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SPI Mathematics · Trigonometric Ratios and Identities

Trigonometric Identities

An identity is an equation that holds for every angle. Three of them, all from Pythagoras, turn long expressions into a single number or a single ratio.

All Trigonometric Ratios and Identities notes

Why this matters

Fourteen questions in the bank, nine EASY and five MODERATE. Four rest on sin²θ + cos²θ = 1, five on its sec and tan or cosec and cot forms, two simplify a square root, and three use an identity as a second equation or to remove the angle. Five of the chapter's nine MODERATE questions are on this page.

Concept 1 of 4: sin²θ + cos²θ = 1 and its rearrangements

In a right triangle, opp2+adj2=hyp2\text{opp}^2 + \text{adj}^2 = \text{hyp}^2. Divide every term by hyp2\text{hyp}^2 and you get sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1. It is true for every angle, so whenever you see sin⁡2\sin^2 and cos⁡2\cos^2 of the same angle together, look for this.

Definition

  • sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1.
  • Rearranged: 1−sin⁡2θ=cos⁡2θ1 - \sin^2\theta = \cos^2\theta and 1−cos⁡2θ=sin⁡2θ1 - \cos^2\theta = \sin^2\theta.
  • Factor first: (1+sin⁡θ)(1−sin⁡θ)=1−sin⁡2θ=cos⁡2θ(1 + \sin\theta)(1 - \sin\theta) = 1 - \sin^2\theta = \cos^2\theta.
  • With unequal coefficients, split off a common part: 5sin⁡2θ+2cos⁡2θ=2(sin⁡2θ+cos⁡2θ)+3sin⁡2θ=2+3sin⁡2θ5\sin^2\theta + 2\cos^2\theta = 2(\sin^2\theta + \cos^2\theta) + 3\sin^2\theta = 2 + 3\sin^2\theta.

The first identity

sin⁡2θ+cos⁡2θ=1,1−sin⁡2θ=cos⁡2θ,1−cos⁡2θ=sin⁡2θ\sin^2\theta + \cos^2\theta = 1, \qquad 1 - \sin^2\theta = \cos^2\theta, \qquad 1 - \cos^2\theta = \sin^2\theta

Worked example

Simplify cos⁡4θ−sin⁡4θ+sin⁡2θ\cos^4\theta - \sin^4\theta + \sin^2\theta.
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Example 1 · Trigonometric Ratios and Identities · Trigonometric Identities

3sin⁡2θ+4cos⁡2θ3\sin^2\theta + 4\cos^2\theta is equal to:

The identity needs the same angle

sin⁡230∘+cos⁡260∘=14+14=12\sin^2 30^\circ + \cos^2 60^\circ = \frac{1}{4} + \frac{1}{4} = \frac{1}{2}, not 1. The two angles must be equal for the sum to be 1.

The leftover goes with the bigger coefficient

5sin⁡2θ+2cos⁡2θ5\sin^2\theta + 2\cos^2\theta is 2+3sin⁡2θ2 + 3\sin^2\theta, not 2+3cos⁡2θ2 + 3\cos^2\theta. Take out the smaller coefficient and the remainder stays on the ratio that had more.
Drill 3 more on sin²θ + cos²θ = 1 and its rearrangements

Concept 2 of 4: 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ

Divide sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 by cos⁡2θ\cos^2\theta and you get tan⁡2θ+1=sec⁡2θ\tan^2\theta + 1 = \sec^2\theta. Divide it by sin⁡2θ\sin^2\theta instead and you get 1+cot⁡2θ=cosec⁡2θ1 + \cot^2\theta = \operatorname{cosec}^2\theta. So sec goes with tan, and cosec goes with cot.

Definition

  • sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, so tan⁡2θ−sec⁡2θ=−1\tan^2\theta - \sec^2\theta = -1.
  • cosec⁡2θ−cot⁡2θ=1\operatorname{cosec}^2\theta - \cot^2\theta = 1.
  • 1cos⁡2θ\frac{1}{\cos^2\theta} is sec⁡2θ\sec^2\theta and 1sin⁡2θ\frac{1}{\sin^2\theta} is cosec⁡2θ\operatorname{cosec}^2\theta: rewrite them first.
  • From one ratio: if cos⁡θ\cos\theta is given, sec⁡2θ=1cos⁡2θ\sec^2\theta = \frac{1}{\cos^2\theta} and tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1.

The second and third identities

1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=cosec⁡2θ1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \operatorname{cosec}^2\theta

Worked example

If tan⁡θ=12\tan\theta = \frac{1}{2}, find sec⁡2θ+cosec⁡2θ\sec^2\theta + \operatorname{cosec}^2\theta.
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Example 2 · Trigonometric Ratios and Identities · Trigonometric Identities

If cos⁡θ=23\cos\theta = \frac{2}{3}, then 2sec⁡2θ+2tan⁡2θ−72\sec^2\theta + 2\tan^2\theta - 7 is equal to

Getting the sign the wrong way round

sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1, so tan⁡2θ−sec⁡2θ=−1\tan^2\theta - \sec^2\theta = -1. Check which term comes first before you write 1.

Pairing sec with cot

The pairs are sec with tan and cosec with cot. sec⁡2θ−cot⁡2θ\sec^2\theta - \cot^2\theta is not 1 for a general angle.
Drill 4 more on 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ

Concept 3 of 4: Simplifying a square root of a ratio

A root like 1+cos⁡A1−cos⁡A\sqrt{\dfrac{1 + \cos A}{1 - \cos A}} clears when the bottom becomes a perfect square. Multiply top and bottom inside the root by the top: the bottom turns into 1−cos⁡2A=sin⁡2A1 - \cos^2 A = \sin^2 A, and the root comes off both.

Definition

  • Multiply inside the root by 1+cos⁡A1+cos⁡A\dfrac{1 + \cos A}{1 + \cos A} (the top over itself).
  • The bottom becomes a squared ratio by the identities: sin⁡2A\sin^2 A, cos⁡2A\cos^2 A, tan⁡2A\tan^2 A.
  • For an acute angle every ratio is positive, so sin⁡2A=sin⁡A\sqrt{\sin^2 A} = \sin A.
  • Check by numbers: put A=60∘A = 60^\circ into the question and each option, and keep the one that matches.

Roots that clear for an acute angle A

1+cos⁡A1−cos⁡A=cosec⁡A+cot⁡A1−cos⁡A1+cos⁡A=cosec⁡A−cot⁡A1−sin⁡2A=cos⁡Asec⁡2A−1=tan⁡A\sqrt{\dfrac{1 + \cos A}{1 - \cos A}} = \operatorname{cosec} A + \cot A \qquad \sqrt{\dfrac{1 - \cos A}{1 + \cos A}} = \operatorname{cosec} A - \cot A \qquad \sqrt{1 - \sin^2 A} = \cos A \qquad \sqrt{\sec^2 A - 1} = \tan A

Worked example

Simplify 1+cos⁡A1−cos⁡A\sqrt{\dfrac{1 + \cos A}{1 - \cos A}} for an acute angle A.
Practice this conceptself-check · 3 quick reps

The same idea in a practice question:

Practice questionModerate

Example 3 · Trigonometric Ratios and Identities · Trigonometric Identities

1−sin⁡A1+sin⁡A=?\sqrt{\frac{1 - \sin A}{1 + \sin A}} = ?

Taking the root of each term separately

1−sin⁡2A\sqrt{1 - \sin^2 A} is cos⁡A\cos A, not 1−sin⁡A1 - \sin A. A root never splits across a plus or minus sign.

Plus and minus look alike in the options

The answer and a wrong option often differ only in one sign, such as cosec⁡A+cot⁡A\operatorname{cosec} A + \cot A and cosec⁡A−cot⁡A\operatorname{cosec} A - \cot A. Put in A=60∘A = 60^\circ to tell them apart.
Drill 1 more on simplifying a square root of a ratio

Concept 4 of 4: Using an identity as a second equation

sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1 factors as (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1. So if you know one bracket, the other is its reciprocal, and two equations give both ratios. The same identities remove the angle when x and y are given as ratios of it.

Definition

  • sec⁡θ+tan⁡θ\sec\theta + \tan\theta and sec⁡θ−tan⁡θ\sec\theta - \tan\theta multiply to 1. So do cosec⁡θ+cot⁡θ\operatorname{cosec}\theta + \cot\theta and cosec⁡θ−cot⁡θ\operatorname{cosec}\theta - \cot\theta.
  • Add the two brackets to get twice sec (or cosec); subtract them to get twice tan (or cot).
  • Removing the angle: if x=asec⁡θx = a\sec\theta and y=atan⁡θy = a\tan\theta, square both and subtract: x2−y2=a2x^2 - y^2 = a^2.

The factored identities

(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1,(cosec⁡θ+cot⁡θ)(cosec⁡θ−cot⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1, \qquad (\operatorname{cosec}\theta + \cot\theta)(\operatorname{cosec}\theta - \cot\theta) = 1

Worked example

If sec⁡θ+tan⁡θ=3\sec\theta + \tan\theta = 3, find sec⁡θ\sec\theta and tan⁡θ\tan\theta.
Practice this conceptself-check · 4 quick reps

The same idea in a practice question:

Practice questionModerate

Example 4 · Trigonometric Ratios and Identities · Trigonometric Identities

If cosec A -cot A = 4/5, then cosec A =

Trying to find the angle first

When sec⁡θ+tan⁡θ=3\sec\theta + \tan\theta = 3, the angle is not a standard one. Use the reciprocal bracket and add; never hunt for θ\theta itself.

Adding when you need tan

Adding the two brackets cancels tan and gives 2sec⁡θ2\sec\theta. Subtracting cancels sec and gives 2tan⁡θ2\tan\theta. Choose the one that keeps the ratio you are asked for.
Drill 2 more on using an identity as a second equation

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 (4)

  • sin²θ + cos²θ = 1 and its rearrangements

    The first identity

    sin⁡2θ+cos⁡2θ=1,1−sin⁡2θ=cos⁡2θ,1−cos⁡2θ=sin⁡2θ\sin^2\theta + \cos^2\theta = 1, \qquad 1 - \sin^2\theta = \cos^2\theta, \qquad 1 - \cos^2\theta = \sin^2\theta
  • 1 + tan²θ = sec²θ and 1 + cot²θ = cosec²θ

    The second and third identities

    1+tan⁡2θ=sec⁡2θ,1+cot⁡2θ=cosec⁡2θ1 + \tan^2\theta = \sec^2\theta, \qquad 1 + \cot^2\theta = \operatorname{cosec}^2\theta
  • Simplifying a square root of a ratio

    Roots that clear for an acute angle A

    1+cos⁡A1−cos⁡A=cosec⁡A+cot⁡A1−cos⁡A1+cos⁡A=cosec⁡A−cot⁡A1−sin⁡2A=cos⁡Asec⁡2A−1=tan⁡A\sqrt{\dfrac{1 + \cos A}{1 - \cos A}} = \operatorname{cosec} A + \cot A \qquad \sqrt{\dfrac{1 - \cos A}{1 + \cos A}} = \operatorname{cosec} A - \cot A \qquad \sqrt{1 - \sin^2 A} = \cos A \qquad \sqrt{\sec^2 A - 1} = \tan A
  • Using an identity as a second equation

    The factored identities

    (sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1,(cosec⁡θ+cot⁡θ)(cosec⁡θ−cot⁡θ)=1(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1, \qquad (\operatorname{cosec}\theta + \cot\theta)(\operatorname{cosec}\theta - \cot\theta) = 1

Watch out for (8)

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