Conversion Relations of Trigonometric Ratios — Table

#Trigonometry
TL;DR
Conversion relations let you write any one of the six trigonometric ratios in terms of any other — for example, expressing sin⁡θ\sin\theta, sec⁡θ\sec\theta, and tan⁡θ\tan\theta all in terms of cot⁡θ\cot\theta. The method chains three engines: the reciprocal relations, the quotient relations, and the Pythagorean identities. This article gives the full conversion table, the step-by-step method, the sign caveat, and six worked examples — including the standard textbook questions.
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Bhanzu TeamLast updated on July 15, 20269 min read

What Are the Conversion Relations of Trigonometric Ratios?

The conversion relations of trigonometric ratios are the rules for expressing any of the six ratios in terms of any other — sine in terms of cosine, all six in terms of tan⁡θ\tan\theta, and so on. The point is interconversion: starting from one known ratio and rebuilding the rest, without going back to the triangle's side lengths.

A quick note on the name, because the phrase is used two ways. Some sources use "conversion relations" for angle transformations — turning sin⁡(90°+θ)\sin(90° + \theta) into cos⁡θ\cos\theta, for instance. This article is about the other, more common classroom meaning: expressing one ratio through another. The angle-transformation idea is covered separately in trigonometric ratios of complementary angles.

The conversions are powered by the basic properties of trigonometric ratios, grouped into three engines.

  • Reciprocal relations — csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}, sec⁡θ=1cos⁡θ\sec\theta = \frac{1}{\cos\theta}, cot⁡θ=1tan⁡θ\cot\theta = \frac{1}{\tan\theta}.

  • Quotient relations — tan⁡θ=sin⁡θcos⁡θ\tan\theta = \frac{\sin\theta}{\cos\theta}, cot⁡θ=cos⁡θsin⁡θ\cot\theta = \frac{\cos\theta}{\sin\theta}.

  • Pythagorean identities — sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1, 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta.

The Full Conversion Table

Here is the complete reference: each of the six ratios written in terms of each base ratio. The table assumes θ\theta is acute (Quadrant I), so every value is positive; the sign caveat for other quadrants comes after.

In terms of →

sin⁡θ\sin\theta

cos⁡θ\cos\theta

tan⁡θ\tan\theta

sin⁡θ\sin\theta

sin⁡θ\sin\theta

1−cos⁡2θ\sqrt{1 - \cos^2\theta}

tan⁡θ1+tan⁡2θ\frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}

cos⁡θ\cos\theta

1−sin⁡2θ\sqrt{1 - \sin^2\theta}

cos⁡θ\cos\theta

11+tan⁡2θ\frac{1}{\sqrt{1 + \tan^2\theta}}

tan⁡θ\tan\theta

sin⁡θ1−sin⁡2θ\frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}

1−cos⁡2θcos⁡θ\frac{\sqrt{1 - \cos^2\theta}}{\cos\theta}

tan⁡θ\tan\theta

csc⁡θ\csc\theta

1sin⁡θ\frac{1}{\sin\theta}

11−cos⁡2θ\frac{1}{\sqrt{1 - \cos^2\theta}}

1+tan⁡2θtan⁡θ\frac{\sqrt{1 + \tan^2\theta}}{\tan\theta}

sec⁡θ\sec\theta

11−sin⁡2θ\frac{1}{\sqrt{1 - \sin^2\theta}}

1cos⁡θ\frac{1}{\cos\theta}

1+tan⁡2θ\sqrt{1 + \tan^2\theta}

cot⁡θ\cot\theta

1−sin⁡2θsin⁡θ\frac{\sqrt{1 - \sin^2\theta}}{\sin\theta}

cos⁡θ1−cos⁡2θ\frac{\cos\theta}{\sqrt{1 - \cos^2\theta}}

1tan⁡θ\frac{1}{\tan\theta}

And the same six ratios in terms of the reciprocal base ratios:

In terms of →

csc⁡θ\csc\theta

sec⁡θ\sec\theta

cot⁡θ\cot\theta

sin⁡θ\sin\theta

1csc⁡θ\frac{1}{\csc\theta}

sec⁡2θ−1sec⁡θ\frac{\sqrt{\sec^2\theta - 1}}{\sec\theta}

11+cot⁡2θ\frac{1}{\sqrt{1 + \cot^2\theta}}

cos⁡θ\cos\theta

csc⁡2θ−1csc⁡θ\frac{\sqrt{\csc^2\theta - 1}}{\csc\theta}

1sec⁡θ\frac{1}{\sec\theta}

cot⁡θ1+cot⁡2θ\frac{\cot\theta}{\sqrt{1 + \cot^2\theta}}

tan⁡θ\tan\theta

1csc⁡2θ−1\frac{1}{\sqrt{\csc^2\theta - 1}}

sec⁡2θ−1\sqrt{\sec^2\theta - 1}

1cot⁡θ\frac{1}{\cot\theta}

sec⁡θ\sec\theta

csc⁡θcsc⁡2θ−1\frac{\csc\theta}{\sqrt{\csc^2\theta - 1}}

sec⁡θ\sec\theta

1+cot⁡2θcot⁡θ\frac{\sqrt{1 + \cot^2\theta}}{\cot\theta}

csc⁡θ\csc\theta

csc⁡θ\csc\theta

sec⁡θsec⁡2θ−1\frac{\sec\theta}{\sqrt{\sec^2\theta - 1}}

1+cot⁡2θ\sqrt{1 + \cot^2\theta}

cot⁡θ\cot\theta

csc⁡2θ−1\sqrt{\csc^2\theta - 1}

1sec⁡2θ−1\frac{1}{\sqrt{\sec^2\theta - 1}}

cot⁡θ\cot\theta

Do not memorise the table. Memorise the three engines and rebuild any cell on demand — the next section shows exactly how.

How Do You Convert One Trigonometric Ratio Into Another?

The method is a fixed three-step chain. Suppose you are given sin⁡θ\sin\theta and want everything else.

  1. Get cosine from the Pythagorean identity. Since sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1:

cos⁡θ=1−sin⁡2θ\cos\theta = \sqrt{1 - \sin^2\theta}

  1. Get tangent from the quotient relation. Now that both sine and cosine are known:

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

  1. Get the reciprocals by flipping. Each of csc⁡θ\csc\theta, sec⁡θ\sec\theta, cot⁡θ\cot\theta is 11 over the matching primary ratio.

csc⁡θ=1sin⁡θ,sec⁡θ=11−sin⁡2θ,cot⁡θ=1−sin⁡2θsin⁡θ\csc\theta = \frac{1}{\sin\theta}, \quad \sec\theta = \frac{1}{\sqrt{1 - \sin^2\theta}}, \quad \cot\theta = \frac{\sqrt{1 - \sin^2\theta}}{\sin\theta}

The same three steps work from any starting ratio. If you start from tan⁡θ\tan\theta, use 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta to get secant first; if you start from cot⁡θ\cot\theta, use 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta. The Pythagorean identity always supplies the "missing partner," and the quotient and reciprocal relations finish the job.

What about the sign?

The Pythagorean step produces a square root, which carries a ±\pm. For acute angles the sign is always positive. For angles beyond 90°90°, the sign is fixed by the quadrant the angle lands in — the ASTC rule from the basic properties decides whether to take the ++ or the −-. Drop the sign check and a Quadrant II answer comes out wrong.

Examples of Conversion Relations of Trigonometric Ratios

Example 1

Express cos⁡θ\cos\theta in terms of sin⁡θ\sin\theta (acute angle).

From the Pythagorean identity sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1:

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

cos⁡θ=1−sin⁡2θ\cos\theta = \sqrt{1 - \sin^2\theta}

Final answer: cos⁡θ=1−sin⁡2θ\cos\theta = \sqrt{1 - \sin^2\theta} (positive, since θ\theta is acute).

Example 2

Express sin⁡θ\sin\theta in terms of tan⁡θ\tan\theta.

A first instinct is to write sin⁡θ=tan⁡θ⋅cos⁡θ\sin\theta = \tan\theta \cdot \cos\theta and stop — but that still contains cos⁡θ\cos\theta, so it is not yet "in terms of tan⁡θ\tan\theta." Test whether the half-finished form is acceptable: substitute θ=45°\theta = 45°, where tan⁡45°=1\tan 45° = 1 and sin⁡45°=12\sin 45° = \frac{1}{\sqrt{2}}. The expression tan⁡θ⋅cos⁡θ=1×12\tan\theta \cdot \cos\theta = 1 \times \frac{1}{\sqrt{2}} only works because we already knew cosine — a genuine conversion must eliminate it.

The correct route uses 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta, so cos⁡θ=1sec⁡θ=11+tan⁡2θ\cos\theta = \frac{1}{\sec\theta} = \frac{1}{\sqrt{1 + \tan^2\theta}}. Then:

sin⁡θ=tan⁡θ⋅cos⁡θ=tan⁡θ⋅11+tan⁡2θ=tan⁡θ1+tan⁡2θ\sin\theta = \tan\theta \cdot \cos\theta = \tan\theta \cdot \frac{1}{\sqrt{1 + \tan^2\theta}} = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}}

Final answer: sin⁡θ=tan⁡θ1+tan⁡2θ\sin\theta = \frac{\tan\theta}{\sqrt{1 + \tan^2\theta}} — now entirely in terms of tan⁡θ\tan\theta.

Example 3

Express the trigonometric ratios sin⁡A\sin A, sec⁡A\sec A, and tan⁡A\tan A in terms of cot⁡A\cot A. (standard exam question)

Start from cot⁡A\cot A and use 1+cot⁡2A=csc⁡2A1 + \cot^2 A = \csc^2 A.

csc⁡A=1+cot⁡2A\csc A = \sqrt{1 + \cot^2 A}

Since sin⁡A=1csc⁡A\sin A = \frac{1}{\csc A}:

sin⁡A=11+cot⁡2A\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}

For tan⁡A\tan A, use the reciprocal relation directly:

tan⁡A=1cot⁡A\tan A = \frac{1}{\cot A}

For sec⁡A\sec A, get cosine from cos⁡A=cot⁡A⋅sin⁡A=cot⁡A1+cot⁡2A\cos A = \cot A \cdot \sin A = \frac{\cot A}{\sqrt{1 + \cot^2 A}}, then flip:

sec⁡A=1cos⁡A=1+cot⁡2Acot⁡A\sec A = \frac{1}{\cos A} = \frac{\sqrt{1 + \cot^2 A}}{\cot A}

Final answer: sin⁡A=11+cot⁡2A\sin A = \frac{1}{\sqrt{1 + \cot^2 A}}, tan⁡A=1cot⁡A\tan A = \frac{1}{\cot A}, sec⁡A=1+cot⁡2Acot⁡A\sec A = \frac{\sqrt{1 + \cot^2 A}}{\cot A}.

Example 4

Write all the other trigonometric ratios of ∠A\angle A in terms of sec⁡A\sec A. (standard exam question)

Use 1+tan⁡2A=sec⁡2A1 + \tan^2 A = \sec^2 A, so tan⁡A=sec⁡2A−1\tan A = \sqrt{\sec^2 A - 1}, and cos⁡A=1sec⁡A\cos A = \frac{1}{\sec A}.

cos⁡A=1sec⁡A\cos A = \frac{1}{\sec A}

tan⁡A=sec⁡2A−1\tan A = \sqrt{\sec^2 A - 1}

sin⁡A=tan⁡A⋅cos⁡A=sec⁡2A−1sec⁡A\sin A = \tan A \cdot \cos A = \frac{\sqrt{\sec^2 A - 1}}{\sec A}

csc⁡A=1sin⁡A=sec⁡Asec⁡2A−1\csc A = \frac{1}{\sin A} = \frac{\sec A}{\sqrt{\sec^2 A - 1}}

cot⁡A=1tan⁡A=1sec⁡2A−1\cot A = \frac{1}{\tan A} = \frac{1}{\sqrt{\sec^2 A - 1}}

Final answer: the five ratios as written above, each expressed purely in sec⁡A\sec A.

Example 5

Given tan⁡θ=34\tan\theta = \frac{3}{4} for an acute angle, find sin⁡θ\sin\theta and cos⁡θ\cos\theta using conversion.

From 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta:

sec⁡2θ=1+916=2516  ⟹  sec⁡θ=54\sec^2\theta = 1 + \frac{9}{16} = \frac{25}{16} \implies \sec\theta = \frac{5}{4}

cos⁡θ=1sec⁡θ=45\cos\theta = \frac{1}{\sec\theta} = \frac{4}{5}

sin⁡θ=tan⁡θ⋅cos⁡θ=34⋅45=35\sin\theta = \tan\theta \cdot \cos\theta = \frac{3}{4} \cdot \frac{4}{5} = \frac{3}{5}

Final answer: sin⁡θ=35\sin\theta = \frac{3}{5}, cos⁡θ=45\cos\theta = \frac{4}{5} — the familiar 33-44-55 triangle, recovered from tangent alone.

Example 6

An angle θ\theta in Quadrant II has sin⁡θ=513\sin\theta = \frac{5}{13}. Convert to find cos⁡θ\cos\theta and tan⁡θ\tan\theta.

The conversion gives the magnitude:

cos⁡θ=±1−25169=±1213\cos\theta = \pm\sqrt{1 - \frac{25}{169}} = \pm\frac{12}{13}

Now apply the sign caveat. In Quadrant II cosine is negative, so:

cos⁡θ=−1213,tan⁡θ=sin⁡θcos⁡θ=5/13−12/13=−512\cos\theta = -\frac{12}{13}, \quad \tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{5/13}{-12/13} = -\frac{5}{12}

Final answer: cos⁡θ=−1213\cos\theta = -\frac{12}{13}, tan⁡θ=−512\tan\theta = -\frac{5}{12}. The conversion supplies the size; the quadrant fixes the sign.

Why Conversion Is the Skill, Not the Table

The conversions matter because they turn one piece of information into all of it — and because the method is reusable in a way the table is not.

  • They solve "given one ratio, find another" instantly. This is one of the most common question shapes in introductory courses and beyond, and the three-engine chain answers every version of it.

  • They are the backbone of identity proofs. To prove an identity, you almost always rewrite everything in terms of sin⁡\sin and cos⁡\cos — which is a conversion. The wider toolkit lives in trigonometric identities.

  • They scale to any quadrant. Once the acute-angle conversions are automatic, extending them is just a sign decision from ASTC — which opens the door to the unit circle and trigonometric functions of any angle.

Where Students Slip on Conversion Relations

Mistake 1: Leaving the answer half-converted

Where it slips in: "Express sin⁡θ\sin\theta in terms of tan⁡θ\tan\theta" type questions.

Don't do this: Writing sin⁡θ=tan⁡θ⋅cos⁡θ\sin\theta = \tan\theta \cdot \cos\theta and calling it done — the right-hand side still contains cos⁡θ\cos\theta.

The correct way: A conversion is complete only when every other ratio has been eliminated. The first-instinct error is stopping at the quotient step; the habit that fixes it is scanning the final expression for any stray ratio that isn't the target base.

Mistake 2: Dropping the ± and ignoring the quadrant

Where it slips in: Any conversion through a Pythagorean identity, especially for non-acute angles.

Don't do this: Taking the positive square root automatically in Quadrant II, III, or IV.

The correct way: The square root gives a ±\pm; the quadrant (ASTC) decides which. The step that looks skippable but isn't is the sign check after the root — for acute angles it's always ++, but the moment the angle passes 90°90° the sign can flip.

Mistake 3: Misusing the wrong Pythagorean identity

Where it slips in: Starting from tan⁡θ\tan\theta or cot⁡θ\cot\theta and reaching for sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1.

Don't do this: Forcing sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1 when the natural partner of tan⁡θ\tan\theta is sec⁡θ\sec\theta.

The correct way: Match the identity to the starting ratio — tan⁡θ\tan\theta pairs with 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta; cot⁡θ\cot\theta pairs with 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta. Choosing the right engine first saves a tangle of substitutions.

Key Takeaways

  • Conversion relations express any trigonometric ratio in terms of any other, at the same angle.

  • The method chains three engines: reciprocal relations, quotient relations, and the three Pythagorean identities.

  • The Pythagorean identity always supplies the missing partner ratio; pick the identity that matches your starting ratio.

  • Conversions through a square root carry a ±\pm — acute angles take the positive sign; other quadrants follow ASTC.

  • The reusable skill is the three-step method, not the memorised table.

Practice Before Moving On

Do each from the engines, not the table.

  1. Express cos⁡θ\cos\theta in terms of sin⁡θ\sin\theta, then check it at θ=30°\theta = 30°.

  2. Given sin⁡θ=817\sin\theta = \frac{8}{17} (acute), find cos⁡θ\cos\theta and tan⁡θ\tan\theta by conversion.

  3. Express sec⁡θ\sec\theta in terms of cot⁡θ\cot\theta for an acute angle.

Answer to Question 1: cos⁡θ=1−sin⁡2θ\cos\theta = \sqrt{1 - \sin^2\theta}; at 30°30°, 1−1/4=32=cos⁡30°\sqrt{1 - 1/4} = \frac{\sqrt{3}}{2} = \cos 30°. Answer to Question 2: cos⁡θ=1517\cos\theta = \frac{15}{17}, tan⁡θ=815\tan\theta = \frac{8}{15} (the 88-1515-1717 triple). Answer to Question 3: sec⁡θ=1+cot⁡2θcot⁡θ\sec\theta = \frac{\sqrt{1 + \cot^2\theta}}{\cot\theta} — re-derive via csc⁡θ=1+cot⁡2θ\csc\theta = \sqrt{1 + \cot^2\theta} and cos⁡θ=cot⁡θcsc⁡θ\cos\theta = \frac{\cot\theta}{\csc\theta} if you didn't land here.

To make the three-engine method automatic with a teacher, explore Bhanzu's trigonometry tutor, the high school math tutor track, or math help online.

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Frequently Asked Questions

How do you convert one trigonometric ratio into another?
The conversion relations of trigonometric ratios use three engines in turn: the Pythagorean identity to get the missing partner ratio, the quotient relation to get tangent or cotangent, and the reciprocal relations to flip to cosecant, secant, or cotangent
How do you express all trigonometric ratios in terms of sin⁡θ\sin\theta?
cos⁡θ=1−sin⁡2θ\cos\theta = \sqrt{1 - \sin^2\theta}, tan⁡θ=sin⁡θ1−sin⁡2θ\tan\theta = \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}}, csc⁡θ=1sin⁡θ\csc\theta = \frac{1}{\sin\theta}, sec⁡θ=11−sin⁡2θ\sec\theta = \frac{1}{\sqrt{1 - \sin^2\theta}}, cot⁡θ=1−sin⁡2θsin⁡θ\cot\theta = \frac{\sqrt{1 - \sin^2\theta}}{\sin\theta}
How do you express tan⁡θ\tan\theta in terms of sin⁡θ\sin\theta?
tan⁡θ=sin⁡θcos⁡θ=sin⁡θ1−sin⁡2θ\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{\sin\theta}{\sqrt{1 - \sin^2\theta}} for an acute angle, using cosine from the Pythagorean identity
Why does the square root appear in conversions, and what about the sign?
Because the Pythagorean identity is squared, undoing it needs a square root, which carries a ±\pm. For acute angles take the positive root; for other angles let the quadrant (ASTC) choose the sign
Which Pythagorean identity should I use?
Match it to your known ratio: sin⁡\sin/cos⁡\cos use sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1; tan⁡\tan/sec⁡\sec use 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta; cot⁡\cot/csc⁡\csc use 1+cot⁡2θ=csc⁡2θ1 + \cot^2\theta = \csc^2\theta
Are the conversion relations of trigonometric ratios the same as cofunction relations like sin⁡(90°−θ)=cos⁡θ\sin(90° - \theta) = \cos\theta?
No. Cofunction relations transform the angle; the conversion relations of trigonometric ratios rewrite one ratio in terms of another at the same angle. The angle-transform topic is covered under complementary angles
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Bhanzu Team
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