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SSC CGL Quantitative Aptitude · Formula sheet

Trigonometric Ratios and Identities formulas

20 formulas, 1 reference table and 44 common traps for SSC CGL Quantitative Aptitude Trigonometric Ratios and Identities, grouped by subtopic.

Full notes with worked examples

Degree, Radian and Standard Values

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The six trigonometric ratios

The six ratios

sin⁡θ=opphypcos⁡θ=adjhyptan⁡θ=oppadj=sin⁡θcos⁡θcosec θ=1sin⁡θsec⁡θ=1cos⁡θcot⁡θ=1tan⁡θ\sin\theta = \frac{\text{opp}}{\text{hyp}} \qquad \cos\theta = \frac{\text{adj}}{\text{hyp}} \qquad \tan\theta = \frac{\text{opp}}{\text{adj}} = \frac{\sin\theta}{\cos\theta} \qquad \text{cosec}\,\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta}

Converting degrees and radians

Degree and radian conversion

θdeg=θrad×180∘πθrad=θdeg×π180∘\theta_{\text{deg}} = \theta_{\text{rad}} \times \frac{180^\circ}{\pi} \qquad \theta_{\text{rad}} = \theta_{\text{deg}} \times \frac{\pi}{180^\circ}

Values of the six ratios at 0°, 30°, 45°, 60° and 90°

Anglesincostan
0°001100
30°12\frac{1}{2}32\frac{\sqrt{3}}{2}13\frac{1}{\sqrt{3}}
45°12\frac{1}{\sqrt{2}}12\frac{1}{\sqrt{2}}11
60°32\frac{\sqrt{3}}{2}12\frac{1}{2}3\sqrt{3}
90°1100Not defined
tan 90° has no value, because cos 90° = 0 sits in the denominator.
cosec, sec and cot are the same entries flipped: cosec 30∘=2\text{cosec}\,30^\circ = 2, sec⁡60∘=2\sec 60^\circ = 2, cot⁡30∘=3\cot 30^\circ = \sqrt{3}. A 0 flips to not defined, so cosec 0°, sec 90° and cot 0° have no value, and cot 90° = 0. Squares are friendlier: sin⁡2\sin^2 runs 0,14,12,34,10, \frac{1}{4}, \frac{1}{2}, \frac{3}{4}, 1 and tan⁡2\tan^2 runs 0,13,1,30, \frac{1}{3}, 1, 3.

Common traps

Pairing cosec with cos

The names mislead: cosec θ\text{cosec}\,\theta is 1sin⁡θ\frac{1}{\sin\theta} and sec⁡θ\sec\theta is 1cos⁡θ\frac{1}{\cos\theta}. Each 'co' ratio is the reciprocal of the ratio without 'co'.

Turning tan the wrong way up

tan⁡θ\tan\theta is opposite over adjacent. Adjacent over opposite is cot⁡θ\cot\theta, and it is always one of the printed wrong options.

Swapping sin 60° and cos 60°

sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2} and cos⁡60∘=12\cos 60^\circ = \frac{1}{2}. The bigger angle has the bigger sine, which is a fast way to check you have not swapped them.

Treating tan 90° as a number

tan⁡90∘\tan 90^\circ and sec⁡90∘\sec 90^\circ are not defined, and neither are cot⁡0∘\cot 0^\circ and cosec 0∘\text{cosec}\,0^\circ. An expression containing one of them has no value at that angle.

Reading the angle off the wrong row

sec⁡θ=2\sec\theta = 2 is 60°, not 30°: sec 30° is 23\frac{2}{\sqrt{3}}. Change a reciprocal ratio back to sin, cos or tan before you look the value up.

Taking the angle as the answer

When the question asks for an expression such as sin⁡A+cos⁡2A\sin A + \cos 2A, the angle is only the middle step. An option that prints the angle itself is a trap: put the values back in before you choose.

Multiplying by the wrong factor

To reach degrees multiply by 180π\frac{180}{\pi}; to reach radians multiply by π180\frac{\pi}{180}. A quick check: degrees should come out as the bigger number, since 1 radian is about 57°.

Rounding 1 radian to 60°

1 radian is about 57.3°, so 1.5 rad is about 85.9° and 2.5 rad about 143.2°. Using 60° per radian lands on a printed wrong option such as 90° or 150°.

Ratios in a Right Triangle

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Ratios from the sides of a right triangle

Sides of a right triangle

a2+b2=c2opposite=hypotenuse×sin⁡θadjacent=hypotenuse×cos⁡θa^2 + b^2 = c^2 \qquad \text{opposite} = \text{hypotenuse} \times \sin\theta \qquad \text{adjacent} = \text{hypotenuse} \times \cos\theta

One ratio gives the other five

From tan θ = a/b

sin⁡θ=aa2+b2cos⁡θ=ba2+b2\sin\theta = \frac{a}{\sqrt{a^2 + b^2}} \qquad \cos\theta = \frac{b}{\sqrt{a^2 + b^2}}

Dividing by cos θ when tan θ is given

Divide through by cos θ

asin⁡θ+bcos⁡θcsin⁡θ+dcos⁡θ=atan⁡θ+bctan⁡θ+d\frac{a\sin\theta + b\cos\theta}{c\sin\theta + d\cos\theta} = \frac{a\tan\theta + b}{c\tan\theta + d}

Common traps

Using the longest printed number as a leg

Find the right angle first. The side that does not touch it is the hypotenuse, whatever order the sides are printed in. Subtract squares to get a leg; add squares only to get the hypotenuse.

Opposite and adjacent belong to one angle

In a right triangle, the side opposite one acute angle is adjacent to the other. Rename the sides every time the question moves to a different angle.

Using the triangle outside the first quadrant

If the angle lies between 90° and 360°, the triangle gives only the size of each ratio. In the second quadrant, for example, cos and tan are negative, so the sign must be fixed by the quadrant.

Rejecting a correct answer that looks different

8113\frac{8}{\sqrt{113}} and 8113113\frac{8\sqrt{113}}{113} are the same number. If no option matches your form, rationalise or un-rationalise before deciding you are wrong.

Dividing when a constant term is present

sin⁡θ+1cos⁡θ\frac{\sin\theta + 1}{\cos\theta} does not become a fraction in tan alone, because the 1 turns into sec⁡θ\sec\theta. The trick needs every term to be a single sin or cos.

Getting tan after dividing by sin

Dividing by sin⁡θ\sin\theta turns cos⁡θ\cos\theta into cot⁡θ\cot\theta, not tan. Divide by the ratio that leaves the one you were given.

Simplifying and Proving Identities

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The three Pythagorean identities

Pythagorean identities

sin⁡2θ+cos⁡2θ=11+tan⁡2θ=sec⁡2θ1+cot⁡2θ=cosec2θ\sin^2\theta + \cos^2\theta = 1 \qquad 1 + \tan^2\theta = \sec^2\theta \qquad 1 + \cot^2\theta = \text{cosec}^2\theta

Rewrite in sin and cos, then simplify

Factor patterns with a² + b² = 1

a3+b3=(a+b)(a2−ab+b2)a3−b3=(a−b)(a2+ab+b2)(1−cos⁡θ)(1+cos⁡θ)=sin⁡2θa^3 + b^3 = (a + b)(a^2 - ab + b^2) \qquad a^3 - b^3 = (a - b)(a^2 + ab + b^2) \qquad (1 - \cos\theta)(1 + \cos\theta) = \sin^2\theta

Eliminating θ from two equations

Removing θ

x=acos⁡θ, y=bsin⁡θ⇒x2a2+y2b2=1x=asec⁡θ, y=btan⁡θ⇒x2a2−y2b2=1(acos⁡θ+bsin⁡θ)2+(asin⁡θ−bcos⁡θ)2=a2+b2x = a\cos\theta,\ y = b\sin\theta \Rightarrow \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \qquad x = a\sec\theta,\ y = b\tan\theta \Rightarrow \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \qquad (a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2

Common traps

A minus sign that is not an identity

cos⁡2θ−sin⁡2θ=1\cos^2\theta - \sin^2\theta = 1 is false: at 45° the left side is 0. Only the sum of the squares of sin and cos is 1.

Moving 1 across with the wrong sign

From 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, tan⁡2θ=sec⁡2θ−1\tan^2\theta = \sec^2\theta - 1, never 1−sec⁡2θ1 - \sec^2\theta. The second would make tan⁡2θ\tan^2\theta negative for every angle.

One passing angle is not a proof

A false statement can still hold at one angle: sin⁡θ=cos⁡θ\sin\theta = \cos\theta is true at 45°. Use a test angle to throw options out, never to accept one on its own.

Cancelling across a plus sign

In sin⁡θ+cos⁡θsin⁡θ\frac{\sin\theta + \cos\theta}{\sin\theta} the sin does not cancel; the fraction is 1+cot⁡θ1 + \cot\theta. Split the fraction or factor first, then cancel.

A test angle where two options agree

At θ = 90°, sin⁡θ1+cos⁡θ\frac{\sin\theta}{1 + \cos\theta} and sin⁡θ1−cos⁡θ\frac{\sin\theta}{1 - \cos\theta} are both 1, yet they are different expressions. If two options tie at your angle, test again at 60°.

Adding where the identity subtracts

sec and tan go together with a minus: sec⁡2θ−tan⁡2θ=1\sec^2\theta - \tan^2\theta = 1. Only sin and cos add their squares to 1. Pair the two given quantities with the right identity before squaring.

Expanding without checking the cross terms

(acos⁡θ+bsin⁡θ)2(a\cos\theta + b\sin\theta)^2 has a middle term 2absin⁡θcos⁡θ2ab\sin\theta\cos\theta. It cancels only if the other expression carries the same term with the opposite sign, so write both middle terms down.

Complementary and Allied Angles

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The co-function rule for complementary angles

Co-function rule

sin⁡(90∘−θ)=cos⁡θtan⁡(90∘−θ)=cot⁡θsec⁡(90∘−θ)=cosec θ\sin(90^\circ - \theta) = \cos\theta \qquad \tan(90^\circ - \theta) = \cot\theta \qquad \sec(90^\circ - \theta) = \text{cosec}\,\theta

Angles that add up to 90°

Partners that make 90°

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

Allied angles, quadrant signs and large angles

Allied angles

sin⁡(180∘−θ)=sin⁡θcos⁡(180∘−θ)=−cos⁡θtan⁡(180∘+θ)=tan⁡θcos⁡(360∘−θ)=cos⁡θsin⁡(−θ)=−sin⁡θ\sin(180^\circ - \theta) = \sin\theta \qquad \cos(180^\circ - \theta) = -\cos\theta \qquad \tan(180^\circ + \theta) = \tan\theta \qquad \cos(360^\circ - \theta) = \cos\theta \qquad \sin(-\theta) = -\sin\theta

Common traps

Pairing angles that add to 180°

The co-function rule needs a sum of exactly 90°. tan⁡50∘tan⁡130∘\tan 50^\circ \tan 130^\circ is not 1: those angles add to 180°, and 130° is handled by the allied-angle rule, not by swapping tan for cot.

Pairing a ratio with itself

sin⁡20∘sin⁡70∘\sin 20^\circ \sin 70^\circ is not 1. Swap one of them first: sin⁡70∘=cos⁡20∘\sin 70^\circ = \cos 20^\circ, so the product is sin⁡20∘cos⁡20∘\sin 20^\circ \cos 20^\circ, and it is tan times cot, or a square plus a square, that gives 1.

Setting the angles equal

sin⁡2x=cos⁡3x\sin 2x = \cos 3x does not mean 2x=3x2x = 3x. Different ratios on the two sides mean the angles ADD to 90°; only the same ratio on both sides means the angles are equal.

Using a non-standard answer as a sign of error

sin⁡x=cos⁡(2x−10∘)\sin x = \cos(2x - 10^\circ) gives x=33.33∘x = 33.33^\circ, which is not a table angle. That is fine: this rule never needs the table, only the sum of 90°.

Dropping the sign in the second quadrant

Between 90° and 180° only sin and cosec are positive. If sin⁡x=35\sin x = \frac{3}{5} there, then cos⁡x=−45\cos x = -\frac{4}{5} and tan⁡x=−34\tan x = -\frac{3}{4}; the positive versions are always among the options.

Taking away 180° from a sine

Only tan and cot repeat every 180°. For sin and cos, sin⁡(θ+180∘)=−sin⁡θ\sin(\theta + 180^\circ) = -\sin\theta, so remove whole turns of 360° instead.

Putting a minus on cos of a negative angle

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos\theta: cos ignores the sign of the angle. sin and tan are the ones that change sign.

Reciprocal Pairs and Given Sums

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sec θ + tan θ and sec θ − tan θ multiply to 1

The reciprocal pairs

(sec⁡θ+tan⁡θ)(sec⁡θ−tan⁡θ)=1(cosec θ+cot⁡θ)(cosec θ−cot⁡θ)=1sec⁡θ+tan⁡θ=p⇒sin⁡θ=p2−1p2+11+sin⁡θ1−sin⁡θ=sec⁡θ+tan⁡θ(\sec\theta + \tan\theta)(\sec\theta - \tan\theta) = 1 \qquad (\text{cosec}\,\theta + \cot\theta)(\text{cosec}\,\theta - \cot\theta) = 1 \qquad \sec\theta + \tan\theta = p \Rightarrow \sin\theta = \frac{p^2 - 1}{p^2 + 1} \qquad \sqrt{\frac{1 + \sin\theta}{1 - \sin\theta}} = \sec\theta + \tan\theta

Squaring a given sum

Squared sums

(sin⁡θ±cos⁡θ)2=1±2sin⁡θcos⁡θ(sin⁡θ+cos⁡θ)2+(sin⁡θ−cos⁡θ)2=2x+1x=k⇒x2+1x2=k2−2tan⁡θ+cot⁡θ=1sin⁡θcos⁡θ(\sin\theta \pm \cos\theta)^2 = 1 \pm 2\sin\theta\cos\theta \qquad (\sin\theta + \cos\theta)^2 + (\sin\theta - \cos\theta)^2 = 2 \qquad x + \frac{1}{x} = k \Rightarrow x^2 + \frac{1}{x^2} = k^2 - 2 \qquad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}

sin⁴θ + cos⁴θ and sin⁶θ + cos⁶θ

Higher powers of sin and cos

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

Substitution chains such as cos θ + cos²θ = 1

The two chains

cos⁡θ+cos⁡2θ=1⇒cos⁡θ=sin⁡2θsin⁡θ+sin⁡2θ=1⇒sin⁡θ=cos⁡2θ\cos\theta + \cos^2\theta = 1 \Rightarrow \cos\theta = \sin^2\theta \qquad \sin\theta + \sin^2\theta = 1 \Rightarrow \sin\theta = \cos^2\theta

Common traps

Taking the partner as p − 1 or −p

If sec⁡θ+tan⁡θ=p\sec\theta + \tan\theta = p, then sec⁡θ−tan⁡θ\sec\theta - \tan\theta is 1p\frac{1}{p}, because the two multiply to 1. It is not p−1p - 1, −p-p or p2−1p^2 - 1.

Swapping the sum and the difference

Adding the pair gives 2sec⁡θ2\sec\theta; subtracting gives 2tan⁡θ2\tan\theta. Halving p+1pp + \frac{1}{p} and calling it tan is the most common slip, and the result is always a printed option.

Forgetting the 2 in the cross term

(sin⁡θ+cos⁡θ)2=1+2sin⁡θcos⁡θ(\sin\theta + \cos\theta)^2 = 1 + 2\sin\theta\cos\theta, not 1+sin⁡θcos⁡θ1 + \sin\theta\cos\theta. Dropping the 2 doubles the product you find, and the doubled value is usually printed.

Answering a given that cannot happen

sin⁡θ+cos⁡θ=3\sin\theta + \cos\theta = \sqrt{3} would need sin⁡θcos⁡θ=1\sin\theta\cos\theta = 1, but sin⁡θcos⁡θ\sin\theta\cos\theta is never more than 12\frac{1}{2}. Any sum above 2\sqrt{2} describes no real angle.

Writing sin⁴θ + cos⁴θ = 1

Only the squares add to 1. The fourth powers add to 1−2sin⁡2θcos⁡2θ1 - 2\sin^2\theta\cos^2\theta, which is 12\frac{1}{2} at 45° and 1 only where sin or cos is 0.

Using θ = 0° when the options contain θ

Putting in one angle finds a constant answer. When the options are expressions in θ, one angle can make two of them equal, so test a second angle before choosing.

Hunting for the angle

cos⁡θ+cos⁡2θ=1\cos\theta + \cos^2\theta = 1 has no table angle: cos θ works out to 5−12\frac{\sqrt{5} - 1}{2}. Trying 30°, 45° and 60° wastes the 36 seconds. Make the swap cos⁡θ=sin⁡2θ\cos\theta = \sin^2\theta instead.

Compound and Multiple Angles

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The addition and subtraction formulas

Compound angles

sin⁡(A±B)=sin⁡Acos⁡B±cos⁡Asin⁡Bcos⁡(A±B)=cos⁡Acos⁡B∓sin⁡Asin⁡Btan⁡(A±B)=tan⁡A±tan⁡B1∓tan⁡Atan⁡B\sin(A \pm B) = \sin A\cos B \pm \cos A\sin B \qquad \cos(A \pm B) = \cos A\cos B \mp \sin A\sin B \qquad \tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A\tan B}

Double, half and triple angle formulas

Multiple angles

sin⁡2A=2sin⁡Acos⁡A=2tan⁡A1+tan⁡2Acos⁡2A=1−2sin⁡2A=2cos⁡2A−1=1−tan⁡2A1+tan⁡2Atan⁡A2=sin⁡A1+cos⁡Asin⁡3A=3sin⁡A−4sin⁡3Acos⁡3A=4cos⁡3A−3cos⁡A\sin 2A = 2\sin A\cos A = \frac{2\tan A}{1 + \tan^2 A} \qquad \cos 2A = 1 - 2\sin^2 A = 2\cos^2 A - 1 = \frac{1 - \tan^2 A}{1 + \tan^2 A} \qquad \tan\frac{A}{2} = \frac{\sin A}{1 + \cos A} \qquad \sin 3A = 3\sin A - 4\sin^3 A \qquad \cos 3A = 4\cos^3 A - 3\cos A

Products and sums of sines and cosines

Product and sum formulas

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2cos⁡Acos⁡B=cos⁡(A−B)+cos⁡(A+B)2sin⁡Acos⁡B=sin⁡(A+B)+sin⁡(A−B)sin⁡C+sin⁡D=2sin⁡C+D2cos⁡C−D2cos⁡C+cos⁡D=2cos⁡C+D2cos⁡C−D22\sin A\sin B = \cos(A - B) - \cos(A + B) \qquad 2\cos A\cos B = \cos(A - B) + \cos(A + B) \qquad 2\sin A\cos B = \sin(A + B) + \sin(A - B) \qquad \sin C + \sin D = 2\sin\tfrac{C + D}{2}\cos\tfrac{C - D}{2} \qquad \cos C + \cos D = 2\cos\tfrac{C + D}{2}\cos\tfrac{C - D}{2}

Common traps

Keeping the sign in cos(A + B)

cos⁡(A+B)=cos⁡Acos⁡B−sin⁡Asin⁡B\cos(A + B) = \cos A\cos B - \sin A\sin B, with a minus. The plus version is cos⁡(A−B)\cos(A - B), and SSC prints both as options.

Splitting tan over a sum

tan⁡(A+B)\tan(A + B) is not tan⁡A+tan⁡B\tan A + \tan B. With A = B = 30°: tan⁡60∘=3≈1.73\tan 60^\circ = \sqrt{3} \approx 1.73, while 2tan⁡30∘≈1.152\tan 30^\circ \approx 1.15.

Doubling the ratio instead of the angle

sin⁡2A\sin 2A is not 2sin⁡A2\sin A. At A = 60°, sin⁡120∘=32\sin 120^\circ = \frac{\sqrt{3}}{2} while 2sin⁡60∘=32\sin 60^\circ = \sqrt{3}.

Turning the half-angle fraction upside down

sin⁡A1+cos⁡A\frac{\sin A}{1 + \cos A} is tan⁡A2\tan\frac{A}{2}, but sin⁡A1−cos⁡A\frac{\sin A}{1 - \cos A} is cot⁡A2\cot\frac{A}{2}. At A = 60° the first is 13\frac{1}{\sqrt{3}} and the second 3\sqrt{3}.

Reversing the order in 2 sin A sin B

2sin⁡Asin⁡B=cos⁡(A−B)−cos⁡(A+B)2\sin A\sin B = \cos(A - B) - \cos(A + B). Written the other way round, cos⁡(A+B)−cos⁡(A−B)\cos(A + B) - \cos(A - B), it is the negative of the right answer, and that form is always offered.

Trigonometric Equations and Maximum Values

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Reduce to one ratio, then read the angle

Mixed squares of sin and cos

asin⁡2θ+bcos⁡2θ=c⇒sin⁡2θ=c−ba−ba\sin^2\theta + b\cos^2\theta = c \Rightarrow \sin^2\theta = \frac{c - b}{a - b}

Greatest and least values of a sin θ + b cos θ and of powers

Range of a sin θ + b cos θ

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

Common traps

Keeping an impossible root

A quadratic in sin⁡θ\sin\theta often gives a root such as sin⁡θ=−2\sin\theta = -2 or 32\frac{3}{2}. No angle has a sine outside −1 to 1, so that root is thrown away, not reported.

Cancelling a factor that can be zero

From cot⁡2θ=2cos⁡2θ\cot^2\theta = 2\cos^2\theta, cancelling cos⁡2θ\cos^2\theta loses θ = 90°, where both sides are 0. Cancel only when the range rules that value out, and say so.

Adding the coefficients

The greatest value of 6sin⁡θ+8cos⁡θ6\sin\theta + 8\cos\theta is not 14, because sin θ and cos θ are never both 1. It is 36+64=10\sqrt{36 + 64} = 10.

Forgetting the least value is negative

The range of asin⁡θ+bcos⁡θa\sin\theta + b\cos\theta is symmetric: if the greatest value is 13, the least is −13, not 0.

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