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CDS Mathematics · Trigonometric Ratios and Identities

Simplifying & Proving Identities

Three Pythagorean identities, and the habit of rewriting everything in sine and cosine, reduce every 'what is this equal to?' expression to a number or a single ratio.

All Trigonometric Ratios and Identities notesCDS Mathematics strategy

Why this matters

Twenty-four PYQs, almost all MODERATE: an expression to simplify, or three statements with 'which are identities?'. None needs a trick the three identities do not supply — what costs marks is algebra done in the wrong order, and a statement that holds at 45° but nowhere else.

Concept 1 of 3: The three Pythagorean identities

Everything starts from sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, which is Pythagoras on a triangle with hypotenuse 11. Divide it by cos⁡2θ\cos^2\theta and you get the secant–tangent identity; divide by sin⁡2θ\sin^2\theta and you get the cosecant–cotangent one. Three identities, one fact.

Definition

  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
  • sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, i.e. 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta
  • cosec⁡2θ−cot⁡2θ=1\operatorname{cosec}^2\theta - \cot^2\theta = 1, i.e. 1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \operatorname{cosec}^2\theta

Use them in every rearranged form: 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta, sec⁡2θ−1=tan⁡2θ\sec^2\theta - 1 = \tan^2\theta, and so on. Products like (1+cos⁡θ)(1−cos⁡θ)(1 + \cos\theta)(1 - \cos\theta) are sin⁡2θ\sin^2\theta in disguise.

Pythagorean identities

sin⁡2θ+cos⁡2θ=1,1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=cosec⁡2θ\sin^2\theta + \cos^2\theta = 1, \quad 1 + \tan^2\theta = \sec^2\theta, \quad 1 + \cot^2\theta = \operatorname{cosec}^2\theta

Worked example

Simplify (1+tan⁡2θ)(1−sin⁡θ)(1+sin⁡θ)(1 + \tan^2\theta)(1 - \sin\theta)(1 + \sin\theta).
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2023 · CDS (I) 2023 — Elementary Mathematics · Q52Easy

Example 1 · Trigonometric Ratios and Identities · Simplifying and Proving Identities

What is the value of (1+cot⁡2θ)(1+cos⁡θ)(1−cos⁡θ)−(1+tan⁡2θ)(1+sin⁡θ)(1−sin⁡θ)(1 + \cot^2\theta)(1 + \cos\theta)(1 - \cos\theta) - (1 + \tan^2\theta)(1 + \sin\theta)(1 - \sin\theta) ?

Watch the sign in the rearranged form

1−cosec⁡2θ1 - \operatorname{cosec}^2\theta is −cot⁡2θ-\cot^2\theta, not cot⁡2θ\cot^2\theta. A statement like (sec⁡2θ−1)(1−cosec⁡2θ)=1(\sec^2\theta - 1)(1 - \operatorname{cosec}^2\theta) = 1 is false only because of that sign — which is exactly why it is on the paper.
Drill 7 more on the three pythagorean identities

Concept 2 of 3: Rewrite in sine and cosine, then factor

When an expression mixes six different ratios, turn every one into sine and cosine. The expression becomes ordinary algebra in two letters, and the familiar factorisations — difference of squares, sum and difference of cubes — do the rest. The answer choices are usually 00, 11, 22 or a single ratio.

Definition

  • Replace tan⁡\tan, cot⁡\cot, sec⁡\sec, cosec⁡\operatorname{cosec} by quotients of sin⁡\sin and cos⁡\cos; put fractions over one denominator.
  • Difference of squares: cos⁡4A−sin⁡4A=(cos⁡2A−sin⁡2A)(cos⁡2A+sin⁡2A)=cos⁡2A−sin⁡2A\cos^4 A - \sin^4 A = (\cos^2 A - \sin^2 A)(\cos^2 A + \sin^2 A) = \cos^2 A - \sin^2 A.
  • Cubes: sin⁡3θ±cos⁡3θ=(sin⁡θ±cos⁡θ)(1∓sin⁡θcos⁡θ)\sin^3\theta \pm \cos^3\theta = (\sin\theta \pm \cos\theta)(1 \mp \sin\theta\cos\theta).
  • A common factor like 2sin⁡2θ−12\sin^2\theta - 1 often cancels between top and bottom — look for it before expanding.

The factorisations that recur

sin⁡3θ±cos⁡3θ=(sin⁡θ±cos⁡θ)(1∓sin⁡θcos⁡θ)\sin^3\theta \pm \cos^3\theta = (\sin\theta \pm \cos\theta)(1 \mp \sin\theta\cos\theta)

Worked example

Simplify sin⁡3θ−cos⁡3θsin⁡θ−cos⁡θ−sin⁡θcos⁡θ\dfrac{\sin^3\theta - \cos^3\theta}{\sin\theta - \cos\theta} - \sin\theta\cos\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (II) 2021 — Elementary Mathematics · Q48Moderate

Example 2 · Trigonometric Ratios and Identities · Simplifying and Proving Identities

What is sin⁡3θ+cos⁡3θsin⁡θ+cos⁡θ+sin⁡3θ−cos⁡3θsin⁡θ−cos⁡θ\dfrac{\sin^3\theta+\cos^3\theta}{\sin\theta+\cos\theta}+\dfrac{\sin^3\theta-\cos^3\theta}{\sin\theta-\cos\theta} equal to ?

Which function you eliminate decides the sign

sin⁡4θ−cos⁡4θ=sin⁡2θ−cos⁡2θ\sin^4\theta - \cos^4\theta = \sin^2\theta - \cos^2\theta, which is 1−2cos⁡2θ1 - 2\cos^2\theta or 2sin⁡2θ−12\sin^2\theta - 1. The option 1−2sin⁡2θ1 - 2\sin^2\theta is the negative of the right answer, and it is always offered.
Drill 8 more on rewrite in sine and cosine, then factor

Concept 3 of 3: Deciding whether a statement is an identity

An identity holds for every allowed angle; an equation holds for some. The fastest test is to try one angle where both sides are easy — not 45∘45^\circ, where sine and cosine are equal and many false statements happen to hold. If the two sides differ at one angle, the statement is not an identity; if they agree, simplify to be sure.

Definition

To judge a statement 'L=RL = R':

  • Disprove with one angle: try 30∘30^\circ or 60∘60^\circ (not 45∘45^\circ). Different values mean it is not an identity.
  • Prove by simplifying one side into the other, or both into a common form.
  • A statement that reduces to something like cos⁡2θ=sin⁡2θ\cos 2\theta = \sin 2\theta holds at one angle only, so it is an equation, not an identity.

Useful forms: sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta (a minus sign), and 1−tan⁡2θ1+tan⁡2θ=cos⁡2θ−sin⁡2θ\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \cos^2\theta - \sin^2\theta.

The test

L(θ0)≠R(θ0) for one θ0  ⇒  not an identityL(\theta_0) \ne R(\theta_0) \text{ for one } \theta_0 \;\Rightarrow\; \text{not an identity}

Worked example

Is 1−tan⁡2θ1+tan⁡2θ=sin⁡2θ−cos⁡2θ\dfrac{1 - \tan^2\theta}{1 + \tan^2\theta} = \sin^2\theta - \cos^2\theta an identity?
Practice this conceptself-check · 4 quick reps

The same idea in a real exam question:

CDS · 2021 · CDS (II) 2021 — Elementary Mathematics · Q53Moderate

Example 3 · Trigonometric Ratios and Identities · Simplifying and Proving Identities

Consider the following : 1. sin⁡4θ−sin⁡2θ=cos⁡4θ−cos⁡2θ\sin^4\theta-\sin^2\theta=\cos^4\theta-\cos^2\theta 2. sin⁡4θ+cos⁡4θ=1+2sin⁡2θcos⁡2θ\sin^4\theta+\cos^4\theta=1+2\sin^2\theta\cos^2\theta 3. tan⁡4θ+tan⁡2θ=sec⁡4θ−sec⁡2θ\tan^4\theta+\tan^2\theta=\sec^4\theta-\sec^2\theta Which of the above are identities ?

Testing at 45° proves nothing

At 45∘45^\circ, sin⁡θ=cos⁡θ\sin\theta = \cos\theta and tan⁡θ=cot⁡θ\tan\theta = \cot\theta, so any statement with the two swapped passes. Test at 30∘30^\circ or 60∘60^\circ instead.
Drill 6 more on deciding whether a statement is an identity

Summary — formulas & gotchas at a glance

A revision cheat-sheet for the formulas and gotchas above. Click any concept name to jump back to its full explanation.

Formulas (3)

  • The three Pythagorean identities

    Pythagorean identities

    sin⁡2θ+cos⁡2θ=1,1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=cosec⁡2θ\sin^2\theta + \cos^2\theta = 1, \quad 1 + \tan^2\theta = \sec^2\theta, \quad 1 + \cot^2\theta = \operatorname{cosec}^2\theta
  • Rewrite in sine and cosine, then factor

    The factorisations that recur

    sin⁡3θ±cos⁡3θ=(sin⁡θ±cos⁡θ)(1∓sin⁡θcos⁡θ)\sin^3\theta \pm \cos^3\theta = (\sin\theta \pm \cos\theta)(1 \mp \sin\theta\cos\theta)
  • Deciding whether a statement is an identity

    The test

    L(θ0)≠R(θ0) for one θ0  ⇒  not an identityL(\theta_0) \ne R(\theta_0) \text{ for one } \theta_0 \;\Rightarrow\; \text{not an identity}

Watch out for (3)

Test yourself on Trigonometric Ratios and Identities

20 past CDS questions from this chapter, timed at 24 minutes and marked the way the exam marks it. You see your score and every answer the moment you finish. Free to start.

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