Trigonometry

Chapter 3 · Mathematics · 4 questions in the paper · 5 min read

Trigonometry

What the syllabus asks for 3
  • Trigonometric equations and general values
  • Inverse trigonometric functions, principal value
  • Properties of triangles, in-centre, ortho-centre and circum-centre, solution of triangles

Trigonometry carries about 4 questions (our estimate). Three kinds dominate: a general solution or a count of solutions in [0,2\pi]; an inverse-trigonometric identity or principal value; and a triangle question (sine rule, cosine rule, area, inradius, circumradius). All three are formula-driven, and the Section B versions simply chain two formulas.

What the paper asks from this chapter
  • General solutions: \sin\theta=\sin\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha; \cos\theta=\cos\alpha\Rightarrow\theta=2n\pi\pm\alpha; \tan\theta=\tan\alpha\Rightarrow\theta=n\pi+\alpha.
  • a\sin\theta+b\cos\theta=c has a solution only if |c|\le\sqrt{a^2+b^2}.
  • Principal-value ranges of \sin^{-1},\cos^{-1},\tan^{-1} and \sin^{-1}x+\cos^{-1}x=\pi/2.
  • \tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} (watch the quadrant when xy>1).
  • Triangle: \frac{a}{\sin A}=2R, cosine rule, \Delta=\sqrt{s(s-a)(s-b)(s-c)}, r=\Delta/s, R=abc/4\Delta.

3.1 Trigonometric equations and general values

Principal solutions lie in [0,2\pi); the general solution adds all periods.

Equation General solution
\sin\theta=0 / \cos\theta=0 / \tan\theta=0 n\pi / (2n+1)\frac{\pi}{2} / n\pi
\sin\theta=\sin\alpha n\pi+(-1)^n\alpha
\cos\theta=\cos\alpha 2n\pi\pm\alpha
\tan\theta=\tan\alpha n\pi+\alpha
\sin^2\theta=\sin^2\alpha (also \cos^2, \tan^2) n\pi\pm\alpha

Method. Reduce to one function (use \sin^2=1-\cos^2 or divide by \cos^2), factorise, solve each factor. Then count solutions in the given interval. For a\sin\theta+b\cos\theta=c write a=r\cos\phi, b=r\sin\phi with r=\sqrt{a^2+b^2} to get r\sin(\theta+\phi)=c; the maximum of the left side is r.

Quadratic in \sin x

Problem. Number of solutions of 2\sin^2x-3\sin x+1=0 in [0,2\pi].

Solution. (2\sin x-1)(\sin x-1)=0: \sin x=\tfrac{1}{2} gives \pi/6,5\pi/6; \sin x=1 gives \pi/2. Three solutions.

Common mistake

When you divide both sides by a trigonometric factor you can lose its zeros: factorise instead. After squaring, check for extra roots.

3.2 Inverse trigonometric functions, principal value

Function Domain Principal range
\sin^{-1}x [-1,1] [-\frac{\pi}{2},\frac{\pi}{2}]
\cos^{-1}x [-1,1] [0,\pi]
\tan^{-1}x \mathrm{R} (-\frac{\pi}{2},\frac{\pi}{2})
\cot^{-1}x \mathrm{R} (0,\pi)
\sec^{-1}x |x|\ge1 [0,\pi]-\{\frac{\pi}{2}\}
\csc^{-1}x |x|\ge1 [-\frac{\pi}{2},\frac{\pi}{2}]-\{0\}

Identities. \sin^{-1}x+\cos^{-1}x=\frac{\pi}{2}; \tan^{-1}x+\cot^{-1}x=\frac{\pi}{2}; \sin^{-1}(-x)=-\sin^{-1}x, \cos^{-1}(-x)=\pi-\cos^{-1}x; \tan^{-1}x+\tan^{-1}\frac{1}{x}=\frac{\pi}{2} for x>0. 2\tan^{-1}x=\sin^{-1}\frac{2x}{1+x^2} (|x|\le1).

\tan^{-1}x\pm\tan^{-1}y=\tan^{-1}\frac{x\pm y}{1\mp xy}

valid when xy<1 (for +); if x,y>0 and xy>1 add \pi. Example: \tan^{-1}1+\tan^{-1}2+\tan^{-1}3=\pi.

\sin(\sin^{-1}x)=x always (for |x|\le1), but \sin^{-1}(\sin x)=x only on [-\frac{\pi}{2},\frac{\pi}{2}]: \sin^{-1}\sin\frac{5\pi}{6}=\frac{\pi}{6}. For \cos(\sin^{-1}\frac{3}{5}) draw the 3-4-5 triangle: \frac{4}{5}.

3.3 Properties of triangles, in-centre, ortho-centre and circum-centre, solution of triangles

With s=\frac{a+b+c}{2} and area \Delta: - Sine rule \dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}=2R. - Cosine rule \cos A=\dfrac{b^2+c^2-a^2}{2bc}, a^2=b^2+c^2-2bc\cos A. - Projection a=b\cos C+c\cos B. Tangent rule \tan\frac{A-B}{2}=\frac{a-b}{a+b}\cot\frac{C}{2}. - Half angles \sin\frac{A}{2}=\sqrt{\dfrac{(s-b)(s-c)}{bc}}, \cos\frac{A}{2}=\sqrt{\dfrac{s(s-a)}{bc}}, \tan\frac{A}{2}=\dfrac{\Delta}{s(s-a)}=\dfrac{r}{s-a}. - Area \Delta=\frac{1}{2}ab\sin C=\sqrt{s(s-a)(s-b)(s-c)}. - Circumradius R=\dfrac{abc}{4\Delta}; inradius r=\dfrac{\Delta}{s}=4R\sin\frac{A}{2}\sin\frac{B}{2}\sin\frac{C}{2}; exradius r_1=\dfrac{\Delta}{s-a}. - \dfrac{1}{r}=\dfrac{1}{r_1}+\dfrac{1}{r_2}+\dfrac{1}{r_3}, r_1r_2+r_2r_3+r_3r_1=s^2.

Centres. Circumcentre: perpendicular bisectors meet (equidistant from vertices; in a right triangle it is the midpoint of the hypotenuse, so R=c/2). In-centre: angle bisectors meet, at distance r from every side. Orthocentre: altitudes meet; its distance from vertex A is 2R\cos A. Centroid divides each median 2 : 1.

Solving triangles. SSS: cosine rule. SAS: cosine rule then sine rule. ASA: third angle and sine rule. SSA is ambiguous: with A acute and a<b, there are two triangles if a>b\sin A, one right triangle if a=b\sin A, none if a<b\sin A.

Formula and fact sheet

Topic Result
\sin\theta=\sin\alpha \theta=n\pi+(-1)^n\alpha
\cos\theta=\cos\alpha \theta=2n\pi\pm\alpha
max of a\sin\theta+b\cos\theta \sqrt{a^2+b^2}
\sin^{-1}x+\cos^{-1}x \pi/2
\tan^{-1}x+\tan^{-1}y \tan^{-1}\frac{x+y}{1-xy}
2R a/\sin A
\Delta (Heron) \sqrt{s(s-a)(s-b)(s-c)}
r, R \Delta/s, abc/4\Delta

Chapter summary

  • Reduce to one trig function, factorise, then write the general solution of each factor.
  • |c|>\sqrt{a^2+b^2} means no solution to a\sin\theta+b\cos\theta=c.
  • Know the three principal ranges; \sin^{-1}(\sin x) folds x into [-\pi/2,\pi/2].
  • For two triangle quantities, find \Delta first (Heron), then r=\Delta/s or R=abc/4\Delta.
  • A right triangle has R= half the hypotenuse.

Practice questions

Set A — Recall (10 questions)

  1. The general solution of \sin\theta=0 isa2n\pibn\pic(2n+1)\pi/2dn\pi/2
  2. The principal value of \sin^{-1}\frac{1}{2} isa\pi/3b\pi/4c\pi/6d5\pi/6
  3. The principal range of \cos^{-1}x isa[0,\pi]b[-\pi/2,\pi/2]c(0,\pi)d[0,2\pi]
  4. The value of \tan^{-1}1+\tan^{-1}2+\tan^{-1}3 isa\pi/2b3\pi/4c2\pid\pi
  5. In a triangle with a=6 and A=30^\circ, the circumradius R isa3b12c6d6\sqrt{3}
  6. The largest angle of the triangle with sides 3, 5, 7 isa90^\circb120^\circc135^\circd150^\circ
  7. The area of the triangle with sides 13, 14, 15 isa84b90c91d105
  8. The number of solutions of \sin x=\frac{1}{2} in [0,2\pi] isa1b3c4d2
  9. The value of \cos\big(\sin^{-1}\frac{3}{5}\big) isa3/4b5/4c4/5d3/5
  10. The inradius of the 3-4-5 right triangle isa1b2c1.5d3
Answers and explanations — Set A

1. (b) — \sin\theta=0 at every multiple of \pi.

2. (c) — \sin\frac{\pi}{6}=\frac{1}{2} and \frac{\pi}{6}\in[-\frac{\pi}{2},\frac{\pi}{2}]; 5\pi/6 is outside the principal range.

3. (a) — [0,\pi]; [-\pi/2,\pi/2] is for \sin^{-1}.

4. (d) — \tan^{-1}1+\tan^{-1}2=\pi-\tan^{-1}3 (since xy=2>1), so the sum is \pi.

5. (c) — 2R=\frac{a}{\sin A}=\frac{6}{1/2}=12, so R=6.

6. (b) — \cos C=\frac{9+25-49}{30}=-\frac{1}{2}, C=120^\circ.

7. (a) — s=21: \sqrt{21\cdot8\cdot7\cdot6}=\sqrt{7056}=84.

8. (d) — x=\pi/6 and 5\pi/6.

9. (c) — draw the 3-4-5 triangle: adjacent side 4, hypotenuse 5.

10. (a) — \Delta=6, s=6, r=\Delta/s=1.

Set B — Understanding (8 questions)

  1. The general solution of \tan\theta=\sqrt{3} isan\pi+\pi/6bn\pi+\pi/3c2n\pi+\pi/3dn\pi\pm\pi/3
  2. The general solution of \cos2\theta=\cos\theta isan\pi/3bn\pic2n\pi/3d2n\pi
  3. The value of \sin^{-1}\big(\sin\frac{5\pi}{6}\big) isa5\pi/6b-\pi/6c\pi/3d\pi/6
  4. In a triangle a=\sqrt{3}, b=1, c=2. The angle C isa30^\circb45^\circc60^\circd90^\circ
  5. The value of \tan^{-1}\frac{1}{2}+\tan^{-1}\frac{1}{3} isa\pi/4b\pi/2c\pi/3d\pi/6
  6. If the sides of a triangle are in the ratio 2:3:4, the cosine of the largest angle isa1/4b11/16c-1/4d-11/16
  7. The circumradius of the triangle with sides 5, 12, 13 isa5b6.5c6d7.5
  8. The number of real solutions of 3\sin x+4\cos x=6 isa0b1c2dinfinite
Answers and explanations — Set B

1. (b) — \tan\frac{\pi}{3}=\sqrt{3}; the general form is n\pi+\alpha (not 2n\pi\pm\alpha, which is for cosine).

2. (c) — 2\theta=2n\pi\pm\theta gives \theta=2n\pi or \theta=\frac{2n\pi}{3}; the first set is contained in the second.

3. (d) — \sin\frac{5\pi}{6}=\frac{1}{2} and the principal value of \sin^{-1}\frac{1}{2} is \frac{\pi}{6}.

4. (d) — \cos C=\frac{3+1-4}{2\sqrt{3}}=0. Indeed a^2+b^2=c^2.

5. (a) — \frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{6}}=\frac{5/6}{5/6}=1, so \frac{\pi}{4}.

6. (c) — \cos C=\frac{4+9-16}{12}=-\frac{1}{4}.

7. (b) — a right triangle: R=\frac{13}{2}.

8. (a) — the maximum of 3\sin x+4\cos x is \sqrt{9+16}=5<6.

Set C — Application (6 questions)

  1. The number of solutions of 2\sin^2x-3\sin x+1=0 in [0,2\pi] isa2b3c4d1
  2. The value of \sin^{-1}\frac{3}{5}+\sin^{-1}\frac{4}{5} isa\pi/2b\pic\pi/3d3\pi/4
  3. The inradius of the triangle with sides 9, 10, 17 isa3b1.5c4d2
  4. The value of \sin\big(2\tan^{-1}\frac{1}{2}\big) isa3/5b2/5c4/5d5/4
  5. The smallest positive x with \sqrt{3}\cos x+\sin x=2 isa\pi/3b\pi/2c\pi/4d\pi/6
  6. In a triangle a=b=2 and C=120^\circ. The side c isa2\sqrt{2}b4c2\sqrt{3}d8
Answers and explanations — Set C

1. (b) — factorise to (2\sin x-1)(\sin x-1)=0: \frac{\pi}{6},\frac{5\pi}{6},\frac{\pi}{2}.

2. (a) — \sin^{-1}\frac{4}{5}=\cos^{-1}\frac{3}{5}, and \sin^{-1}u+\cos^{-1}u=\frac{\pi}{2}.

3. (d) — s=18, \Delta=\sqrt{18\cdot9\cdot8\cdot1}=36, r=36/18=2.

4. (c) — \sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{1}{5/4}=\frac{4}{5}.

5. (d) — \sqrt{3}\cos x+\sin x=2\cos(x-\frac{\pi}{6})=2, so x=\frac{\pi}{6}.

6. (c) — c^2=4+4-2\cdot4\cos120^\circ=12.