a reason. Fix Write one line: "\(\sec x=-\tfrac13\) is impossible since \(|\sec x|\ge1\); reject." Then continue with the other root.
• "\(t=\tfrac12\) and \(t=-1\)" offered as the final answer. Why it fails The question asked for \(\theta\), and \(t\) was a substitution introduced for convenience. Values of \(t\) are working, not an answer. Fix Make the last line of every substitution question a list of angles with the correct unit and accuracy. If the letter in your final line is not the letter in the question, you have not finished.
• Proving an identity by working on the left and the right simultaneously until they meet. Why it fails Both chains start from the statement being proved, so the argument assumes its own conclusion. It also cannot be read as a derivation in either direction, which is what a proof has to be. Fix Choose one side, write "LHS \(=\)", and transform only that until it is literally the other side. Then write "\(=\) RHS".
• "Multiply both sides by \(\sin\theta\), then it is obviously true." Why it fails Multiplying both sides of the target statement uses the statement as a premise. It is the same error as the two-sided approach, wearing different clothes. It also risks multiplying by zero, which turns a false statement into a true one. Fix Do the same manipulation on one side only, by combining that side over a common denominator instead of multiplying the whole statement through.
• Cancelling \(\left(1+\cos\theta\right)\) from a fraction without comment. Why it fails Cancellation is division, and division by zero is not allowed. \(1+\cos\theta=0\) at \(\theta=180^\circ\), so unless that angle is already excluded, the step is invalid there. Fix Justify it in half a line: "valid since \(1+\cos\theta\ne0\) on the domain, because it appears as a denominator on the original left-hand side". That sentence is what makes the cancellation legitimate rather than lucky.
• A completed proof with no mention of where it holds. Why it fails An identity is an equality on a common domain. Without naming that domain the statement is over-claimed: as written it asserts truth at angles where one side does not even exist. Fix Scan both original sides for \(\sec\), \(\operatorname{cosec}\), \(\cot\), \(\tan\) and for any denominator, collect the conditions, and finish with one line such as "valid for \(\cos x\ne0\)".
• "Give exact answers" answered with \(0.524\) and \(2.618\). Why it fails A rounded decimal is by definition not exact. \(0.524\) is an approximation to \(\dfrac{\pi}{6}\), and what was asked for is the surd or the multiple of \(\pi\), not a good decimal. Fix If the reference angle is one of \(30^\circ,45^\circ,60^\circ\) or their radian equivalents, the answer is available exactly — give it as \(\dfrac{\pi}{6}\), \(\dfrac{\sqrt3}{2}\) and so on, and never reach for the calculator at all.

• Both papers are written papers of two hours, each worth 80 marks, and both are set on any part of the content. There is no separate topic paper, so trigonometry can appear anywhere in either. They differ in one decisive way: Paper 1 is a non-calculator paper — calculators are not allowed — while Paper 2 requires a scientific calculator.
• What the non-calculator paper means for this topic. On Paper 1 you cannot evaluate \(\tan^{-1}(0.5)\). Every angle a Topic 10 question asks for there must therefore be reachable exactly, which in practice means it is built from \(0^\circ,30^\circ,45^\circ,60^\circ,90^\circ\) and their radian equivalents, placed in the correct quadrant. So a Paper 1 equation reduces to a value such as \(\sin\theta=-\dfrac{\sqrt3}{2}\), \(\cos\theta=\dfrac12\) or \(\tan\theta=-1\), never to \(\cos x=0.4\). The exact-value table in the prerequisite check is not background reading for Paper 1 — it is the method. Wherever this chapter says "set the calculator to degree mode", read it as Paper 2 advice.
• "Write down the amplitude and period of…" — often the opening part of a longer graph question, and the quickest part of it, provided you never give an amplitude for a tangent.
• What has to be right: the number of complete cycles, the maximum and minimum values, the centre line, and — for tangent — labelled asymptotes with separated branches.
• What has to be right: the choice of identity, the quadratic it produces, each trigonometric value that comes out of it, and then the complete solution set. Extra angles and missing angles are both errors.
• What has to be right: the working stays on one side throughout, and the common domain is stated. Two-sided "manipulation until they meet" is not a proof, whatever it arrives at.