Conversion Relations of Trigonometric Ratios — Table

Conversion Relations of Trigonometric Ratios — Table

TL;DR

Conversion relations let you write any one of the six trigonometric ratios in terms of any other — for example, expressing sin⁡θ, sec⁡θ, and tan⁡θ all in terms of cot⁡θ. 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.

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, 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) into cos⁡θ, 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.

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 is acute (Quadrant I), so every value is positive; the sign caveat for other quadrants comes after.

In terms of → sin⁡θ cos⁡θ tan⁡θ
sin⁡θ sin⁡θ 1−cos²θ tan⁡θ/
cos⁡θ 1−sin²θ cos⁡θ 1/tan²θ
tan⁡θ sin⁡θ/1−sin²θ 1/cos²θ tan⁡θ
csc⁡θ 1/sin⁡θ 1/cos²θ 1+tan²θ/tan⁡θ
sec⁡θ 1/sin²θ 1/cos⁡θ 1+tan²θ
cot⁡θ 1−sin²θ/sin⁡θ cos⁡θ/(1−cos²θ) 1/tan⁡θ

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

In terms of → csc⁡θ sec⁡θ cot⁡θ
sin⁡θ 1/csc⁡θ sec²θ−1/sec⁡θ 1+cot²θ
cos⁡θ csc²θ−1/csc⁡θ 1/sec⁡θ cotθ/1+cot²θ
tan⁡θ 1/csc²θ−1 sec²θ−1 1/cot⁡θ
sec⁡θ cscθ/csc²θ−1 sec⁡θ 1+cot²θ
csc⁡θ csc⁡θ sec²θ−1/sec²θ 1+cot²θ
cot⁡θ 1/csc²θ−1 1/sec²θ−1 cot⁡θ

How Do You Convert One Trigonometric Ratio Into Another?

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

  1. Get cosine from the Pythagorean identity. Since sin²θ+cos²θ=1:

    cos⁡θ=√(1−sin²θ)

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

    tan⁡θ=sin⁡θ/cos⁡θ=sin⁡θ/√(1−sin²θ)

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

csc⁡θ=1/sin⁡θ, sec⁡θ=1/√(1−sin²θ), cot⁡θ=√(1−sin²θ)/sin⁡θ.

The same three steps work from any starting ratio. If you start from tan⁡θ, use 1+tan²θ=sec²θ to get secant first; if you start from cot⁡θ, use 1+cot²θ=csc²θ. 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 ±. For acute angles the sign is always positive. For angles beyond 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 in terms of sin⁡θ (acute angle).

From the Pythagorean identity sin²θ+cos²θ=1:

cos²θ=1−sin²θ

cos⁡θ=√(1−sin²θ)

Final answer: cos⁡θ=√(1−sin²θ) (positive, since θ is acute).

Example 2

Express sin⁡θ in terms of tan⁡θ.

The first instinct is to write sin⁡θ=tan⁡θ·cos⁡θ and stop, but that still contains cos⁡θ, so it is not yet "in terms of tan⁡θ." The correct route uses 1+tan²θ=sec²θ:

cos⁡θ=1/sec⁡θ=1/√(1+tan²θ);

then:

sin⁡θ=tan⁡θ·cos⁡θ=tan⁡θ·1/√(1+tan²θ).

Final answer: sin⁡θ=tan⁡θ/√(1+tan²θ) — now entirely in terms of tan⁡θ.

Example 3

Express the trigonometric ratios sin⁡A, sec⁡A, and tan⁡A in terms of cot⁡A.

Start from cot⁡A and use 1+cot²A=csc²A:

csc⁡A=√(1+cot²A);

since sin⁡A=1/csc⁡A:

sin⁡A=1/√(1+cot²A);

for tan⁡A, use the reciprocal relation directly:

tan⁡A=1/cot⁡A;

for sec⁡A, get cosine from cos⁡A=cot⁡A·sin⁡A=cot⁡A/√(1+cot²A), then flip:

sec⁡A=1/cos⁡A=√(1+cot²A)/cot⁡A.

Final answer: sin⁡A=1/√(1+cot²A), tan⁡A=1/cot⁡A, sec⁡A=√(1+cot²A)/cot⁡A.

Example 4

Write all the other trigonometric ratios of ∠A in terms of sec⁡A.

Use 1+tan²A=sec²A, so tan⁡A=√(sec²A−1);

cos⁡A=1/sec⁡A;

sin⁡A=tan⁡A·cos⁡A=√(sec²A−1)/sec⁡A;

csc⁡A=1/sin⁡A=sec⁡A/√(sec²A−1);

cot⁡A=1/tan⁡A=1/√(sec²A−1).

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

Example 5

Given tan⁡θ=34 for an acute angle, find sin⁡θ and cos⁡θ using conversion.

From 1+tan²θ=sec²θ:

sec²θ=1 + 916=2516 ⟹ sec⁡θ=54;

cos⁡θ=1/sec⁡θ=45;

sin⁡θ=tan⁡θ·cos⁡θ=34·45=35.

Final answer: sin⁡θ=35, cos⁡θ=45 — the familiar 333-444-555 triangle, recovered from tangent alone.

Example 6

An angle θ in Quadrant II has sin⁡θ=513. Convert to find cos⁡θ and tan⁡θ.

The conversion gives the magnitude:

cos⁡θ=±√(1−25/169)=±12/13;

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

cos⁡θ=−12/13, tan⁡θ=sin⁡θ/cos⁡θ=5/13−12/13=−5/12.

Final answer: cos⁡θ=−12/13, tan⁡θ=−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.