For an acute angle all six ratios are pos­i­tive, because they are ratios of lengths. For larger angles the point P(x,y)P(x, y) on the ter­mi­nal side moves into other quad­rants, where xx or yy (or both) is neg­a­tive, and some ratios become neg­a­tive. Know­ing which ratios are pos­i­tive in which quad­rant lets us find all six ratios of any angle from just one of them. This les­son sets out the rule and a sim­ple way to remem­ber it.

Where the signs come from

In every def­i­n­i­tion the dis­tance rr is pos­i­tive. So the sign of each ratio depends only on the signs of xx and yy:

  • sin⁡θ=yr\displaystyle \sin\theta = \frac{y}{r} and csc⁡θ=ry\displaystyle \csc\theta = \frac{r}{y} have the sign of yy;
  • cos⁡θ=xr\displaystyle \cos\theta = \frac{x}{r} and sec⁡θ=rx\displaystyle \sec\theta = \frac{r}{x} have the sign of xx;
  • tan⁡θ=yx\displaystyle \tan\theta = \frac{y}{x} and cot⁡θ=xy\displaystyle \cot\theta = \frac{x}{y} are pos­i­tive when xx and yy have the same sign.

The four quad­rants

Quad­rant Angle Signs of x,yx, y Pos­i­tive ratios Neg­a­tive ratios
I 0∘<θ<90∘0^\circ < \theta < 90^\circ +,++, + all six none
II 90∘<θ<180∘90^\circ < \theta < 180^\circ −,+-, + sin⁡\sin, csc⁡\csc the other four
III 180∘<θ<270∘180^\circ < \theta < 270^\circ −,−-, - tan⁡\tan, cot⁡\cot the other four
IV 270∘<θ<360∘270^\circ < \theta < 360^\circ +,−+, - cos⁡\cos, sec⁡\sec the other four

Remem­ber: "All Sil­ver Tea Cups". Going anti­clock­wise from the first quad­rant: All are pos­i­tive, then Sine, then Tangent, then Cosine (each together with its rec­i­p­ro­cal).

Find­ing all ratios from one, with signs

Use a right tri­an­gle (or Pythago­ras' the­o­rem) for the sizes, and the quad­rant for the signs.

Exam­ple 1

If sin⁡θ=−513\displaystyle \sin\theta = -\frac{5}{13} and θ\theta lies in the third quad­rant, find the other ratios.

Sizes: oppo­site 55, hypotenuse 1313, so adja­cent =169−25=12= \sqrt{169 - 25} = 12. In quad­rant III only tan⁡\tan and cot⁡\cot are pos­i­tive.

cos⁡θ=−1213,tan⁡θ=512,cot⁡θ=125,sec⁡θ=−1312,csc⁡θ=−135\displaystyle \cos\theta = -\frac{12}{13}, \quad \tan\theta = \frac{5}{12}, \quad \cot\theta = \frac{12}{5}, \quad \sec\theta = -\frac{13}{12}, \quad \csc\theta = -\frac{13}{5}

Exam­ple 2

If csc⁡θ=−6516\displaystyle \csc\theta = -\frac{65}{16} and θ\theta is not in the fourth quad­rant, find the remain­ing ratios.

csc⁡θ\csc\theta is neg­a­tive, so sin⁡θ=−1665\displaystyle \sin\theta = -\frac{16}{65} is neg­a­tive: θ\theta is in quad­rant III or IV. It is not in IV, so it is in III.

Sizes: oppo­site 1616, hypotenuse 6565, adja­cent =652−162=4225−256=3969=63= \sqrt{65^2 - 16^2} = \sqrt{4225 - 256} = \sqrt{3969} = 63.

cos⁡θ=−6365,tan⁡θ=1663,cot⁡θ=6316,sec⁡θ=−6563\displaystyle \cos\theta = -\frac{63}{65}, \quad \tan\theta = \frac{16}{63}, \quad \cot\theta = \frac{63}{16}, \quad \sec\theta = -\frac{65}{63}

Exam­ple 3

If sec⁡θ+tan⁡θ=1.5\sec\theta + \tan\theta = 1.5, in which quad­rant does θ\theta lie?

sec⁡θ+tan⁡θ=32\displaystyle \sec\theta + \tan\theta = \frac{3}{2}, so sec⁡θ−tan⁡θ=23\displaystyle \sec\theta - \tan\theta = \frac{2}{3}. Adding: 2sec⁡θ=136\displaystyle 2\sec\theta = \frac{13}{6}, so sec⁡θ=1312\displaystyle \sec\theta = \frac{13}{12}. Sub­tract­ing: 2tan⁡θ=56\displaystyle 2\tan\theta = \frac{5}{6}, so tan⁡θ=512\displaystyle \tan\theta = \frac{5}{12}.

Both are pos­i­tive, so cos⁡θ>0\cos\theta > 0 and tan⁡θ>0\tan\theta > 0: θ\theta is in the first quad­rant.

Exam­ple 4

In tri­an­gle ABCABC, 5cos⁡A+3=05\cos A + 3 = 0. Find the qua­dratic equa­tion whose roots are sin⁡A\sin A and tan⁡A\tan A.

cos⁡A=−35\displaystyle \cos A = -\frac{3}{5} is neg­a­tive. An angle of a tri­an­gle lies between 0∘0^\circ and 180∘180^\circ, so AA is obtuse (quad­rant II), where sine is pos­i­tive and tan­gent neg­a­tive.

sin⁡A=45,tan⁡A=−43\displaystyle \sin A = \frac{4}{5}, \qquad \tan A = -\frac{4}{3}

Sum of roots =45−43=12−2015=−815\displaystyle = \frac{4}{5} - \frac{4}{3} = \frac{12 - 20}{15} = -\frac{8}{15}. Prod­uct of roots =45×(−43)=−1615\displaystyle = \frac{4}{5} \times \left(-\frac{4}{3}\right) = -\frac{16}{15}.

The equa­tion is x2−(sum)x+product=0x^2 - (\text{sum})x + \text{product} = 0:

x2+815x−1615=0  ⟹  15x2+8x−16=0\displaystyle x^2 + \frac{8}{15}x - \frac{16}{15} = 0 \implies 15x^2 + 8x - 16 = 0

Prac­tice

  1. Give the sign of (a) sin⁡200∘\sin 200^\circ (b) cos⁡300∘\cos 300^\circ (c) tan⁡100∘\tan 100^\circ (d) sec⁡250∘\sec 250^\circ.
  2. In which quad­rant does θ\theta lie if (a) sin⁡θ<0\sin\theta < 0 and tan⁡θ>0\tan\theta > 0 (b) cos⁡θ<0\cos\theta < 0 and csc⁡θ>0\csc\theta > 0?
  3. If cos⁡θ=−35\displaystyle \cos\theta = -\frac{3}{5} and θ\theta is in the sec­ond quad­rant, find the other five ratios.
  4. If tan⁡θ=−815\displaystyle \tan\theta = -\frac{8}{15} and θ\theta is in the fourth quad­rant, find sin⁡θ\sin\theta and cos⁡θ\cos\theta.
  5. If sec⁡θ=−257\displaystyle \sec\theta = -\frac{25}{7} and θ\theta is not in the sec­ond quad­rant, find sin⁡θ\sin\theta, cos⁡θ\cos\theta and tan⁡θ\tan\theta.
  6. If sin⁡θ=1213\displaystyle \sin\theta = \frac{12}{13} and θ\theta is in the sec­ond quad­rant, find cos⁡θ\cos\theta and tan⁡θ\tan\theta.
  7. If sec⁡θ+tan⁡θ=13\displaystyle \sec\theta + \tan\theta = \frac{1}{3}, in which quad­rant does θ\theta lie?
  8. In tri­an­gle ABCABC, 13cos⁡A+5=013\cos A + 5 = 0. Find the qua­dratic equa­tion whose roots are sin⁡A\sin A and tan⁡A\tan A.

Answers

Show answers
  1. (a) neg­a­tive (III) (b) pos­i­tive (IV) (c) neg­a­tive (II) (d) neg­a­tive (III)
  2. (a) III (b) II
  3. sin⁡θ=45\displaystyle \sin\theta = \frac{4}{5}, tan⁡θ=−43\displaystyle \tan\theta = -\frac{4}{3}, cot⁡θ=−34\displaystyle \cot\theta = -\frac{3}{4}, sec⁡θ=−53\displaystyle \sec\theta = -\frac{5}{3}, csc⁡θ=54\displaystyle \csc\theta = \frac{5}{4}
  4. Hypotenuse 1717: sin⁡θ=−817\displaystyle \sin\theta = -\frac{8}{17}, cos⁡θ=1517\displaystyle \cos\theta = \frac{15}{17}
  5. cos⁡θ=−725\displaystyle \cos\theta = -\frac{7}{25} is neg­a­tive, so II or III; not II, so III. sin⁡θ=−2425\displaystyle \sin\theta = -\frac{24}{25}, tan⁡θ=247\displaystyle \tan\theta = \frac{24}{7}.
  6. cos⁡θ=−513\displaystyle \cos\theta = -\frac{5}{13}, tan⁡θ=−125\displaystyle \tan\theta = -\frac{12}{5}
  7. sec⁡θ−tan⁡θ=3\sec\theta - \tan\theta = 3, so sec⁡θ=53\displaystyle \sec\theta = \frac{5}{3} and tan⁡θ=−43\displaystyle \tan\theta = -\frac{4}{3}. Cosine pos­i­tive, tan­gent neg­a­tive: quad­rant IV.
  8. cos⁡A=−513\displaystyle \cos A = -\frac{5}{13}, so sin⁡A=1213\displaystyle \sin A = \frac{12}{13} and tan⁡A=−125\displaystyle \tan A = -\frac{12}{5}. Sum −9665\displaystyle -\frac{96}{65}, prod­uct −14465\displaystyle -\frac{144}{65}: 65x2+96x−144=065x^2 + 96x - 144 = 0.