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10. Trigonometry

Syllabus
0606–2028–2029
Topic
10
Level
—

Learning objectives

10.1Use the six trigonometric functions• Know and use sine, cosine, tangent, secant, cosecant and cotangent for angles of any magnitude.10.2Use amplitude and period• Understand and use the amplitude and period of a trigonometric function.• Understand relationships between graphs of related trigonometric functions, such as y = sin x and y = 3 sin 2x.• The period may be in degrees or radians.10.3Draw and use trigonometric graphs• Draw and use graphs of y = a sin bx + c, y = a cos bx + c and y = a tan bx + c.• Graphs are drawn over a given domain, in degrees or radians.• For y = a tan bx + c, clearly label the x-coordinate of any asymptote.• Use a as a positive integer, b as a simple fraction or integer, and c as an integer; simple fractions have denominator 2, 3, 4, 6 or 8 only.10.4Use trigonometric identities• Use the relationships sin^2 A + cos^2 A = 1, sec^2 A = 1 + tan^2 A and cosec^2 A = 1 + cot^2 A.• Trigonometric identities are given in the List of formulas.10.5Solve trigonometric equations• Solve trigonometric equations over a given domain involving the six trigonometric functions.• Includes use of the relationships in 10.4.10.6Prove trigonometric relationships• Prove trigonometric relationships involving the six trigonometric functions.• Includes use of the relationships in 10.4.

Connect all six trigonometric functions

Sine and cosine locate an angle on the unit circle; the other four functions are ratios or reciprocals of them. This connects all six functions for angles of any magnitude, not only acute angles.

Function In terms of sin⁡θ\sin\theta and cos⁡θ\cos\theta Reciprocal pair
sin⁡θ\sin\theta sin⁡θ\sin\theta cosec⁡θ=1/sin⁡θ\cosec\theta=1/\sin\theta
cos⁡θ\cos\theta cos⁡θ\cos\theta sec⁡θ=1/cos⁡θ\sec\theta=1/\cos\theta
tan⁡θ\tan\theta sin⁡θ/cos⁡θ\sin\theta/\cos\theta cot⁡θ=1/tan⁡θ=cos⁡θ/sin⁡θ\cot\theta=1/\tan\theta=\cos\theta/\sin\theta

Reduce a large or negative angle to a coterminal angle, find its reference angle, then attach the quadrant sign. In quadrant I all six are positive; in II only sine and cosecant are positive; in III only tangent and cotangent are positive; in IV only cosine and secant are positive.

If cos⁡θ=−5/6\cos\theta=-\sqrt{5/6} and θ\theta is in quadrant III, then sin⁡θ=−1−5/6=−1/6\sin\theta=-\sqrt{1-5/6}=-\sqrt{1/6}. Hence tan⁡θ=1/5\tan\theta=1/\sqrt5, sec⁡θ=−6/5\sec\theta=-\sqrt{6/5}, cosec⁡θ=−6\cosec\theta=-\sqrt6 and cot⁡θ=5\cot\theta=\sqrt5. The quadrant fixes signs that a square root alone cannot decide.

A reciprocal is undefined when its denominator is zero: secant fails where cosine is zero, and cosecant and cotangent fail where sine is zero. Keep the calculator angle mode consistent with the question's degrees or radians.

Read amplitude and period from a trig rule

In y=asin⁡(bx)+cy=a\sin(bx)+c or y=acos⁡(bx)+cy=a\cos(bx)+c, ∣a∣|a| is the amplitude, cc is the midline, and bb controls how quickly the cycle repeats. Tangent has no amplitude because it is unbounded.

Family Amplitude Period in radians Period in degrees
asin⁡(bx)+ca\sin(bx)+c ∣a∣|a| 2π/∣b∣2\pi/|b| 360∘/∣b∣360^\circ/|b|
acos⁡(bx)+ca\cos(bx)+c ∣a∣|a| 2π/∣b∣2\pi/|b| 360∘/∣b∣360^\circ/|b|
atan⁡(bx)+ca\tan(bx)+c none π/∣b∣\pi/|b| 180∘/∣b∣180^\circ/|b|

Multiplying the function by aa scales vertical distances from the midline; multiplying xx by bb divides every horizontal feature by ∣b∣|b|; adding cc shifts the entire graph vertically. A negative aa also reflects the graph in its midline.

For y=2cos⁡(x/3)−1y=2\cos(x/3)-1, the amplitude is 22, the midline is y=−1y=-1, and the period is 2π/(1/3)=6π2\pi/(1/3)=6\pi. For y=5tan⁡(x/4)+1y=5\tan(x/4)+1, there is no amplitude and the period is π/(1/4)=4π\pi/(1/4)=4\pi.

Do not use the sine/cosine period rule for tangent, and do not call the vertical multiplier of tangent an amplitude. State whether the horizontal unit is degrees or radians before calculating a period.

Sketch transformed trigonometric graphs

A reliable trig sketch starts from one standard cycle, transforms its horizontal and vertical features, and repeats them only as far as the stated domain requires.

Graph One-cycle anchors in radians Shape feature
y=sin⁡xy=\sin x (0,0),(π/2,1),(π,0),(3π/2,−1),(2π,0)(0,0),(\pi/2,1),(\pi,0),(3\pi/2,-1),(2\pi,0) crosses the midline at multiples of π\pi
y=cos⁡xy=\cos x (0,1),(π/2,0),(π,−1),(3π/2,0),(2π,1)(0,1),(\pi/2,0),(\pi,-1),(3\pi/2,0),(2\pi,1) begins a cycle at a maximum
y=tan⁡xy=\tan x (−π/4,−1),(0,0),(π/4,1)(-\pi/4,-1),(0,0),(\pi/4,1) increasing branch between x=−π/2x=-\pi/2 and x=π/2x=\pi/2

For y=asin⁡(bx)+cy=a\sin(bx)+c or y=acos⁡(bx)+cy=a\cos(bx)+c, draw the midline y=cy=c, calculate the period, mark quarter-period anchors and scale their heights by aa. For y=atan⁡(bx)+cy=a\tan(bx)+c, solve bx=π/2+kπbx=\pi/2+k\pi for every asymptote in the domain, label each xx-coordinate, then place the centre crossing where bx=kπbx=k\pi and y=cy=c.

For y=5tan⁡(x/4)+1y=5\tan(x/4)+1 on −π≤x≤5π-\pi\le x\le5\pi, the period is 4π4\pi. Its asymptotes satisfy x=2π+4kπx=2\pi+4k\pi, so the only one strictly inside this domain is x=2πx=2\pi. The branch crosses its midline at (0,1)(0,1) and (4π,1)(4\pi,1); label the asymptote before drawing the increasing branches.

A curve must approach but never touch a tangent asymptote. Apply the factor bb to the input before locating roots, turning points or asymptotes, and plot only the requested degree or radian domain.

Choose and use a trigonometric identity

The three Pythagorean identities express the same unit-circle relationship in different function families. Choose the identity containing the function you know and the function you need.

\sin^2 A+\cos^2 A=1,\qquad \sec^2 A=1+\tan^2 A,\qquad \cosec^2 A=1+\cot^2 A

The second identity comes from dividing sin⁡2A+cos⁡2A=1\sin^2A+\cos^2A=1 by cos⁡2A\cos^2A; the third comes from dividing by sin⁡2A\sin^2A. Use sec⁡A=1/cos⁡A\sec A=1/\cos A, cosec⁡A=1/sin⁡A\cosec A=1/\sin A, tan⁡A=sin⁡A/cos⁡A\tan A=\sin A/\cos A and cot⁡A=cos⁡A/sin⁡A\cot A=\cos A/\sin A to move between families.

Suppose sec⁡x=α\sec x=\alpha and xx is in quadrant IV. Then cos⁡x=1/α\cos x=1/\alpha and tan⁡2x=α2−1\tan^2x=\alpha^2-1. Tangent is negative in quadrant IV, so tan⁡x=−α2−1\tan x=-\sqrt{\alpha^2-1} and sin⁡x=tan⁡xcos⁡x=−α2−1/α\sin x=\tan x\cos x=-\sqrt{\alpha^2-1}/\alpha.

An identity gives a squared value before it gives a sign. Use the stated domain or quadrant to choose ++ or −-, and note that reciprocal functions require a non-zero denominator.

Solve trigonometric equations over a domain

A trigonometric equation is complete only when every solution in the stated domain has been found and checked. Work with the transformed angle first, then convert back to the original variable.

  1. Rewrite reciprocal functions or use an identity until the equation is in one trig function. 2. Factor or solve the resulting algebraic equation. 3. Translate the original domain into the domain of the transformed angle. 4. Find every reference-angle and quadrant solution in that interval. 5. Convert back, list in order and substitute to reject invalid values.

Solve sec⁡2(3x)+tan⁡(3x)−3=0\sec^2(3x)+\tan(3x)-3=0 for 0∘≤x≤120∘0^\circ\le x\le120^\circ. Since sec⁡2(3x)=1+tan⁡2(3x)\sec^2(3x)=1+\tan^2(3x), let t=tan⁡(3x)t=\tan(3x): t2+t−2=0t^2+t-2=0, so t=1t=1 or t=−2t=-2. The angle 3x3x ranges from 0∘0^\circ to 360∘360^\circ, giving x=15∘,38.9∘,75∘,98.9∘x=15^\circ,38.9^\circ,75^\circ,98.9^\circ (to 11 d.p.).

For secant or cosecant equations, first use the reciprocal relationship and reject values outside the possible sine/cosine range. For cotangent, convert to tangent when convenient. If the equation contains squares, both signs may produce branches that must be tested over the whole domain.

Do not divide by sin⁡x\sin x, cos⁡x\cos x or another expression until checking whether zero makes the original equation true; division can lose solutions. Also exclude angles where the original reciprocal function is undefined.

Prove trigonometric relationships from one side

To prove a trigonometric identity, transform one side until it has exactly the form of the other. Each line must follow from an identity or valid algebra; checking a few angles is not a proof.

Start with the more complicated side. Replace sec⁡\sec, cosec⁡\cosec, tan⁡\tan and cot⁡\cot by sine and cosine when this reduces the number of function types. Then factor, use a common denominator, or create sin⁡2x+cos⁡2x\sin^2x+\cos^2x. Stop as soon as the required expression appears; do not manipulate both sides toward an unproved middle statement.

For cos⁡x1−sin⁡x+1−sin⁡xcos⁡x\dfrac{\cos x}{1-\sin x}+\dfrac{1-\sin x}{\cos x}, rationalise the first fraction: cos⁡x(1+sin⁡x)1−sin⁡2x=1+sin⁡xcos⁡x\dfrac{\cos x(1+\sin x)}{1-\sin^2x}=\dfrac{1+\sin x}{\cos x}. Adding the second fraction gives 1+sin⁡x+1−sin⁡xcos⁡x=2cos⁡x=2sec⁡x\dfrac{1+\sin x+1-\sin x}{\cos x}=\dfrac2{\cos x}=2\sec x.

Look for a structure before expanding: 1−sin⁡2x1-\sin^2x suggests cos⁡2x\cos^2x; sec⁡2x−1\sec^2x-1 suggests tan⁡2x\tan^2x; a sum of fractions suggests a common denominator. The useful identity is the one that removes, rather than creates, complexity.

A proof is valid only where every expression used is defined. Cancelling a factor assumes it is non-zero, so preserve domain restrictions from denominators and reciprocal functions even when the final expression looks defined more widely.