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CDS Mathematics · Formula sheet

Trigonometric Ratios and Identities formulas

32 formulas, 1 reference table and 34 common traps for CDS Mathematics Trigonometric Ratios and Identities, grouped by subtopic.

Full notes with worked examples

Degree, Radian & Standard Values

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Degree and radian measure, and arc length

Degree–radian conversion and arc length

π rad=180∘,s=rθ  (θ in radians)\pi \text{ rad} = 180^\circ, \qquad s = r\theta \;(\theta \text{ in radians})

What values a ratio can take, and its sign

Ranges

∣sin⁡θ∣,∣cos⁡θ∣≤1,∣sec⁡θ∣,∣cosec⁡θ∣≥1|\sin\theta|, |\cos\theta| \le 1, \qquad |\sec\theta|, |\operatorname{cosec}\theta| \ge 1

Ratios of the standard angles

Anglesincostan
0∘0^\circ001100
30∘30^\circ12\dfrac{1}{2}32\dfrac{\sqrt3}{2}13\dfrac{1}{\sqrt3}
45∘45^\circ12\dfrac{1}{\sqrt2}12\dfrac{1}{\sqrt2}11
60∘60^\circ32\dfrac{\sqrt3}{2}12\dfrac{1}{2}3\sqrt3
90∘90^\circ1100not defined
tan⁡90∘\tan 90^\circ and sec⁡90∘\sec 90^\circ are not defined — they are not 0 and not 1.
Sine climbs as n/2\sqrt{n}/2, cosine is the same list reversed, and tangent is their quotient.

Common traps

Arc length needs the angle in radians

s=rθs = r\theta is only true with θ\theta in radians. Putting 6060 (degrees) into it gives 360360 cm for a 66 cm circle — an arc longer than the whole circumference. Convert first.

One radian is not a small angle

Students read 1c1^c as 'about one degree'. It is about 57.3∘57.3^\circ, so sin⁡1c≈0.84\sin 1^c \approx 0.84 while sin⁡1∘≈0.017\sin 1^\circ \approx 0.017. Every comparison between the two turns on this.

Rationalise before comparing

2−12+1\dfrac{\sqrt2 - 1}{\sqrt2 + 1} and 3−223 - 2\sqrt2 are the same number. When two expressions in a question look different, rationalise both before deciding they differ.

A larger cosine means a smaller angle

On 0∘0^\circ to 90∘90^\circ the cosine falls as the angle rises. So cos⁡θ<cos⁡ϕ\cos\theta < \cos\phi means θ>ϕ\theta > \phi. Reading it the way sine behaves reverses the answer.

Ratios in a Right Triangle

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Sine, cosine and tangent as quotients of sides

The three primary ratios

sin⁡θ=opphyp,cos⁡θ=adjhyp,tan⁡θ=oppadj\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}

From one given ratio to every other ratio

Third side from a Pythagorean pair

sin⁡θ=m2−n2m2+n2  ⇒  cos⁡θ=2mnm2+n2(θ acute)\sin\theta = \frac{m^2-n^2}{m^2+n^2} \;\Rightarrow\; \cos\theta = \frac{2mn}{m^2+n^2} \quad (\theta \text{ acute})

Right triangles hidden in other figures

Two tools that recur

chord=2rsin⁡θ,Area=12absin⁡C\text{chord} = 2r\sin\theta, \qquad \text{Area} = \tfrac12 ab\sin C

Common traps

Name the sides from the angle asked about

The side opposite AA is adjacent to BB. Questions that ask about the other acute angle, or that label the right angle at CC instead of BB, are built to catch a triangle labelled once and read the wrong way.

The triangle gives sizes; the quadrant gives signs

A triangle only ever produces positive lengths. If the angle is in the second quadrant, the cosine you read off the triangle must be made negative before you use it.

The area formula can hide an obtuse angle

12absin⁡C\dfrac12 ab\sin C fixes sin⁡C\sin C, and sin⁡C=23\sin C = \dfrac23 fits both an acute and an obtuse CC. A question that asks for cos⁡C\cos C from the area is quietly assuming the acute one; if both signs are offered, the question cannot decide.

Complementary Angles

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The co-function rule: sin(90° − θ) = cos θ

Co-function rule

sin⁡(90∘−θ)=cos⁡θ,tan⁡(90∘−θ)=cot⁡θ,sec⁡(90∘−θ)=cosec⁡θ\sin(90^\circ-\theta) = \cos\theta, \quad \tan(90^\circ-\theta) = \cot\theta, \quad \sec(90^\circ-\theta) = \operatorname{cosec}\theta

Pairing the terms of a long product or sum

Pairs that collapse

tan⁡θ tan⁡(90∘−θ)=1,sin⁡2θ+sin⁡2(90∘−θ)=1\tan\theta\,\tan(90^\circ-\theta) = 1, \qquad \sin^2\theta + \sin^2(90^\circ-\theta) = 1

tan A = cot B, and angles of a triangle

Complementary condition

tan⁡A=cot⁡B  ⇒  A+B=90∘(A,B acute)\tan A = \cot B \;\Rightarrow\; A + B = 90^\circ \quad (A, B \text{ acute})

Common traps

Check the angles really add to 90°

cos⁡61∘\cos 61^\circ and sin⁡29∘\sin 29^\circ are equal; cos⁡61∘\cos 61^\circ and sin⁡31∘\sin 31^\circ are not. Statement questions plant a pair that misses by two degrees. Add the angles before you cancel.

Count the terms before pairing

With an odd number of terms one is left unpaired, and it is not always 45∘45^\circ: in sin⁡26∘+⋯+sin⁡290∘\sin^2 6^\circ + \cdots + \sin^2 90^\circ the stray term is sin⁡290∘=1\sin^2 90^\circ = 1. Write the first and last angles and the step, count, then pair.

The rule needs acute angles — check the range

tan⁡Ptan⁡Q=1\tan P \tan Q = 1 in general gives P+Q=90∘+180∘kP + Q = 90^\circ + 180^\circ k. A question with a range like 0≤x<30∘0 \le x < 30^\circ is there to make you pick the one value that fits, and to reject the edge value that the range excludes.

Simplifying & Proving Identities

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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}

Common traps

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.

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.

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.

Reciprocal Pairs: sec ± tan, cosec ± cot

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sec θ + tan θ and sec θ − tan θ are reciprocals

The reciprocal pair

sec⁡θ+tan⁡θ=k  ⇒  sec⁡θ−tan⁡θ=1k,sec⁡θ=12(k+1k)\sec\theta + \tan\theta = k \;\Rightarrow\; \sec\theta - \tan\theta = \frac1k, \quad \sec\theta = \frac12\left(k + \frac1k\right)

(1 + sin θ)/cos θ is sec θ + tan θ

The disguised pair

1±sin⁡θcos⁡θ=sec⁡θ±tan⁡θ,1±cos⁡θsin⁡θ=cosec⁡θ±cot⁡θ\frac{1 \pm \sin\theta}{\cos\theta} = \sec\theta \pm \tan\theta, \qquad \frac{1 \pm \cos\theta}{\sin\theta} = \operatorname{cosec}\theta \pm \cot\theta

Replacing the 1 in (tan θ + sec θ − 1)/(tan θ − sec θ + 1)

The standing result

tan⁡θ+sec⁡θ−1tan⁡θ−sec⁡θ+1=sec⁡θ+tan⁡θ=1+sin⁡θcos⁡θ\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}

Common traps

Half the sum gives sec, half the difference gives tan — not the other way

sec⁡θ=12(k+1k)\sec\theta = \dfrac12\left(k + \dfrac1k\right) is at least 11, as a secant must be. If your 'secant' comes out below 11, you have swapped the sum and the difference.

A square root returns a modulus

cos⁡2θ=∣cos⁡θ∣\sqrt{\cos^2\theta} = |\cos\theta|. The simplification to sec⁡θ−tan⁡θ\sec\theta - \tan\theta is right only where cos⁡θ>0\cos\theta > 0. When the stem gives a range, check it; when it does not, the intended range is the first quadrant.

Cross-multiplying works, but costs three minutes

Every member of this family can be proved by clearing denominators and expanding, and every one of them then takes a page. Replacing the 11 takes two lines. If a fraction has a bare ±1\pm 1 beside tan⁡\tan and sec⁡\sec (or cot⁡\cot and cosec⁡\operatorname{cosec}), reach for the replacement first.

Power Identities & Given Sums

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Square the given sum to get sin θ cos θ

The product from the sum

sin⁡θ+cos⁡θ=k  ⇒  sin⁡θcos⁡θ=k2−12,tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ\sin\theta + \cos\theta = k \;\Rightarrow\; \sin\theta\cos\theta = \frac{k^2 - 1}{2}, \quad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}

sin⁴ + cos⁴ and sin⁶ + cos⁶ in terms of sin θ cos θ

Power identities

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ,sin⁡6θ+cos⁡6θ=1−3sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta, \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta

a sin θ + b cos θ and its partner a cos θ − b sin θ

Partner identity

(asin⁡θ+bcos⁡θ)2+(acos⁡θ−bsin⁡θ)2=a2+b2(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2

Common traps

Squaring loses the sign

From (sin⁡θ−cos⁡θ)2(\sin\theta - \cos\theta)^2 you get ±\pm. Above 45∘45^\circ the sine is larger and the difference is positive; below, it is negative. When the stem gives no range, both signs occur, and the paper will print only one of them.

It is minus, not plus

sin⁡4θ+cos⁡4θ=1−2sin⁡2θcos⁡2θ\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta. A statement with +2+2 is planted in identity questions and looks right at a glance. Check it at 45∘45^\circ: the true value is 12\dfrac12, and the plus version gives 32\dfrac32.

When only one sign is printed, it is not the only answer

3sin⁡θ+5cos⁡θ=53\sin\theta + 5\cos\theta = 5 gives 5sin⁡θ−3cos⁡θ=±35\sin\theta - 3\cos\theta = \pm 3, and both signs genuinely happen (θ=0∘\theta = 0^\circ gives −3-3). The paper printed only −3-3. Choose the value offered; do not conclude your +3+3 was wrong.

Trigonometric Equations

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Reduce to one ratio and solve the quadratic

The substitutions

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

Equations in a ratio and its reciprocal

Clearing the reciprocal

t+1t=c  ⇒  t2−ct+1=0t + \frac1t = c \;\Rightarrow\; t^2 - ct + 1 = 0

Linear equations a sin θ + b cos θ = c

Substitute into the identity

asin⁡θ+bcos⁡θ=c,sin⁡2θ+cos⁡2θ=1a\sin\theta + b\cos\theta = c, \quad \sin^2\theta + \cos^2\theta = 1

Systems in two or three angles

Sum of the three combinations

(B+C−A)+(C+A−B)+(A+B−C)=A+B+C(B + C - A) + (C + A - B) + (A + B - C) = A + B + C

Common traps

A strict interval can exclude the only root

If the only surviving root is sin⁡θ=1\sin\theta = 1, that is θ=90∘\theta = 90^\circ. On 0<θ<π20 < \theta < \dfrac{\pi}{2} — a strict inequality — it is excluded, and the equation has no solution. Read the inequality signs of the range before answering.

Both roots of tan θ + cot θ = c are real angles

tan⁡θ=43\tan\theta = \dfrac43 and tan⁡θ=34\tan\theta = \dfrac34 both solve 12(tan⁡θ+cot⁡θ)=2512(\tan\theta + \cot\theta) = 25; they are complementary angles. Only the stated range decides which one the question means — so a range like 45∘<θ<90∘45^\circ < \theta < 90^\circ is not decoration.

The quadratic always offers a root that fails

Substituting and squaring doubles the solutions. In the first quadrant both sine and cosine must be positive; the rejected root almost always gives a negative cosine. The paper prints its value among the options.

Use the range to choose the angle, not the calculator's first answer

sin⁡(A+B)=32\sin(A + B) = \dfrac{\sqrt3}{2} allows A+B=60∘A + B = 60^\circ or 120∘120^\circ. The condition that AA and BB are acute decides which, and the wrong one is always among the options.

Compound & Multiple Angles

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The addition formulas

Addition formulas

sin⁡(A+B)=sin⁡Acos⁡B+cos⁡Asin⁡B,cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\sin(A+B) = \sin A\cos B + \cos A\sin B, \quad \cos(A+B) = \cos A\cos B - \sin A\sin B

Double-angle forms

sin⁡2θ=2sin⁡θcos⁡θ,cos⁡2θ=1−2sin⁡2θ,tan⁡2θ=2tan⁡θ1−tan⁡2θ\sin 2\theta = 2\sin\theta\cos\theta, \quad \cos 2\theta = 1 - 2\sin^2\theta, \quad \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}

Common traps

cos(A + B) has a minus sign

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B. The sign in the cosine formula is the opposite of the sign inside the bracket, and writing it with a plus reproduces cos⁡(A−B)\cos(A - B) instead.

tan 2θ can be negative for an acute θ

If θ\theta is above 45∘45^\circ, then 2θ2\theta is above 90∘90^\circ and tan⁡2θ<0\tan 2\theta < 0. A positive option with the right size is the planted answer.

Maximum, Minimum & Impossible Values

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Expressions linear in sin²θ or sin θ

The two ends

asin⁡2θ+bcos⁡2θ∈[min⁡(a,b), max⁡(a,b)]a\sin^2\theta + b\cos^2\theta \in [\min(a,b),\, \max(a,b)]

t + 1/t ≥ 2, and weighted forms by AM–GM

The bounds

t+1t≥2,asec⁡2θ+bcosec⁡2θ≥(a+b)2t + \frac1t \ge 2, \qquad a\sec^2\theta + b\operatorname{cosec}^2\theta \ge (\sqrt a + \sqrt b)^2

Quadratics in sin θ or cos²θ

Vertex, then ends

c2−c+1=(c−12)2+34,c∈[0,1]c^2 - c + 1 = \left(c - \tfrac12\right)^2 + \tfrac34, \quad c \in [0, 1]

a sin θ + b cos θ is at most √(a² + b²)

Amplitude bound

∣asin⁡θ+bcos⁡θ∣≤a2+b2|a\sin\theta + b\cos\theta| \le \sqrt{a^2 + b^2}

Equations that can never hold

The key bound

∣x+1x∣≥2,∣sin⁡θ∣,∣cos⁡θ∣≤1\left|x + \frac1x\right| \ge 2, \qquad |\sin\theta|, |\cos\theta| \le 1

Common traps

A restricted range moves the ends

On 0≤θ≤π20 \le \theta \le \dfrac{\pi}{2}, sin⁡θ\sin\theta never goes below 00. The minimum of 6+4sin⁡θ6 + 4\sin\theta there is 66, not 22. Always take the ends of the ratio over the stated range.

On an open interval the bound may not be reached

For 0<x<π20 < x < \dfrac{\pi}{2}, sin⁡x+cosec⁡x\sin x + \operatorname{cosec} x would equal 22 only at sin⁡x=1\sin x = 1, which the open interval excludes. So the correct statement is '>2> 2', and the option '≥2\ge 2' is the trap.

The vertex may lie outside the variable's range

For s2+4ss^2 + 4s with s=sin⁡θs = \sin\theta, the vertex s=−2s = -2 is not a possible sine. The minimum is then at the nearer end, s=−1s = -1, giving −3-3 — not the vertex value −4-4.

Check whether the given value IS the maximum

Before squaring and solving a quadratic, compute a2+b2\sqrt{a^2 + b^2}. If it equals the right-hand side, the sine and cosine are read off immediately; solving the long way wastes minutes and invites a spurious root.

A secant CAN exceed 1 — the bound runs the other way

a2+b22ab≥1\dfrac{a^2 + b^2}{2ab} \ge 1 can never be a sine or cosine, but it is a perfectly good secant or cosecant. Match the bound to the ratio: sine and cosine live inside [−1,1][-1, 1], secant and cosecant outside it.

Eliminating θ & Substitution Chains

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Square and add

The cancelling squares

(acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2(a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2

Rewrite each quantity in sine and cosine, then combine

The two reductions that recur

cosec⁡θ−sin⁡θ=cos⁡2θsin⁡θ,sec⁡θ−cos⁡θ=sin⁡2θcos⁡θ\operatorname{cosec}\theta - \sin\theta = \frac{\cos^2\theta}{\sin\theta}, \qquad \sec\theta - \cos\theta = \frac{\sin^2\theta}{\cos\theta}

Substitution chains: sin x + sin²x = 1

The swap and the grouping

sin⁡x+sin⁡2x=1⇒sin⁡x=cos⁡2x,cos⁡2x(cos⁡2x+1)=1\sin x + \sin^2 x = 1 \Rightarrow \sin x = \cos^2 x, \qquad \cos^2 x(\cos^2 x + 1) = 1

Solving p sin²α + q cos²α = m for tan²α

The weighted-average solution

psin⁡2α+qcos⁡2α=m  ⇒  tan⁡2α=m−qp−mp\sin^2\alpha + q\cos^2\alpha = m \;\Rightarrow\; \tan^2\alpha = \frac{m - q}{p - m}

Common traps

Add for sine–cosine, subtract for secant–tangent

With sine and cosine the useful identity is a sum (sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1); with secant and tangent it is a difference (sec⁡2−tan⁡2=1\sec^2 - \tan^2 = 1). Choosing the wrong operation leaves the cross terms doubled instead of cancelled.

Square roots need the sign of θ's quadrant

cot⁡2θcos⁡2θ\sqrt{\cot^2\theta\cos^2\theta} is cot⁡θcos⁡θ\cot\theta\cos\theta only when that product is positive. The stems that use it restrict θ\theta to 0<θ<90∘0 < \theta < 90^\circ for exactly this reason.

Swap the square, not the first power

sin⁡x+sin⁡2x=1\sin x + \sin^2 x = 1 gives sin⁡x=cos⁡2x\sin x = \cos^2 x — it does not give sin⁡2x=cos⁡x\sin^2 x = \cos x. That second form belongs to the cosine version of the question, and the two are set in alternate years.

Keep the sign pattern of the fraction

tan⁡2α=m−qp−m\tan^2\alpha = \dfrac{m - q}{p - m}. The options include m−pq−m\dfrac{m - p}{q - m} and m−qm−p\dfrac{m - q}{m - p}, which differ by a sign in the denominator. A squared tangent cannot be negative, so check your answer's sign with sample values (p>m>qp > m > q).

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