Angles such as −θ-\theta, 90°±θ90° \pm \theta, 180°±θ180° \pm \theta, 270°±θ270° \pm \theta and 360°−θ360° - \theta are called allied angles of θ\theta. Their trigono­met­ric ratios can always be writ­ten in terms of the ratios of θ\theta itself. This is what lets you find sin⁡150°\sin 150° or cos⁡240°\cos 240° from the table of stan­dard val­ues. This les­son proves the key results, gives the full list and, most use­fully, a short rule that pro­duces any of them with­out mem­o­ris­ing.

Neg­a­tive angles

Result.

sin⁡(−θ)=−sin⁡θ,cos⁡(−θ)=cos⁡θ,tan⁡(−θ)=−tan⁡θ\sin(-\theta) = -\sin\theta, \quad \cos(-\theta) = \cos\theta, \quad \tan(-\theta) = -\tan\theta

cot⁡(−θ)=−cot⁡θ,sec⁡(−θ)=sec⁡θ,csc⁡(−θ)=−csc⁡θ\cot(-\theta) = -\cot\theta, \quad \sec(-\theta) = \sec\theta, \quad \csc(-\theta) = -\csc\theta

Rea­son. Let P(x,y)P(x, y) be on the arm of θ\theta, at dis­tance rr from OO. Turn­ing by −θ-\theta (the same amount, clock­wise) reaches the mir­ror image of PP in the xx-axis, the point P′(x,−y)P'(x, -y), also at dis­tance rr. So

sin⁡(−θ)=−yr=−sin⁡θ,cos⁡(−θ)=xr=cos⁡θ\displaystyle \sin(-\theta) = \frac{-y}{r} = -\sin\theta, \qquad \cos(-\theta) = \frac{x}{r} = \cos\theta

The rest fol­low from these two. Only cos⁡\cos and its rec­i­p­ro­cal sec⁡\sec keep their sign.

Com­ple­men­tary angle 90°−θ90° - \theta

Result.

sin⁡(90°−θ)=cos⁡θ,cos⁡(90°−θ)=sin⁡θ,tan⁡(90°−θ)=cot⁡θ\sin(90° - \theta) = \cos\theta, \quad \cos(90° - \theta) = \sin\theta, \quad \tan(90° - \theta) = \cot\theta

cot⁡(90°−θ)=tan⁡θ,sec⁡(90°−θ)=csc⁡θ,csc⁡(90°−θ)=sec⁡θ\cot(90° - \theta) = \tan\theta, \quad \sec(90° - \theta) = \csc\theta, \quad \csc(90° - \theta) = \sec\theta

Rea­son. In the right tri­an­gle with ver­tices OO, (x,0)(x, 0) and P(x,y)P(x, y), the angle at OO is θ\theta and the angle at PP is 90°−θ90° - \theta. The side oppo­site 90°−θ90° - \theta is xx, and the side next to it is yy. So

sin⁡(90°−θ)=xr=cos⁡θ,cos⁡(90°−θ)=yr=sin⁡θ\displaystyle \sin(90° - \theta) = \frac{x}{r} = \cos\theta, \qquad \cos(90° - \theta) = \frac{y}{r} = \sin\theta

This is where the names come from: "co-sine" is the sine of the com­ple­ment.

The other allied angles

Work­ing the same way with points in the other quad­rants gives the fol­low­ing. Here θ\theta is thought of as an acute angle.

Angle sin⁡\sin cos⁡\cos tan⁡\tan
90°+θ90° + \theta cos⁡θ\cos\theta −sin⁡θ-\sin\theta −cot⁡θ-\cot\theta
180°−θ180° - \theta sin⁡θ\sin\theta −cos⁡θ-\cos\theta −tan⁡θ-\tan\theta
180°+θ180° + \theta −sin⁡θ-\sin\theta −cos⁡θ-\cos\theta tan⁡θ\tan\theta
270°−θ270° - \theta −cos⁡θ-\cos\theta −sin⁡θ-\sin\theta cot⁡θ\cot\theta
270°+θ270° + \theta −cos⁡θ-\cos\theta sin⁡θ\sin\theta −cot⁡θ-\cot\theta
360°−θ360° - \theta −sin⁡θ-\sin\theta cos⁡θ\cos\theta −tan⁡θ-\tan\theta

The rec­i­p­ro­cals fol­low. For exam­ple csc⁡(90°+θ)=sec⁡θ\csc(90° + \theta) = \sec\theta, cot⁡(90°+θ)=−tan⁡θ\cot(90° + \theta) = -\tan\theta, sec⁡(90°+θ)=−csc⁡θ\sec(90° + \theta) = -\csc\theta, sec⁡(180°+θ)=−sec⁡θ\sec(180° + \theta) = -\sec\theta, csc⁡(360°−θ)=−csc⁡θ\csc(360° - \theta) = -\csc\theta and cot⁡(360°−θ)=−cot⁡θ\cot(360° - \theta) = -\cot\theta.

The rule that replaces the table

You do not need to mem­o­rise the table. Three steps pro­duce any entry. Let AA be the allied angle.

  1. If A=90°±θA = 90° \pm \theta or 270°±θ270° \pm \theta, the ratio changes: sin⁡↔cos⁡\sin \leftrightarrow \cos, tan⁡↔cot⁡\tan \leftrightarrow \cot, sec⁡↔csc⁡\sec \leftrightarrow \csc.
  2. If A=180°±θA = 180° \pm \theta or 360°±θ360° \pm \theta, the ratio does not change.
  3. The sign (++ or −-) is the sign of the orig­i­nal ratio in the quad­rant where AA lies, treat­ing θ\theta as acute.

In short: angles on the ver­ti­cal axis (90°90°, 270°270°) change the ratio; angles on the hor­i­zon­tal axis (180°180°, 360°360°) keep it.

Exam­ple 1

Write each in terms of θ\theta: (a) csc⁡(90°+θ)\csc(90° + \theta) (b) sec⁡(180°−θ)\sec(180° - \theta) (c) cot⁡(270°+θ)\cot(270° + \theta) (d) tan⁡(360°−θ)\tan(360° - \theta).

(a) 90°90° is on the ver­ti­cal axis, so csc⁡\csc changes to sec⁡\sec. The angle 90°+θ90° + \theta is in the sec­ond quad­rant, where csc⁡\csc is pos­i­tive. So csc⁡(90°+θ)=+sec⁡θ\csc(90° + \theta) = +\sec\theta.

(b) 180°180° keeps the ratio. 180°−θ180° - \theta is in the sec­ond quad­rant, where sec⁡\sec is neg­a­tive. So sec⁡(180°−θ)=−sec⁡θ\sec(180° - \theta) = -\sec\theta.

(c) 270°270° changes cot⁡\cot to tan⁡\tan. 270°+θ270° + \theta is in the fourth quad­rant, where cot⁡\cot is neg­a­tive. So cot⁡(270°+θ)=−tan⁡θ\cot(270° + \theta) = -\tan\theta.

(d) 360°360° keeps the ratio. 360°−θ360° - \theta is in the fourth quad­rant, where tan⁡\tan is neg­a­tive. So tan⁡(360°−θ)=−tan⁡θ\tan(360° - \theta) = -\tan\theta.

Whole turns

Adding a whole num­ber of com­plete turns brings the arm back to the same place, so noth­ing changes. If n∈Zn \in \mathbb{Z},

sin⁡(n⋅360°+θ)=sin⁡θ,cos⁡(n⋅360°+θ)=cos⁡θ,tan⁡(n⋅360°+θ)=tan⁡θ\sin(n \cdot 360° + \theta) = \sin\theta, \quad \cos(n \cdot 360° + \theta) = \cos\theta, \quad \tan(n \cdot 360° + \theta) = \tan\theta

and, with a minus sign,

sin⁡(n⋅360°−θ)=sin⁡(−θ)=−sin⁡θ\sin(n \cdot 360° - \theta) = \sin(-\theta) = -\sin\theta

cos⁡(n⋅360°−θ)=cos⁡(−θ)=cos⁡θ\cos(n \cdot 360° - \theta) = \cos(-\theta) = \cos\theta

tan⁡(n⋅360°−θ)=tan⁡(−θ)=−tan⁡θ\tan(n \cdot 360° - \theta) = \tan(-\theta) = -\tan\theta

Exam­ple 2

Sim­plify sin⁡(180°+θ)cos⁡(90°−θ)cos⁡(−θ)tan⁡(180°−θ)\displaystyle \frac{\sin(180° + \theta)\cos(90° - \theta)}{\cos(-\theta)\tan(180° - \theta)}.

Replace each ratio:

(−sin⁡θ)(sin⁡θ)(cos⁡θ)(−tan⁡θ)=−sin⁡2θ−cos⁡θ⋅sin⁡θcos⁡θ=−sin⁡2θ−sin⁡θ=sin⁡θ\displaystyle \frac{(-\sin\theta)(\sin\theta)}{(\cos\theta)(-\tan\theta)} = \frac{-\sin^2\theta}{-\cos\theta \cdot \frac{\sin\theta}{\cos\theta}} = \frac{-\sin^2\theta}{-\sin\theta} = \sin\theta

Exam­ple 3

Sim­plify tan⁡(90°+θ)tan⁡(180°−θ)\tan(90° + \theta)\tan(180° - \theta).

(−cot⁡θ)(−tan⁡θ)=cot⁡θtan⁡θ=1(-\cot\theta)(-\tan\theta) = \cot\theta\tan\theta = 1

Prac­tice

Write each in terms of a ratio of θ\theta.

  1. sin⁡(270°+θ)\sin(270° + \theta)
  2. cos⁡(180°+θ)\cos(180° + \theta)
  3. tan⁡(90°−θ)\tan(90° - \theta)
  4. sec⁡(270°−θ)\sec(270° - \theta)
  5. cot⁡(180°+θ)\cot(180° + \theta)
  6. csc⁡(720°−θ)\csc(720° - \theta)
  7. Sim­plify sin⁡(180°−θ)sin⁡(−θ)\displaystyle \frac{\sin(180° - \theta)}{\sin(-\theta)}.
  8. Sim­plify sin⁡(180°+θ)cos⁡(90°−θ)cos⁡(−θ)tan⁡(180°−θ)×cot⁡θ\displaystyle \frac{\sin(180° + \theta)\cos(90° - \theta)}{\cos(-\theta)\tan(180° - \theta)} \times \cot\theta.
  9. Sim­plify cot⁡(90°+θ)cot⁡(180°−θ)\cot(90° + \theta)\cot(180° - \theta).
  10. Sim­plify cos⁡(90°+θ)sec⁡(−θ)tan⁡(180°−θ)sec⁡(360°−θ)sin⁡(180°+θ)cot⁡(90°−θ)\displaystyle \frac{\cos(90° + \theta)\sec(-\theta)\tan(180° - \theta)}{\sec(360° - \theta)\sin(180° + \theta)\cot(90° - \theta)}.

Answers

Show answers
  1. 270°270° changes sin⁡\sin to cos⁡\cos; 270°+θ270° + \theta is in IV where sin⁡\sin is neg­a­tive: −cos⁡θ-\cos\theta.
  2. No change; 180°+θ180° + \theta is in III where cos⁡\cos is neg­a­tive: −cos⁡θ-\cos\theta.
  3. cot⁡θ\cot\theta.
  4. sec⁡\sec changes to csc⁡\csc; 270°−θ270° - \theta is in III where sec⁡\sec is neg­a­tive: −csc⁡θ-\csc\theta.
  5. No change; in III cot⁡\cot is pos­i­tive: cot⁡θ\cot\theta.
  6. 720°=2×360°720° = 2 \times 360°, so csc⁡(720°−θ)=csc⁡(−θ)=−csc⁡θ\csc(720° - \theta) = \csc(-\theta) = -\csc\theta.
  7. sin⁡θ−sin⁡θ=−1\displaystyle \frac{\sin\theta}{-\sin\theta} = -1.
  8. By Exam­ple 2 the frac­tion is sin⁡θ\sin\theta, so the whole is sin⁡θcot⁡θ=cos⁡θ\sin\theta \cot\theta = \cos\theta.
  9. (−tan⁡θ)(−cot⁡θ)=1(-\tan\theta)(-\cot\theta) = 1.
  10. The top is (−sin⁡θ)(sec⁡θ)(−tan⁡θ)=sin⁡θsec⁡θtan⁡θ(-\sin\theta)(\sec\theta)(-\tan\theta) = \sin\theta\sec\theta\tan\theta; the bot­tom is (sec⁡θ)(−sin⁡θ)(tan⁡θ)(\sec\theta)(-\sin\theta)(\tan\theta). The value is −1-1.