Every primary trig ratio has a reciprocal ratio : flip the fraction and you get cosecant, secant, or cotangent. They’re especially useful for simplifying expressions and proving identities . All angles are in degrees .
For an acute angle θ \theta θ
Ratio Name In a right triangle Reciprocal of csc θ \csc\theta csc θ cosecant hypotenuse opposite \dfrac{\text{hypotenuse}}{\text{opposite}} opposite hypotenuse sin θ \sin\theta sin θ sec θ \sec\theta sec θ secant hypotenuse adjacent \dfrac{\text{hypotenuse}}{\text{adjacent}} adjacent hypotenuse cos θ \cos\theta cos θ cot θ \cot\theta cot θ cotangent adjacent opposite \dfrac{\text{adjacent}}{\text{opposite}} opposite adjacent tan θ \tan\theta tan θ
So:
csc θ = 1 sin θ sec θ = 1 cos θ cot θ = 1 tan θ \csc\theta = \frac{1}{\sin\theta} \qquad \sec\theta = \frac{1}{\cos\theta} \qquad \cot\theta = \frac{1}{\tan\theta} csc θ = sin θ 1 sec θ = cos θ 1 cot θ = tan θ 1 A way to remember the pairs: each pair has exactly one “co”: s ine with co secant, co sine with s ecant, tangent with co tangent.
With a point ( x , y ) (x, y) ( x , y ) r = x 2 + y 2 r = \sqrt{x^2 + y^2} r = x 2 + y 2
csc θ = r y sec θ = r x cot θ = x y \csc\theta = \frac{r}{y} \qquad \sec\theta = \frac{r}{x} \qquad \cot\theta = \frac{x}{y} csc θ = y r sec θ = x r cot θ = y x A reciprocal ratio is undefined whenever its denominator is 0 0 0 csc 0 ∘ \csc 0^\circ csc 0 ∘ sin 0 ∘ = 0 \sin 0^\circ = 0 sin 0 ∘ = 0
Calculators don’t have csc \csc csc sec \sec sec cot \cot cot sec 52 ∘ = 1 cos 52 ∘ \sec 52^\circ = \dfrac{1}{\cos 52^\circ} sec 5 2 ∘ = cos 5 2 ∘ 1
Don’t use the sin − 1 \sin^{-1} sin − 1 inverse sine, which finds an angle, not a reciprocal.
In a right triangle, angle A A A 5 5 5 12 12 12 13 13 13 csc A \csc A csc A sec A \sec A sec A cot A \cot A cot A
Solution.
csc A = 13 5 , sec A = 13 12 , cot A = 12 5 \csc A = \frac{13}{5}, \qquad \sec A = \frac{13}{12}, \qquad \cot A = \frac{12}{5} csc A = 5 13 , sec A = 12 13 , cot A = 5 12 Find the exact values of csc 30 ∘ \csc 30^\circ csc 3 0 ∘ sec 45 ∘ \sec 45^\circ sec 4 5 ∘ cot 60 ∘ \cot 60^\circ cot 6 0 ∘
Solution. Flip the special angle values:
csc 30 ∘ = 1 1 2 = 2 , sec 45 ∘ = 1 2 2 = 2 2 = 2 , cot 60 ∘ = 1 3 = 3 3 \csc 30^\circ = \frac{1}{\frac{1}{2}} = 2, \qquad \sec 45^\circ = \frac{1}{\frac{\sqrt{2}}{2}} = \frac{2}{\sqrt{2}} = \sqrt{2}, \qquad \cot 60^\circ = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3} csc 3 0 ∘ = 2 1 1 = 2 , sec 4 5 ∘ = 2 2 1 = 2 2 = 2 , cot 6 0 ∘ = 3 1 = 3 3 Evaluate sec 52 ∘ \sec 52^\circ sec 5 2 ∘ csc 200 ∘ \csc 200^\circ csc 20 0 ∘
Solution.
sec 52 ∘ = 1 cos 52 ∘ ≈ 1.624 , csc 200 ∘ = 1 sin 200 ∘ ≈ − 2.924 \sec 52^\circ = \frac{1}{\cos 52^\circ} \approx 1.624, \qquad \csc 200^\circ = \frac{1}{\sin 200^\circ} \approx -2.924 sec 5 2 ∘ = cos 5 2 ∘ 1 ≈ 1.624 , csc 20 0 ∘ = sin 20 0 ∘ 1 ≈ − 2.924 csc 200 ∘ \csc 200^\circ csc 20 0 ∘ sin 200 ∘ \sin 200^\circ sin 20 0 ∘
Solve sec θ = − 2 \sec\theta = -2 sec θ = − 2 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \le \theta \le 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
Solution. Flip it: cos θ = − 1 2 \cos\theta = -\tfrac{1}{2} cos θ = − 2 1 find the angles : β = 60 ∘ \beta = 60^\circ β = 6 0 ∘
θ = 120 ∘ or θ = 240 ∘ \theta = 120^\circ \qquad \text{or} \qquad \theta = 240^\circ θ = 12 0 ∘ or θ = 24 0 ∘ Mixing up the pairs. Secant goes with cosine , and cosecant goes with sine . It’s easy to guess the other way round.
Using the inverse key. sin − 1 ( 0.5 ) = 30 ∘ \sin^{-1}(0.5) = 30^\circ sin − 1 ( 0.5 ) = 3 0 ∘ csc 30 ∘ = 1 sin 30 ∘ = 2 \csc 30^\circ = \tfrac{1}{\sin 30^\circ} = 2 csc 3 0 ∘ = s i n 3 0 ∘ 1 = 2
Forgetting where they’re undefined. sec 90 ∘ \sec 90^\circ sec 9 0 ∘ csc 180 ∘ \csc 180^\circ csc 18 0 ∘ cos 90 ∘ = 0 \cos 90^\circ = 0 cos 9 0 ∘ = 0 sin 180 ∘ = 0 \sin 180^\circ = 0 sin 18 0 ∘ = 0
Changing the sign. sec θ \sec\theta sec θ cos θ \cos\theta cos θ
1. (Warm-up) A right triangle has sides 8 8 8 θ \theta θ 15 15 15 θ \theta θ 17 17 17 csc θ \csc\theta csc θ sec θ \sec\theta sec θ cot θ \cot\theta cot θ
Solution csc θ = 17 8 \csc\theta = \tfrac{17}{8} csc θ = 8 17 sec θ = 17 15 \sec\theta = \tfrac{17}{15} sec θ = 15 17 cot θ = 15 8 \cot\theta = \tfrac{15}{8} cot θ = 8 15
2. (Warm-up) Find the exact values of csc 90 ∘ \csc 90^\circ csc 9 0 ∘ sec 60 ∘ \sec 60^\circ sec 6 0 ∘ cot 45 ∘ \cot 45^\circ cot 4 5 ∘
Solution csc 90 ∘ = 1 \csc 90^\circ = 1 csc 9 0 ∘ = 1 sec 60 ∘ = 2 \sec 60^\circ = 2 sec 6 0 ∘ = 2 cot 45 ∘ = 1 \cot 45^\circ = 1 cot 4 5 ∘ = 1
3. (Warm-up) Explain why csc 0 ∘ \csc 0^\circ csc 0 ∘
Solution csc 0 ∘ = 1 sin 0 ∘ = 1 0 \csc 0^\circ = \tfrac{1}{\sin 0^\circ} = \tfrac{1}{0} csc 0 ∘ = s i n 0 ∘ 1 = 0 1
4. (Core) Evaluate to three decimal places: csc 35 ∘ \csc 35^\circ csc 3 5 ∘ cot 75 ∘ \cot 75^\circ cot 7 5 ∘ sec 140 ∘ \sec 140^\circ sec 14 0 ∘
Solution csc 35 ∘ ≈ 1.743 \csc 35^\circ \approx 1.743 csc 3 5 ∘ ≈ 1.743 cot 75 ∘ = 1 tan 75 ∘ ≈ 0.268 \cot 75^\circ = \tfrac{1}{\tan 75^\circ} \approx 0.268 cot 7 5 ∘ = t a n 7 5 ∘ 1 ≈ 0.268 sec 140 ∘ ≈ − 1.305 \sec 140^\circ \approx -1.305 sec 14 0 ∘ ≈ − 1.305
5. (Core) The point ( − 8 , 15 ) (-8, 15) ( − 8 , 15 ) θ \theta θ csc θ \csc\theta csc θ sec θ \sec\theta sec θ cot θ \cot\theta cot θ
Solution r = 64 + 225 = 17 r = \sqrt{64 + 225} = 17 r = 64 + 225 = 17
csc θ = 17 15 \csc\theta = \tfrac{17}{15} csc θ = 15 17 sec θ = − 17 8 \sec\theta = -\tfrac{17}{8} sec θ = − 8 17 cot θ = − 8 15 \cot\theta = -\tfrac{8}{15} cot θ = − 15 8
6. (Core) Solve csc θ = 2 \csc\theta = 2 csc θ = 2 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \le \theta \le 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
Solution sin θ = 1 2 \sin\theta = \tfrac{1}{2} sin θ = 2 1 θ = 30 ∘ \theta = 30^\circ θ = 3 0 ∘ 150 ∘ 150^\circ 15 0 ∘
7. (Core) Solve cot θ = − 1 \cot\theta = -1 cot θ = − 1 0 ∘ ≤ θ ≤ 360 ∘ 0^\circ \le \theta \le 360^\circ 0 ∘ ≤ θ ≤ 36 0 ∘
Solution tan θ = − 1 \tan\theta = -1 tan θ = − 1 β = 45 ∘ \beta = 45^\circ β = 4 5 ∘ θ = 135 ∘ \theta = 135^\circ θ = 13 5 ∘ 315 ∘ 315^\circ 31 5 ∘
8. (Challenge) Show that sec 2 45 ∘ − tan 2 45 ∘ = 1 \sec^2 45^\circ - \tan^2 45^\circ = 1 sec 2 4 5 ∘ − tan 2 4 5 ∘ = 1
Solution sec 45 ∘ = 2 \sec 45^\circ = \sqrt{2} sec 4 5 ∘ = 2 tan 45 ∘ = 1 \tan 45^\circ = 1 tan 4 5 ∘ = 1 ( 2 ) 2 − 1 2 = 2 − 1 = 1 (\sqrt{2})^2 - 1^2 = 2 - 1 = 1 ( 2 ) 2 − 1 2 = 2 − 1 = 1
9. (Challenge) Explain the difference between sin − 1 ( 0.5 ) \sin^{-1}(0.5) sin − 1 ( 0.5 ) ( sin 30 ∘ ) − 1 (\sin 30^\circ)^{-1} ( sin 3 0 ∘ ) − 1
Solution sin − 1 ( 0.5 ) \sin^{-1}(0.5) sin − 1 ( 0.5 ) inverse sine : the angle whose sine is 0.5 0.5 0.5 30 ∘ 30^\circ 3 0 ∘
( sin 30 ∘ ) − 1 (\sin 30^\circ)^{-1} ( sin 3 0 ∘ ) − 1 reciprocal of sin 30 ∘ \sin 30^\circ sin 3 0 ∘ 1 0.5 = 2 \tfrac{1}{0.5} = 2 0.5 1 = 2 csc 30 ∘ \csc 30^\circ csc 3 0 ∘
The notation looks similar, but one gives an angle and the other gives a number.
