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Trigonometric Ratios and Identities formulas

11 formulas, 1 reference table and 24 common traps for SPI Mathematics Trigonometric Ratios and Identities, grouped by subtopic.

Full notes with worked examples

Trigonometric Ratios of an Acute Angle

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The six ratios of a right triangle

The six trigonometric ratios

sin⁡A=opphyp, cos⁡A=adjhyp, tan⁡A=oppadj, cosec⁡A=hypopp, sec⁡A=hypadj, cot⁡A=adjopp\sin A = \frac{\text{opp}}{\text{hyp}},\ \cos A = \frac{\text{adj}}{\text{hyp}},\ \tan A = \frac{\text{opp}}{\text{adj}},\ \operatorname{cosec} A = \frac{\text{hyp}}{\text{opp}},\ \sec A = \frac{\text{hyp}}{\text{adj}},\ \cot A = \frac{\text{adj}}{\text{opp}}
  • oppthe side opposite the angle
  • adjthe side next to the angle (not the hypotenuse)
  • hypthe hypotenuse, across from the right angle

Finding every ratio from one given ratio

Reciprocal and quotient relations

sin⁡θcosec⁡θ=1,cos⁡θsec⁡θ=1,tan⁡θcot⁡θ=1,tan⁡θ=sin⁡θcos⁡θ,cot⁡θ=cos⁡θsin⁡θ\sin\theta \operatorname{cosec}\theta = 1,\quad \cos\theta \sec\theta = 1,\quad \tan\theta \cot\theta = 1,\quad \tan\theta = \frac{\sin\theta}{\cos\theta},\quad \cot\theta = \frac{\cos\theta}{\sin\theta}

Dividing by cos to turn a fraction into tan

A sin and cos fraction in terms of tan

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

Opposite and adjacent change with the angle

In a right triangle ABC with the right angle at B, side BC is opposite A but adjacent to C. Name the sides again every time the question moves to the other acute angle.

Reading sin A as sin times A

sin⁡A\sin A is one number, the value of a ratio for the angle A. You cannot cancel the A in sin⁡AA\frac{\sin A}{A} or split sin⁡(A+B)\sin(A + B) into sin⁡A+sin⁡B\sin A + \sin B.

Pairing sec with sin

The names cross over: sec is the reciprocal of cos, and cosec is the reciprocal of sin. If cos⁡θ=35\cos\theta = \frac{3}{5}, then sec⁡θ=53\sec\theta = \frac{5}{3}.

Forgetting the square root

With hypotenuse 13 and one side 5, the third side is 169−25=12\sqrt{169 - 25} = 12, not 144. Take the root before you write the ratio.

Dividing only the top

Divide the numerator and the denominator by cos⁡θ\cos\theta. Dividing only one of them changes the value of the fraction.

A plain number breaks the trick

In sin⁡θ+1cos⁡θ\dfrac{\sin\theta + 1}{\cos\theta} the 1 has no sin or cos, so dividing by cos⁡θ\cos\theta leaves 1cos⁡θ\frac{1}{\cos\theta} behind. Then you need the actual values of sin⁡θ\sin\theta and cos⁡θ\cos\theta from the sides.

Values at Standard Angles

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Reading the angle back from a value

Values that give the angle at once (0° to 90°)

sin⁡x=12⇒x=30∘cos⁡x=12⇒x=60∘tan⁡x=1⇒x=45∘tan⁡x=3⇒x=60∘tan⁡x=13⇒x=30∘\sin x = \tfrac{1}{2} \Rightarrow x = 30^\circ \qquad \cos x = \tfrac{1}{2} \Rightarrow x = 60^\circ \qquad \tan x = 1 \Rightarrow x = 45^\circ \qquad \tan x = \sqrt{3} \Rightarrow x = 60^\circ \qquad \tan x = \tfrac{1}{\sqrt{3}} \Rightarrow x = 30^\circ

Sides of a right triangle from one side and an angle

Legs from the hypotenuse

opp=hyp×sin⁡θ,adj=hyp×cos⁡θ,opp=adj×tan⁡θ\text{opp} = \text{hyp} \times \sin\theta, \qquad \text{adj} = \text{hyp} \times \cos\theta, \qquad \text{opp} = \text{adj} \times \tan\theta

The table of standard values

Ratio0°30°45°60°
sin0012\frac{1}{2}12\frac{1}{\sqrt{2}}32\frac{\sqrt{3}}{2}
cos1132\frac{\sqrt{3}}{2}12\frac{1}{\sqrt{2}}12\frac{1}{2}
tan0013\frac{1}{\sqrt{3}}113\sqrt{3}
cosecnot defined222\sqrt{2}23\frac{2}{\sqrt{3}}
sec1123\frac{2}{\sqrt{3}}2\sqrt{2}22
cotnot defined3\sqrt{3}1113\frac{1}{\sqrt{3}}
At 90∘90^\circ, tan⁡90∘\tan 90^\circ and sec⁡90∘\sec 90^\circ are not defined. They are not 0 and not 1.
At 90°: sin 90° = 1, cos 90° = 0, cosec 90° = 1, cot 90° = 0, and tan 90° and sec 90° are not defined. Sine rises from 0 to 1, cosine falls from 1 to 0, and tangent is their quotient.

Common traps

Doubling the angle does not double the value

sin⁡60∘=32\sin 60^\circ = \frac{\sqrt{3}}{2}, but 2sin⁡30∘=12\sin 30^\circ = 1. The ratio of twice an angle is not twice the ratio.

Squaring the angle instead of the value

cos⁡260∘\cos^2 60^\circ is (12)2=14\left(\frac{1}{2}\right)^2 = \frac{1}{4}. The square belongs to the value of the ratio, never to the angle.

Stopping at the angle in the bracket

From tan⁡2θ=3\tan 2\theta = \sqrt{3} you get 2θ=60∘2\theta = 60^\circ, so θ=30∘\theta = 30^\circ. The bracket's angle, 60∘60^\circ, is usually printed as a wrong option.

Answering the angle when a part of it is asked

If the question wants θ2\frac{\theta}{2} or θ3\frac{\theta}{3}, finish the last division. Writing θ\theta itself is the most common slip on these questions.

Using sin for the side next to the angle

The side touching the angle (not the hypotenuse) is hypotenuse ×cos⁡θ\times \cos\theta. Sin gives the side across from the angle. Swapping them swaps the two legs.

Taking the hypotenuse as a leg

The hypotenuse is always across from the right angle. In a right triangle PQR with hypotenuse PR, the right angle is at Q, and the area uses the two legs PQ and QR.

Trigonometric Identities

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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

Common traps

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.

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.

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.

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.

Ratios of Complementary Angles

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Switching a ratio with 90° minus the angle

Complementary angle rules

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 terms of a long sum or product

Number of terms in a list of angles

terms=last−firststep+1\text{terms} = \frac{\text{last} - \text{first}}{\text{step}} + 1

Common traps

Keeping the same ratio

sin⁡(90∘−A)\sin(90^\circ - A) is cos⁡A\cos A, not sin⁡A\sin A. The ratio always changes to its partner when the angle becomes 90∘90^\circ minus it.

Switching when the angles do not add to 90°

sin⁡40∘\sin 40^\circ equals cos⁡50∘\cos 50^\circ, not cos⁡40∘\cos 40^\circ. Check that the two angles add to 90∘90^\circ before you swap the ratio.

Forgetting the unpaired terms

In a sum of sin⁡2\sin^2 running up to 90∘90^\circ, the 45∘45^\circ term pairs with itself and the 90∘90^\circ term has no partner. Add 12\frac{1}{2} and 1 for them after counting the pairs.

Miscounting the terms

From 10∘10^\circ to 80∘80^\circ in steps of 10∘10^\circ there are 8 terms, not 7. Use last minus first, divided by the step, plus 1.

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