Trigonometrical Equation formulas

Every formula JEE Advanced expects you to know for Trigonometrical Equation, with the variables spelled out. Free, no sign-up. Read it before a mock, not during one.

Practise these formulas 319 practice questions

Formulas: 28
Subject: Mathematics
Exam: JEE Advanced
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Trigonometrical Equation, every formula

28 formulas, typeset and free. Print it, or keep it open beside your practice.

General solution of sin θ = k
$sin θ = sin α \Rightarrow θ = nπ + (- 1)^{n} α, n \in Z$
General solution of cos θ = k
$cos θ = cos α \Rightarrow θ = 2 nπ \pm α, n \in Z$
General solution of tan θ = k
$tan θ = tan α \Rightarrow θ = nπ + α, n \in Z$
sin θ = 0
$sin θ = 0 \Rightarrow θ = nπ, n \in Z$
cos θ = 0
$cos θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{2}, n \in Z$
tan θ = 0
$tan θ = 0 \Rightarrow θ = nπ, n \in Z$
sin θ = 1
$sin θ = 1 \Rightarrow θ = (4 n + 1) \frac{π}{2}$
cos θ = 1
$cos θ = 1 \Rightarrow θ = 2 nπ$
cos θ = -1
$cos θ = - 1 \Rightarrow θ = (2 n + 1) π$
sin²θ = sin²α
$sin^{2} θ = sin^{2} α \Rightarrow θ = nπ \pm α$
same for cos²,tan²
Principal value range of sin⁻¹
$sin θ = k \Rightarrow α \in [- \frac{π}{2}, \frac{π}{2}]$
Principal value range of cos⁻¹
$cos θ = k \Rightarrow α \in [0, π]$
Principal value range of tan⁻¹
$tan θ = k \Rightarrow α \in (- \frac{π}{2}, \frac{π}{2})$
Solvability of a cosθ+b sinθ=c
$a cos θ + b sin θ = c solvable \Leftrightarrow c^{2} \leq a^{2} + b^{2}$
No solution condition
$∣ c ∣ > a^{2} + b^{2} \Rightarrow no solution$
Check |k|≤1
$sin θ = k or cos θ = k needs ∣ k ∣ \leq 1$
Common solution of two equations
$sin θ = sin α, cos θ = cos α \Rightarrow θ = 2 nπ + α$
cosec θ = k
$csc θ = csc α \Rightarrow θ = nπ + (- 1)^{n} α, ∣ k ∣ \geq 1$
sec θ = k
$sec θ = sec α \Rightarrow θ = 2 nπ \pm α, ∣ k ∣ \geq 1$
cot θ = k
$cot θ = cot α \Rightarrow θ = nπ + α$
tan 15°
$tan 1 5^{\circ} = 2 - 3, cot 1 5^{\circ} = 2 + 3$
sin 18°
$sin 1 8^{\circ} = \frac{5 - 1}{4}$
cos 36°
$cos 3 6^{\circ} = \frac{5 + 1}{4}$
LHS≤k≤RHS extremum trick
$if LHS \leq k \leq RHS, solve LHS = RHS = k$
Reject undefined values
$discard θ making any term/denominator undefined$
Avoid squaring
$squaring may introduce extraneous roots — verify$
Don't cancel common factor
$cancelling f (θ) can lose solutions f (θ) = 0$
a cosθ+b sinθ as R form
$a cos θ + b sin θ = R cos (θ - ϕ), R = a^{2} + b^{2}$

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Trigonometrical Equation formulas

Trigonometrical Equation, every formula

General solution of sin θ = k

General solution of cos θ = k

General solution of tan θ = k

sin θ = 0

cos θ = 0

tan θ = 0

sin θ = 1

cos θ = 1

cos θ = -1

sin²θ = sin²α

Principal value range of sin⁻¹

Principal value range of cos⁻¹

Principal value range of tan⁻¹

Solvability of a cosθ+b sinθ=c

No solution condition

Check |k|≤1

Common solution of two equations

cosec θ = k

sec θ = k

cot θ = k

tan 15°

sin 18°

cos 36°

LHS≤k≤RHS extremum trick

Reject undefined values

Avoid squaring

Don't cancel common factor

a cosθ+b sinθ as R form

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