Trigonometry — CDS Elementary Mathematics

Formulas (2)

• Degree and radian measure, and arc length · Degree–radian conversion and arc length
  $$\pi \text{ rad} = 180^\circ, \qquad s = r\theta \;(\theta \text{ in radians})$$
• What values a ratio can take, and its sign · Ranges
  $$|\sin\theta|, |\cos\theta| \le 1, \qquad |\sec\theta|, |\operatorname{cosec}\theta| \ge 1$$

Reference tables (1)

Watch out for (4)

• Arc length needs the angle in radians
  → Degree and radian measure, and arc length
• One radian is not a small angle
  → Degree and radian measure, and arc length
• Rationalise before comparing
  → Ratios of the standard angles
• A larger cosine means a smaller angle
  → What values a ratio can take, and its sign

Formulas (3)

• Sine, cosine and tangent as quotients of sides · The three primary ratios
  $$\sin\theta = \frac{\text{opp}}{\text{hyp}}, \quad \cos\theta = \frac{\text{adj}}{\text{hyp}}, \quad \tan\theta = \frac{\text{opp}}{\text{adj}}$$
• From one given ratio to every other ratio · Third side from a Pythagorean pair
  $$\sin\theta = \frac{m^2-n^2}{m^2+n^2} \;\Rightarrow\; \cos\theta = \frac{2mn}{m^2+n^2} \quad (\theta \text{ acute})$$
• Right triangles hidden in other figures · Two tools that recur
  $$\text{chord} = 2r\sin\theta, \qquad \text{Area} = \tfrac12 ab\sin C$$

Watch out for (3)

• Name the sides from the angle asked about
  → Sine, cosine and tangent as quotients of sides
• The triangle gives sizes; the quadrant gives signs
  → From one given ratio to every other ratio
• The area formula can hide an obtuse angle
  → Right triangles hidden in other figures

Formulas (3)

• The co-function rule: sin(90° − θ) = cos θ · Co-function rule
  $$\sin(90^\circ-\theta) = \cos\theta, \quad \tan(90^\circ-\theta) = \cot\theta, \quad \sec(90^\circ-\theta) = \operatorname{cosec}\theta$$
• Pairing the terms of a long product or sum · Pairs that collapse
  $$\tan\theta\,\tan(90^\circ-\theta) = 1, \qquad \sin^2\theta + \sin^2(90^\circ-\theta) = 1$$
• tan A = cot B, and angles of a triangle · Complementary condition
  $$\tan A = \cot B \;\Rightarrow\; A + B = 90^\circ \quad (A, B \text{ acute})$$

Watch out for (3)

• Check the angles really add to 90°
  → The co-function rule: sin(90° − θ) = cos θ
• Count the terms before pairing
  → Pairing the terms of a long product or sum
• The rule needs acute angles — check the range
  → tan A = cot B, and angles of a triangle

Formulas (3)

• The three Pythagorean identities · Pythagorean identities
  $$\sin^2\theta + \cos^2\theta = 1, \quad 1 + \tan^2\theta = \sec^2\theta, \quad 1 + \cot^2\theta = \operatorname{cosec}^2\theta$$
• Rewrite in sine and cosine, then factor · The factorisations that recur
  $$\sin^3\theta \pm \cos^3\theta = (\sin\theta \pm \cos\theta)(1 \mp \sin\theta\cos\theta)$$
• Deciding whether a statement is an identity · The test
  $$L(\theta_0) \ne R(\theta_0) \text{ for one } \theta_0 \;\Rightarrow\; \text{not an identity}$$

Watch out for (3)

• Watch the sign in the rearranged form
  → The three Pythagorean identities
• Which function you eliminate decides the sign
  → Rewrite in sine and cosine, then factor
• Testing at 45° proves nothing
  → Deciding whether a statement is an identity

Formulas (3)

• sec θ + tan θ and sec θ − tan θ are reciprocals · The reciprocal pair
  $$\sec\theta + \tan\theta = k \;\Rightarrow\; \sec\theta - \tan\theta = \frac1k, \quad \sec\theta = \frac12\left(k + \frac1k\right)$$
• (1 + sin θ)/cos θ is sec θ + tan θ · The disguised pair
  $$\frac{1 \pm \sin\theta}{\cos\theta} = \sec\theta \pm \tan\theta, \qquad \frac{1 \pm \cos\theta}{\sin\theta} = \operatorname{cosec}\theta \pm \cot\theta$$
• Replacing the 1 in (tan θ + sec θ − 1)/(tan θ − sec θ + 1) · The standing result
  $$\frac{\tan\theta + \sec\theta - 1}{\tan\theta - \sec\theta + 1} = \sec\theta + \tan\theta = \frac{1 + \sin\theta}{\cos\theta}$$

Watch out for (3)

• Half the sum gives sec, half the difference gives tan — not the other way
  → sec θ + tan θ and sec θ − tan θ are reciprocals
• A square root returns a modulus
  → (1 + sin θ)/cos θ is sec θ + tan θ
• Cross-multiplying works, but costs three minutes
  → Replacing the 1 in (tan θ + sec θ − 1)/(tan θ − sec θ + 1)

Formulas (3)

• Square the given sum to get sin θ cos θ · The product from the sum
  $$\sin\theta + \cos\theta = k \;\Rightarrow\; \sin\theta\cos\theta = \frac{k^2 - 1}{2}, \quad \tan\theta + \cot\theta = \frac{1}{\sin\theta\cos\theta}$$
• sin⁴ + cos⁴ and sin⁶ + cos⁶ in terms of sin θ cos θ · Power identities
  $$\sin^4\theta + \cos^4\theta = 1 - 2\sin^2\theta\cos^2\theta, \qquad \sin^6\theta + \cos^6\theta = 1 - 3\sin^2\theta\cos^2\theta$$
• a sin θ + b cos θ and its partner a cos θ − b sin θ · Partner identity
  $$(a\sin\theta + b\cos\theta)^2 + (a\cos\theta - b\sin\theta)^2 = a^2 + b^2$$

Watch out for (3)

• Squaring loses the sign
  → Square the given sum to get sin θ cos θ
• It is minus, not plus
  → sin⁴ + cos⁴ and sin⁶ + cos⁶ in terms of sin θ cos θ
• When only one sign is printed, it is not the only answer
  → a sin θ + b cos θ and its partner a cos θ − b sin θ

Formulas (4)

• Reduce to one ratio and solve the quadratic · The substitutions
  $$\cos^2\theta = 1 - \sin^2\theta, \qquad \tan^2\theta = \sec^2\theta - 1$$
• Equations in a ratio and its reciprocal · Clearing the reciprocal
  $$t + \frac1t = c \;\Rightarrow\; t^2 - ct + 1 = 0$$
• Linear equations a sin θ + b cos θ = c · Substitute into the identity
  $$a\sin\theta + b\cos\theta = c, \quad \sin^2\theta + \cos^2\theta = 1$$
• Systems in two or three angles · Sum of the three combinations
  $$(B + C - A) + (C + A - B) + (A + B - C) = A + B + C$$

Watch out for (4)

• A strict interval can exclude the only root
  → Reduce to one ratio and solve the quadratic
• Both roots of tan θ + cot θ = c are real angles
  → Equations in a ratio and its reciprocal
• The quadratic always offers a root that fails
  → Linear equations a sin θ + b cos θ = c
• Use the range to choose the angle, not the calculator's first answer
  → Systems in two or three angles

Formulas (2)

• The addition formulas · Addition formulas
  $$\sin(A+B) = \sin A\cos B + \cos A\sin B, \quad \cos(A+B) = \cos A\cos B - \sin A\sin B$$
• Double-angle forms · Double-angle forms
  $$\sin 2\theta = 2\sin\theta\cos\theta, \quad \cos 2\theta = 1 - 2\sin^2\theta, \quad \tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$

Watch out for (2)

• cos(A + B) has a minus sign
  → The addition formulas
• tan 2θ can be negative for an acute θ
  → Double-angle forms

Formulas (5)

• Expressions linear in sin²θ or sin θ · The two ends
  $$a\sin^2\theta + b\cos^2\theta \in [\min(a,b),\, \max(a,b)]$$
• t + 1/t ≥ 2, and weighted forms by AM–GM · The bounds
  $$t + \frac1t \ge 2, \qquad a\sec^2\theta + b\operatorname{cosec}^2\theta \ge (\sqrt a + \sqrt b)^2$$
• Quadratics in sin θ or cos²θ · Vertex, then ends
  $$c^2 - c + 1 = \left(c - \tfrac12\right)^2 + \tfrac34, \quad c \in [0, 1]$$
• a sin θ + b cos θ is at most √(a² + b²) · Amplitude bound
  $$|a\sin\theta + b\cos\theta| \le \sqrt{a^2 + b^2}$$
• Equations that can never hold · The key bound
  $$\left|x + \frac1x\right| \ge 2, \qquad |\sin\theta|, |\cos\theta| \le 1$$

Watch out for (5)

• A restricted range moves the ends
  → Expressions linear in sin²θ or sin θ
• On an open interval the bound may not be reached
  → t + 1/t ≥ 2, and weighted forms by AM–GM
• The vertex may lie outside the variable's range
  → Quadratics in sin θ or cos²θ
• Check whether the given value IS the maximum
  → a sin θ + b cos θ is at most √(a² + b²)
• A secant CAN exceed 1 — the bound runs the other way
  → Equations that can never hold

Formulas (4)

• Square and add · The cancelling squares
  $$(a\cos\theta + b\sin\theta)^2 + (a\sin\theta - b\cos\theta)^2 = a^2 + b^2$$
• Rewrite each quantity in sine and cosine, then combine · The two reductions that recur
  $$\operatorname{cosec}\theta - \sin\theta = \frac{\cos^2\theta}{\sin\theta}, \qquad \sec\theta - \cos\theta = \frac{\sin^2\theta}{\cos\theta}$$
• Substitution chains: sin x + sin²x = 1 · The swap and the grouping
  $$\sin x + \sin^2 x = 1 \Rightarrow \sin x = \cos^2 x, \qquad \cos^2 x(\cos^2 x + 1) = 1$$
• Solving p sin²α + q cos²α = m for tan²α · The weighted-average solution
  $$p\sin^2\alpha + q\cos^2\alpha = m \;\Rightarrow\; \tan^2\alpha = \frac{m - q}{p - m}$$

Watch out for (4)

• Add for sine–cosine, subtract for secant–tangent
  → Square and add
• Square roots need the sign of θ's quadrant
  → Rewrite each quantity in sine and cosine, then combine
• Swap the square, not the first power
  → Substitution chains: sin x + sin²x = 1
• Keep the sign pattern of the fraction
  → Solving p sin²α + q cos²α = m for tan²α