Mathematics guide, Foundations, step 8 of 33Previous step Next step

Step 8: Trigonometric Identities and the Harmonic Form

By the EconTutorials team. 10-minute read with 6 practice questions.

Syllabus9709 P2/P3 2.3, 3.3 · IB SL 3.6, HL 3.9–3.10

In 30 seconds

The reciprocal ratios are $ \sec\theta = \tfrac{1}{\cos\theta} $, $ \operatorname{cosec}\theta = \tfrac{1}{\sin\theta} $ and $ \cot\theta = \tfrac{\cos\theta}{\sin\theta} $. Dividing $ \sin^2\theta + \cos^2\theta = 1 $ by $ \cos^2\theta $ or $ \sin^2\theta $ gives $ 1 + \tan^2\theta = \sec^2\theta $ and $ 1 + \cot^2\theta = \operatorname{cosec}^2\theta $.
$ \sin(A + B) = \sin A\cos B + \cos A\sin B $ and $ \cos(A + B) = \cos A\cos B-\sin A\sin B $. Put $ B = A $ to get the double angle formulae.
$ \cos 2A $ has three forms. Pick the one that leaves a single ratio, and use $ \sin^2 x = \tfrac12(1-\cos 2x) $ to remove a square.
Exam point: $ a\sin\theta + b\cos\theta = R\sin(\theta + \alpha) $ with $ R = \sqrt{a^2 + b^2} $ and $ \tan\alpha = \tfrac{b}{a} $. The maximum is $ R $, not $ a + b $.

Video lesson coming soonA short animated lesson for this step will appear here.

Reciprocal ratios and two more Pythagorean identities

Step 7 gave you sine, cosine, tangent and the identity $ \sin^2\theta + \cos^2\theta = 1 $. This step turns them into a working algebra. You will need it constantly in calculus: the derivative of $ \tan x $ is $ \sec^2 x $, and the only way to integrate $ \sin^2 x $ by hand uses a double angle formula. In Cambridge 9709 this is topic 2.3 (Paper 2) and 3.3 (Paper 3). In IB Analysis and Approaches the double angle formulae are in the SL core (3.6), while the reciprocal ratios and compound angles are HL (3.9 to 3.10).

Each basic ratio has a reciprocal, “one over” it. They add no new information, but they shorten many expressions.

\[ \sec\theta = \frac{1}{\cos\theta} \qquad \operatorname{cosec}\theta = \frac{1}{\sin\theta} \qquad \cot\theta = \frac{\cos\theta}{\sin\theta} \]

A memory hook: look at the third letter. sec goes with cos, cosec with sin, cot with tan. Calculators have no buttons for them, so always work through the parent ratio.

Key termSecant: $ \sec\theta = \dfrac{1}{\cos\theta} $, defined wherever $ \cos\theta \neq 0 $. Cosecant and cotangent are defined in the same way wherever $ \sin\theta \neq 0 $.

Each reciprocal graph is built from its parent. Where the parent is $ \pm1 $, so is the reciprocal, and the graphs touch. Where the parent is small, the reciprocal is large. Where the parent is zero, the reciprocal has a vertical asymptote (Figure 1).

y = sec x and its parent y = cos x

y = cos xy = sec x

Figure 1. Where cos x = ±1, sec x = ±1 and the graphs touch. Where cos x = 0 (x = 90°, 270°) sec x has vertical asymptotes. sec x never lies between −1 and 1.

Now divide every term of $ \sin^2\theta + \cos^2\theta = 1 $ by $ \cos^2\theta $. You get $ \tan^2\theta + 1 = \sec^2\theta $. Divide by $ \sin^2\theta $ instead and you get $ 1 + \cot^2\theta = \operatorname{cosec}^2\theta $. So there is really only one Pythagorean identity, seen through three different ratios.

\[ 1 + \tan^2\theta \equiv \sec^2\theta \qquad\qquad 1 + \cot^2\theta \equiv \operatorname{cosec}^2\theta \]

Worked example

Given that $ \tan\theta = -2 $ and $ 90^\circ < \theta < 180^\circ $, find the exact values of $ \sec\theta $ and $ \operatorname{cosec}\theta $.

Step 1$ \sec^2\theta = 1 + \tan^2\theta = 1 + 4 = 5 $, so $ \sec\theta = \pm\sqrt5 $.

Step 2In the second quadrant cosine is negative, so $ \sec\theta $ is too: $ \sec\theta = -\sqrt5 $ and $ \cos\theta = -\tfrac{1}{\sqrt5} $.

Step 3$ \sin\theta = \tan\theta\cos\theta = (-2)\left(-\tfrac{1}{\sqrt5}\right) = \tfrac{2}{\sqrt5} $, positive as it should be.

Step 4So $ \operatorname{cosec}\theta = \tfrac{\sqrt5}{2} $.

Checking the sign of the ratio you found against CAST is a free way to catch an error.

The identities earn their keep in equations. When an equation mixes $ \tan^2\theta $ with $ \sec\theta $, replace the square so that only one ratio is left, and you have a quadratic.

Worked example

Solve $ 2\tan^2\theta = 3\sec\theta $ for $ 0^\circ \le \theta \le 360^\circ $.

Step 1Replace $ \tan^2\theta $ by $ \sec^2\theta-1 $: $ 2\sec^2\theta-3\sec\theta-2 = 0 $.

Step 2Factorise: $ (2\sec\theta + 1)(\sec\theta-2) = 0 $, so $ \sec\theta = 2 $ or $ \sec\theta = -\tfrac12 $.

Step 3$ \sec\theta = -\tfrac12 $ is impossible, because $ |\sec\theta| \ge 1 $ always.

Step 4$ \sec\theta = 2 $ means $ \cos\theta = \tfrac12 $, so $ \theta = 60^\circ $ or $ 300^\circ $.

Figure 1 shows why the second root fails: the secant graph never enters the band between $ -1 $ and 1.

Common mistakeWriting $ \tan^2\theta = \sec^2\theta + 1 $. Check with $ \theta = 0 $: $ \tan 0 = 0 $ and $ \sec 0 = 1 $, so it must be $ 1 + \tan^2\theta = \sec^2\theta $. The “1 +” goes with the ratio that can be zero.

Compound angle formulae

If you know the sine and cosine of $ A $ and of $ B $, what is $ \sin(A + B) $? The tempting answer, $ \sin A + \sin B $, is wrong. Try $ A = B = 45^\circ $: $ \sin 90^\circ = 1 $, but $ \sin 45^\circ + \sin 45^\circ = \sqrt2 $. The true answer mixes sines and cosines.

\[ \sin(A \pm B) \equiv \sin A\cos B \pm \cos A\sin B \] \[ \cos(A \pm B) \equiv \cos A\cos B \mp \sin A\sin B \] \[ \tan(A \pm B) \equiv \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B} \]

Top signs go together and bottom signs go together. Notice that cosine swaps the sign: $ \cos(A + B) $ has a minus. The sine formula comes from reaching the point at angle $ A + B $ on the unit circle in two steps, first along a line at angle $ A $, then at right angles to it, and adding up the heights. Replacing $ B $ by $ -B $ gives the difference formulae, and dividing sine by cosine gives the tangent one.

Worked example

(a) Find the exact value of $ \cos 105^\circ $. (b) Given $ \tan A = 3 $ and $ \tan B = \tfrac12 $, find $ \tan(A-B) $.

Step 1(a) $ 105^\circ = 60^\circ + 45^\circ $, so $ \cos 105^\circ = \cos 60^\circ\cos 45^\circ-\sin 60^\circ\sin 45^\circ $.

Step 2$ = \tfrac12 \cdot \tfrac{\sqrt2}{2}-\tfrac{\sqrt3}{2} \cdot \tfrac{\sqrt2}{2} = \tfrac{\sqrt2-\sqrt6}{4} $.

Step 3(b) $ \tan(A-B) = \dfrac{3-\tfrac12}{1 + 3 \cdot \tfrac12} = \dfrac{5/2}{5/2} = 1 $.

The answer to (a) is negative, as it must be: $ 105^\circ $ is in the second quadrant, where cosine is negative.

Common mistakeSplitting a function across a bracket. $ \sin(A + B) \neq \sin A + \sin B $, $ \cos 2x \neq 2\cos x $ and $ \tan(x + 45^\circ) \neq \tan x + 1 $. A sine is a function of the bracket, not a number multiplying it. When unsure, test a value.

Double angle formulae

Put $ B = A $ in the compound angle formulae and you get formulae for twice an angle.

\[ \sin 2A \equiv 2\sin A\cos A \qquad \tan 2A \equiv \frac{2\tan A}{1-\tan^2 A} \] \[ \cos 2A \equiv \cos^2 A-\sin^2 A \] \[ \equiv 2\cos^2 A-1 \equiv 1-2\sin^2 A \]

Why does $ \cos 2A $ have three forms? From $ \cos(A + A) $ you get $ \cos^2 A-\sin^2 A $, which contains both ratios. Using $ \sin^2 A + \cos^2 A = 1 $ you can swap out either one. That is a choice, not three facts to learn: pick the form that matches the rest of the problem so only one ratio is left.

Worked example

Given $ \tan A = \tfrac34 $ with $ 180^\circ < A < 270^\circ $, find the exact values of $ \sin 2A $, $ \cos 2A $ and $ \tan 2A $.

Step 1A 3, 4, 5 triangle gives the sizes. In the third quadrant sine and cosine are both negative: $ \sin A = -\tfrac35 $, $ \cos A = -\tfrac45 $.

Step 2$ \sin 2A = 2\left(-\tfrac35\right)\left(-\tfrac45\right) = \tfrac{24}{25} $.

Step 3$ \cos 2A = \cos^2 A-\sin^2 A = \tfrac{16}{25}-\tfrac{9}{25} = \tfrac{7}{25} $.

Step 4$ \tan 2A = \dfrac{24/25}{7/25} = \tfrac{24}{7} $. Check: $ \dfrac{2 \cdot \tfrac34}{1-\tfrac{9}{16}} = \dfrac{3/2}{7/16} = \tfrac{24}{7} $.

Working the tangent two ways is a quick check on the signs.

To solve an equation with a double angle, rewrite it so every term uses the single angle.

Worked example

Solve $ \cos 2x + 3\cos x + 2 = 0 $ for $ 0^\circ \le x \le 360^\circ $.

Step 1The other term is in cosine, so use $ \cos 2x = 2\cos^2 x-1 $: $ 2\cos^2 x + 3\cos x + 1 = 0 $.

Step 2Factorise: $ (2\cos x + 1)(\cos x + 1) = 0 $.

Step 3$ \cos x = -\tfrac12 $: $ x = 120^\circ $ or $ 240^\circ $. $ \cos x = -1 $: $ x = 180^\circ $.

So $ x = 120^\circ, 180^\circ, 240^\circ $. Choosing $ 1-2\sin^2 x $ here would have left a mix of sine and cosine.

Rearranging the last two forms of $ \cos 2x $ trades a square for a double angle. These power-reducing forms are what make $ \int \sin^2 x\,dx $ possible later in the course.

\[ \cos^2 x \equiv \tfrac12(1 + \cos 2x) \] \[ \sin^2 x \equiv \tfrac12(1-\cos 2x) \]

y = sin²x = ½ − ½ cos 2x

y = sin²x

Figure 2. Squaring the sine wave gives a wave of period 180° that rocks between 0 and 1 about y = ½. That is exactly what ½ − ½ cos 2x describes.

An economics example: why revenue has a half-year rhythm. Suppose a seasonal good has price $ P = 8 + 2\sin(30t)^\circ $ dollars and sales $ Q = 50 + 10\sin(30t)^\circ $ units, with $ t $ in months, so both peak together. Revenue is $ PQ = 400 + 180\sin(30t)^\circ + 20\sin^2(30t)^\circ $. Using $ \sin^2\theta = \tfrac12(1-\cos 2\theta) $, this becomes $ PQ = 410 + 180\sin(30t)^\circ-10\cos(60t)^\circ $. Average revenue over a year is \$410, not $ 8 \times 50 = \$400 $, because high prices coincide with high sales. And a six-month cycle appears, even though nothing in the market has a six-month rhythm: multiplying two cycles doubles the angle.

Proving identities

A question that says “prove that … $ \equiv $ …” has no $ x $ to find. You must turn one side into the other through steps that are each clearly true.

The method: start with one side, usually the more complicated one. Rewrite everything in sine and cosine, use the double angle formulae, combine fractions and factorise. Stop when you reach the other side exactly.

Worked example

Prove that (a) $ \dfrac{1-\cos 2x}{\sin 2x} \equiv \tan x $ and (b) $ \dfrac{1}{1-\sin x} + \dfrac{1}{1 + \sin x} \equiv 2\sec^2 x $.

Step 1(a) The “1 −” tells you which form to use: $ 1-\cos 2x = 1-(1-2\sin^2 x) = 2\sin^2 x $.

Step 2LHS $ = \dfrac{2\sin^2 x}{2\sin x\cos x} = \dfrac{\sin x}{\cos x} = \tan x = $ RHS.

Step 3(b) Combine over a common denominator: LHS $ = \dfrac{(1 + \sin x) + (1-\sin x)}{(1-\sin x)(1 + \sin x)} = \dfrac{2}{1-\sin^2 x} $.

Step 4$ = \dfrac{2}{\cos^2 x} = 2\sec^2 x = $ RHS.

Each line follows from the one before. That chain is what earns the marks.

Exam tipNever work on both sides at once or cross-multiply; that assumes what you are trying to prove and loses the final mark. Write “LHS =” at the start and “= RHS” at the end. If you get stuck, you may simplify each side separately to the same expression, as long as you say so. And choose the form of $ \cos 2x $ that cancels the constant: $ 1 + \cos 2x = 2\cos^2 x $, $ 1-\cos 2x = 2\sin^2 x $.

The harmonic form $ R\sin(\theta \pm \alpha) $

What does $ 5\sin x + 12\cos x $ look like? Both parts have period $ 360^\circ $ but peak in different places. Their sum is a single regular wave, taller than either part but only 13 high, not 17 (Figure 3).

Here is why. Write the coefficients as a point $ (5, 12) $. Its distance from the origin is $ R = \sqrt{5^2 + 12^2} = 13 $, and its angle $ \alpha $ satisfies $ 5 = R\cos\alpha $ and $ 12 = R\sin\alpha $. Then $ 5\sin x + 12\cos x = R(\sin x\cos\alpha + \cos x\sin\alpha) = R\sin(x + \alpha) $, by the compound angle formula read backwards. The coefficients only set the height and the shift; the shape is still a sine wave.

Key termHarmonic form: an expression $ a\sin\theta + b\cos\theta $ rewritten as $ R\sin(\theta + \alpha) $ (or with $ \cos $, or with $ \theta-\alpha $), where $ R = \sqrt{a^2 + b^2} > 0 $. $ R $ is the amplitude of the combined wave.

Worked example

Express $ 5\sin x + 12\cos x $ as $ R\sin(x + \alpha) $, with $ R > 0 $ and $ 0^\circ < \alpha < 90^\circ $. Hence find its maximum value, and solve $ 5\sin x + 12\cos x = 6.5 $ for $ 0^\circ \le x \le 360^\circ $.

Step 1Expand: $ R\sin(x + \alpha) = R\sin x\cos\alpha + R\cos x\sin\alpha $. Match: $ R\cos\alpha = 5 $, $ R\sin\alpha = 12 $.

Step 2Square and add: $ R^2 = 25 + 144 = 169 $, so $ R = 13 $. Divide: $ \tan\alpha = \tfrac{12}{5} $, so $ \alpha = 67.38^\circ $.

Step 3The maximum is 13, when $ x + 67.38^\circ = 90^\circ $, at $ x = 22.62^\circ $.

Step 4$ 13\sin(x + 67.38^\circ) = 6.5 $ gives $ \sin(x + 67.38^\circ) = 0.5 $, with $ x + 67.38^\circ $ between $ 67.38^\circ $ and $ 427.38^\circ $.

Step 5So $ x + 67.38^\circ = 150^\circ $ or $ 390^\circ $, giving $ x = 82.62^\circ $ or $ 322.62^\circ $.

Check at $ x = 0 $: $ 13\sin 67.38^\circ = 12 $, which matches $ 5\sin 0 + 12\cos 0 = 12 $.

5 sin x + 12 cos x is one wave of height 13

5 sin x12 cos x13 sin(x + 67.38°)

Figure 3. Adding 5 sin x (gold) and 12 cos x (teal) gives a single sine wave (indigo) with amplitude √(5² + 12²) = 13, not 5 + 12 = 17, because the two parts do not peak together.

An economics example: two cycles, one wave. A resort’s seasonal bookings have a holiday cycle $ 15\sin(30t)^\circ $ and a conference cycle $ 8\cos(30t)^\circ $, in hundreds, with $ t $ in months from the start of January. The combined swing is $ 17\sin(30t + 28.07)^\circ $, since $ R = \sqrt{15^2 + 8^2} = 17 $ and $ \tan\alpha = \tfrac{8}{15} $. It peaks at 1,700 extra bookings, not 2,300, because the cycles do not peak together, and the peak comes when $ 30t + 28.07 = 90 $, at $ t \approx 2.06 $: early March, about a month before the holiday cycle alone would peak. Analysts combine overlapping seasonal patterns in exactly this way.

Exam tipQuestions often end “hence find the greatest value of $ \dfrac{26}{20-5\sin x-12\cos x} $”. The denominator is $ 20-13\sin(x + \alpha) $, which is least when the sine is 1, at $ 20-13 = 7 $. A fraction with a positive denominator is greatest when the denominator is least, so the answer is $ \tfrac{26}{7} $. The harmonic form is a Cambridge 9709 topic; IB does not list it by name, though HL students can derive it from the compound angle formulae.

Practise: 6 questions

Try each one on paper first, then check the worked answer.

EasyQ1. Find the exact values of $ \sec 135^\circ $ and $ \cot 120^\circ $. Show worked answer

$ \cos 135^\circ = -\tfrac{\sqrt2}{2} $, so $ \sec 135^\circ = -\tfrac{2}{\sqrt2} = -\sqrt2 $.
$ \tan 120^\circ = -\sqrt3 $, so $ \cot 120^\circ = -\tfrac{1}{\sqrt3} = -\tfrac{\sqrt3}{3} $.

EasyQ2. Use $ 15^\circ = 45^\circ-30^\circ $ to find the exact value of $ \sin 15^\circ $. Show worked answer

$ \sin 15^\circ = \sin 45^\circ\cos 30^\circ-\cos 45^\circ\sin 30^\circ $
$ = \tfrac{\sqrt2}{2} \cdot \tfrac{\sqrt3}{2}-\tfrac{\sqrt2}{2} \cdot \tfrac12 = \tfrac{\sqrt6-\sqrt2}{4} $.

MediumQ3. Solve $ 3\operatorname{cosec}^2\theta-5\cot\theta-5 = 0 $ for $ 0^\circ \le \theta \le 360^\circ $, to 1 decimal place. Show worked answer

Use $ \operatorname{cosec}^2\theta = 1 + \cot^2\theta $: $ 3 + 3\cot^2\theta-5\cot\theta-5 = 0 $, so $ 3\cot^2\theta-5\cot\theta-2 = 0 $.
$ (3\cot\theta + 1)(\cot\theta-2) = 0 $.
$ \cot\theta = 2 $: $ \tan\theta = \tfrac12 $, so $ \theta = 26.6^\circ $ or $ 206.6^\circ $.
$ \cot\theta = -\tfrac13 $: $ \tan\theta = -3 $, so $ \theta = 108.4^\circ $ or $ 288.4^\circ $.

MediumQ4. Solve $ \cos 2x = 5\sin x-2 $ for $ 0 \le x \le 2\pi $. Show worked answer

Use $ \cos 2x = 1-2\sin^2 x $: $ 1-2\sin^2 x = 5\sin x-2 $.
So $ 2\sin^2 x + 5\sin x-3 = 0 $, which factorises as $ (2\sin x-1)(\sin x + 3) = 0 $.
$ \sin x = -3 $ is impossible, so $ \sin x = \tfrac12 $: $ x = \tfrac{\pi}{6} $ or $ \tfrac{5\pi}{6} $.

MediumQ5. Prove that $ \tan x + \cot x \equiv 2\operatorname{cosec} 2x $. Show worked answer

LHS $ = \frac{\sin x}{\cos x} + \frac{\cos x}{\sin x} = \frac{\sin^2 x + \cos^2 x}{\sin x\cos x} $
$ = \frac{1}{\sin x\cos x} = \frac{2}{2\sin x\cos x} = \frac{2}{\sin 2x} = 2\operatorname{cosec} 2x = $ RHS.

ExamQ6. (a) Express $ 4\cos\theta-3\sin\theta $ in the form $ R\cos(\theta + \alpha) $, where $ R > 0 $ and $ 0^\circ < \alpha < 90^\circ $, giving $ \alpha $ to 2 decimal places. [3] (b) Hence solve $ 4\cos\theta-3\sin\theta = 2 $ for $ 0^\circ < \theta < 360^\circ $. [3] (c) State the greatest value of $ \dfrac{1}{7-4\cos\theta + 3\sin\theta} $. [2] Show worked answer

(a) $ R\cos(\theta + \alpha) = R\cos\theta\cos\alpha-R\sin\theta\sin\alpha $, so $ R\cos\alpha = 4 $, $ R\sin\alpha = 3 $ [M1].
$ R = \sqrt{16 + 9} = 5 $ [A1]; $ \tan\alpha = \tfrac34 $, $ \alpha = 36.87^\circ $ [A1].
(b) $ \cos(\theta + 36.87^\circ) = 0.4 $ [M1], with $ 36.87^\circ < \theta + 36.87^\circ < 396.87^\circ $.
$ \theta + 36.87^\circ = 66.42^\circ $ or $ 293.58^\circ $ [A1], so $ \theta = 29.55^\circ $ or $ 256.71^\circ $ [A1].
(c) The denominator is $ 7-5\cos(\theta + 36.87^\circ) $, least value $ 7-5 = 2 $ [M1], so the greatest value is $ \tfrac12 $ [A1].
Total: 8 marks.

Check yourself

1. Which of these is an identity?

Dividing $ \sin^2\theta + \cos^2\theta = 1 $ by $ \sin^2\theta $ gives $ 1 + \cot^2\theta = \operatorname{cosec}^2\theta $. The first has the sign wrong; the second should be $ 2\sin\theta\cos\theta $.

2. Which form of $ \cos 2A $ contains only sine?

Replace $ \cos^2 A $ in $ \cos^2 A-\sin^2 A $ by $ 1-\sin^2 A $ to get $ 1-2\sin^2 A $. Test with $ A = 0 $: both sides equal 1.

3. What is the maximum value of $ 6\sin x + 8\cos x $?

$ R = \sqrt{6^2 + 8^2} = 10 $. The two parts do not peak at the same $ x $, so the maximum is less than $ 6 + 8 $.

Your score: 0 out of 3

The short version

$ \sec $, $ \operatorname{cosec} $ and $ \cot $ are one over $ \cos $, $ \sin $ and $ \tan $; their graphs have asymptotes where the parent is zero.
$ 1 + \tan^2\theta = \sec^2\theta $ and $ 1 + \cot^2\theta = \operatorname{cosec}^2\theta $ turn mixed equations into a quadratic in one ratio.
Compound angle formulae mix sines and cosines; cosine swaps the sign. Setting $ B = A $ gives $ \sin 2A $, $ \tan 2A $ and the three forms of $ \cos 2A $.
Prove identities in one chain from one side to the other, never working on both sides at once.
$ a\sin\theta + b\cos\theta = R\sin(\theta + \alpha) $ with $ R = \sqrt{a^2 + b^2} $: one wave whose maximum is $ R $.
The full chapter in Mathematics Made Simple has many more worked examples and a full set of practice questions with a complete answer key.

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Next: step 9 of 33Limits from Graphs, Tables and Algebra

Step 8: Trigonometric Identities and the Harmonic Form

Reciprocal ratios and two more Pythagorean identities

Compound angle formulae

Double angle formulae

Proving identities

The harmonic form \( R\sin(\theta \pm \alpha) \)

Practise: 6 questions

Check yourself

The short version

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