Maharashtra State Board · Class 12 · Mathematics · Part I Chapter 3

Trigonometric Functions — Formula Sheet

Board Formulas

18 formulas

1.Principal Solutions
$0 \leq θ < 2 π$
Solutions of a trigonometric equation lying in [0, 2π) are its principal solutions. For sin θ = 1/2 they are π/6 and 5π/6.
2.Equations with Zero Right Side
$sin θ = 0 \Rightarrow θ = nπ, cos θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{2}, tan θ = 0 \Rightarrow θ = nπ$
$n$ : Any integer (n ∈ Z)
Here n ∈ Z. Factorise the equation first and set each factor equal to zero.
3.General Solution of sin θ = sin α★
$sin θ = sin α \Rightarrow θ = nπ + (- 1)^{n} α, n \in Z$
$α$ : A known angle with the same sine (radians)
Choose α in [−π/2, π/2]. For n even the term is +α, for n odd it is −α.
4.General Solution of cos θ = cos α★
$cos θ = cos α \Rightarrow θ = 2 nπ \pm α, n \in Z$
Choose α in [0, π]. Do not mix this up with the sine result.
5.General Solution of tan θ = tan α
$tan θ = tan α \Rightarrow θ = nπ + α, n \in Z$
Choose α in (−π/2, π/2). The period of tan is π, so no ± and no (−1)ⁿ.
6.Squared Forms
$sin^{2} θ = sin^{2} α, cos^{2} θ = cos^{2} α, tan^{2} θ = tan^{2} α \Rightarrow θ = nπ \pm α$
All three squared equations have the same general solution. Useful when the equation contains sin²θ = 1/4 or tan²θ = 3.
7.Polar Co-ordinates
$x = r cos θ, y = r sin θ, r = x^{2} + y^{2}, tan θ = \frac{y}{x}$
$r$ : Radius vector: distance OP of the point from the pole (origin), r > 0 — the pole itself has no polar co-ordinates · $θ$ : Vectorial angle from the polar axis (positive x-axis), 0 ≤ θ < 2π
Use the signs of x and y to decide the quadrant of θ — tan θ alone does not fix it.
8.Sine Rule★
$\frac{a}{s i n A} = \frac{b}{s i n B} = \frac{c}{s i n C} = 2 R$
$a, b, c$ : Sides opposite angles A, B, C · $R$ : Circumradius of the triangle
Writing a = k sin A, b = k sin B, c = k sin C (k = 2R) converts any side expression into angles.
9.Cosine Rule★
$a^{2} = b^{2} + c^{2} - 2 b c cos A, cos A = \frac{b ^{2} + c ^{2} - a ^{2}}{2 b c}$
Similarly cos B = (c² + a² − b²)/(2ca) and cos C = (a² + b² − c²)/(2ab). A negative cosine means an obtuse angle.
10.Projection Rule
$a = b cos C + c cos B, b = c cos A + a cos C, c = a cos B + b cos A$
Each side equals the sum of the projections of the other two sides on it.
11.Half-Angle Formulas★
$sin \frac{A}{2} = \frac{( s - b ) ( s - c )}{b c}, cos \frac{A}{2} = \frac{s ( s - a )}{b c}, tan \frac{A}{2} = \frac{( s - b ) ( s - c )}{s ( s - a )}$
$s$ : Semi-perimeter, s = (a + b + c)/2
Positive square roots are taken because A/2 is acute. Write the formulas for B/2 and C/2 by cycling a → b → c.
12.Area of a Triangle
$Δ = \frac{1}{2} b c sin A = s (s - a) (s - b) (s - c), tan \frac{A}{2} = \frac{Δ}{s ( s - a )}$
$Δ$ : Area of triangle ABC (square units)
Use ½bc sin A when two sides and the included angle are known, Heron's form when all three sides are known.
13.Napier's Analogy
$tan \frac{B - C}{2} = \frac{b - c}{b + c} cot \frac{A}{2}$
Cyclic forms: tan((C − A)/2) = ((c − a)/(c + a)) cot(B/2) and tan((A − B)/2) = ((a − b)/(a + b)) cot(C/2).
14.Principal Value Branches
$sin^{- 1} x \in [- \frac{π}{2}, \frac{π}{2}], cos^{- 1} x \in [0, π], tan^{- 1} x \in (- \frac{π}{2}, \frac{π}{2})$
Also cot⁻¹x ∈ (0, π), sec⁻¹x ∈ [0, π] − {π/2}, cosec⁻¹x ∈ [−π/2, π/2] − {0}. Domain of sin⁻¹ and cos⁻¹ is [−1, 1].
15.Inverse of Negative Arguments
$sin^{- 1} (- x) = - sin^{- 1} x, cos^{- 1} (- x) = π - cos^{- 1} x, tan^{- 1} (- x) = - tan^{- 1} x$
cos⁻¹ and cot⁻¹ follow the 'π minus' rule because their range is not symmetric about 0: cot⁻¹(−x) = π − cot⁻¹x.
16.Complementary Sums
$sin^{- 1} x + cos^{- 1} x = \frac{π}{2}, tan^{- 1} x + cot^{- 1} x = \frac{π}{2}, sec^{- 1} x + cosec^{- 1} x = \frac{π}{2}$
Valid for |x| ≤ 1 (first), all real x (second) and |x| ≥ 1 (third).
17.Sum and Difference of tan⁻¹★
$tan^{- 1} x + tan^{- 1} y = tan^{- 1} \frac{x + y}{1 - x y} (x y < 1), tan^{- 1} x - tan^{- 1} y = tan^{- 1} \frac{x - y}{1 + x y} (x y > - 1)$
Check the condition before using it. If x, y > 0 and xy > 1, the sum is π + tan⁻¹((x + y)/(1 − xy)).
18.Composition with Inverse Functions
$sin (sin^{- 1} x) = x (∣ x ∣ \leq 1), sin^{- 1} (sin θ) = θ (- \frac{π}{2} \leq θ \leq \frac{π}{2})$
Outside the principal range, sin⁻¹(sin θ) is NOT θ: sin⁻¹(sin 5π/6) = π/6.