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Trigonometrical Equation formulas

Master Trigonometrical Equation through 28 JEE Advanced-level formulas, systematically structured with every variable spelled out. Revise concept-wise, identify the areas where you need improvement, and focus your preparation with greater precision.

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Trigonometrical Equation, every formula

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General solution of sin θ = k
$sin θ = sin α \Rightarrow θ = nπ + (- 1)^{n} α, n \in Z$
Q1MCQGeneral sin
The general solution of $\sin\theta=\dfrac12$ is:
- A
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{6}$
- B
  $\theta=2n\pi\pm\dfrac{\pi}{6}$
- C
  $\theta=n\pi+\dfrac{\pi}{6}$
- D
  $\theta=n\pi\pm\dfrac{\pi}{6}$
General solution of cos θ = k
$cos θ = cos α \Rightarrow θ = 2 nπ \pm α, n \in Z$
Q1MCQGeneral cos
The general solution of $\cos\theta=\dfrac12$ is:
- A
  $\theta=2n\pi\pm\dfrac{\pi}{3}$
- B
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{3}$
- C
  $\theta=n\pi\pm\dfrac{\pi}{3}$
- D
  $\theta=2n\pi+\dfrac{\pi}{3}$
General solution of tan θ = k
$tan θ = tan α \Rightarrow θ = nπ + α, n \in Z$
Q1MCQGeneral tan
The general solution of $\tan\theta=1$ is:
- A
  $\theta=n\pi+\dfrac{\pi}{4}$
- B
  $\theta=2n\pi\pm\dfrac{\pi}{4}$
- C
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{4}$
- D
  $\theta=n\pi\pm\dfrac{\pi}{4}$
sin θ = 0
$sin θ = 0 \Rightarrow θ = nπ, n \in Z$
Q1MCQsin=0
The general solution of $\sin\theta=0$ is:
- A
  $\theta=n\pi$
- B
  $\theta=(2n+1)\dfrac{\pi}{2}$
- C
  $\theta=2n\pi$
- D
  $\theta=n\pi+\dfrac{\pi}{2}$
cos θ = 0
$cos θ = 0 \Rightarrow θ = (2 n + 1) \frac{π}{2}, n \in Z$
Q1MCQcos=0
The general solution of $\cos\theta=0$ is:
- A
  $\theta=(2n+1)\dfrac{\pi}{2}$
- B
  $\theta=n\pi$
- C
  $\theta=2n\pi$
- D
  $\theta=n\pi+\dfrac{\pi}{4}$
tan θ = 0
$tan θ = 0 \Rightarrow θ = nπ, n \in Z$
Q1MCQtan=0
The general solution of $\tan\theta=0$ is:
- A
  $\theta=n\pi$
- B
  $\theta=(2n+1)\dfrac{\pi}{2}$
- C
  $\theta=2n\pi$
- D
  $\theta=n\pi+\dfrac{\pi}{2}$
sin θ = 1
$sin θ = 1 \Rightarrow θ = (4 n + 1) \frac{π}{2}$
Q1MCQsin=1
The general solution of $\sin\theta=1$ is:
- A
  $\theta=(4n+1)\dfrac{\pi}{2}$
- B
  $\theta=2n\pi$
- C
  $\theta=n\pi+\dfrac{\pi}{2}$
- D
  $\theta=(2n+1)\pi$
cos θ = 1
$cos θ = 1 \Rightarrow θ = 2 nπ$
Q1MCQcos=1
The general solution of $\cos\theta=1$ is:
- A
  $\theta=2n\pi$
- B
  $\theta=n\pi$
- C
  $\theta=(2n+1)\pi$
- D
  $\theta=(4n+1)\dfrac{\pi}{2}$
cos θ = -1
$cos θ = - 1 \Rightarrow θ = (2 n + 1) π$
Q1MCQcos=-1
The general solution of $\cos\theta=-1$ is:
- A
  $\theta=(2n+1)\pi$
- B
  $\theta=2n\pi$
- C
  $\theta=n\pi$
- D
  $\theta=(4n+1)\dfrac{\pi}{2}$
sin²θ = sin²α
$sin^{2} θ = sin^{2} α \Rightarrow θ = nπ \pm α$
same for cos²,tan²
Q1MCQsin²=sin²
The general solution of $\sin^{2}\theta=\dfrac34$ (i.e. $\sin^{2}\theta=\sin^{2}\dfrac{\pi}{3}$) is:
- A
  $\theta=n\pi\pm\dfrac{\pi}{3}$
- B
  $\theta=2n\pi\pm\dfrac{\pi}{3}$
- C
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{3}$
- D
  $\theta=n\pi+\dfrac{\pi}{3}$
Principal value range of sin⁻¹
$sin θ = k \Rightarrow α \in [- \frac{π}{2}, \frac{π}{2}]$
Q1MCQPrincipal sin range
The principal value $\alpha$ for $\sin\theta=k$ lies in:
- A
  $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
- B
  $[0,\pi]$
- C
  $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
- D
  $(0,\pi)$
Principal value range of cos⁻¹
$cos θ = k \Rightarrow α \in [0, π]$
Q1MCQPrincipal cos range
The principal value $\alpha$ for $\cos\theta=k$ lies in:
- A
  $[0,\pi]$
- B
  $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
- C
  $(0,\pi)$
- D
  $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
Principal value range of tan⁻¹
$tan θ = k \Rightarrow α \in (- \frac{π}{2}, \frac{π}{2})$
Q1MCQPrincipal tan range
The principal value $\alpha$ for $\tan\theta=k$ lies in:
- A
  $\left(-\dfrac{\pi}{2},\dfrac{\pi}{2}\right)$
- B
  $[0,\pi]$
- C
  $\left[-\dfrac{\pi}{2},\dfrac{\pi}{2}\right]$
- D
  $(0,\pi)$
Solvability of a cosθ+b sinθ=c
$a cos θ + b sin θ = c solvable \Leftrightarrow c^{2} \leq a^{2} + b^{2}$
Q1MCQSolvability
$3\cos\theta+4\sin\theta=c$ has a solution iff:
- A
  $c^{2}\le25$
- B
  $c^{2}\le7$
- C
  $c^{2}\ge25$
- D
  $c\le5$ only
No solution condition
$∣ c ∣ > a^{2} + b^{2} \Rightarrow no solution$
Q1MCQNo solution
$\cos\theta+\sin\theta=3$ has:
- A
  no solution
- B
  one solution
- C
  two solutions
- D
  infinitely many
Check |k|≤1
$sin θ = k or cos θ = k needs ∣ k ∣ \leq 1$
Q1MCQ|k|≤1 check
$\sin\theta=2$ has:
- A
  no solution
- B
  $\theta=n\pi+(-1)^{n}2$
- C
  $\theta=2n\pi$
- D
  one solution
Common solution of two equations
$sin θ = sin α, cos θ = cos α \Rightarrow θ = 2 nπ + α$
Q1MCQCommon solution
If $\sin\theta=\sin\alpha$ and $\cos\theta=\cos\alpha$, then:
- A
  $\theta=2n\pi+\alpha$
- B
  $\theta=n\pi+\alpha$
- C
  $\theta=2n\pi\pm\alpha$
- D
  $\theta=n\pi+(-1)^{n}\alpha$
cosec θ = k
$csc θ = csc α \Rightarrow θ = nπ + (- 1)^{n} α, ∣ k ∣ \geq 1$
Q1MCQcosec
The general solution of $\csc\theta=2$ is:
- A
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{6}$
- B
  $\theta=2n\pi\pm\dfrac{\pi}{6}$
- C
  $\theta=n\pi+\dfrac{\pi}{6}$
- D
  no solution
sec θ = k
$sec θ = sec α \Rightarrow θ = 2 nπ \pm α, ∣ k ∣ \geq 1$
Q1MCQsec
The general solution of $\sec\theta=2$ is:
- A
  $\theta=2n\pi\pm\dfrac{\pi}{3}$
- B
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{3}$
- C
  $\theta=n\pi\pm\dfrac{\pi}{3}$
- D
  no solution
cot θ = k
$cot θ = cot α \Rightarrow θ = nπ + α$
Q1MCQcot
The general solution of $\cot\theta=1$ is:
- A
  $\theta=n\pi+\dfrac{\pi}{4}$
- B
  $\theta=2n\pi\pm\dfrac{\pi}{4}$
- C
  $\theta=n\pi+(-1)^{n}\dfrac{\pi}{4}$
- D
  $\theta=n\pi\pm\dfrac{\pi}{4}$
tan 15°
$tan 1 5^{\circ} = 2 - 3, cot 1 5^{\circ} = 2 + 3$
Q1MCQtan 15°
$\tan15^{\circ}$ equals:
- A
  $2-\sqrt3$
- B
  $2+\sqrt3$
- C
  $\sqrt3-1$
- D
  $\dfrac{1}{\sqrt3}$
sin 18°
$sin 1 8^{\circ} = \frac{5 - 1}{4}$
Q1MCQsin 18°
$\sin18^{\circ}$ equals:
- A
  $\dfrac{\sqrt5-1}{4}$
- B
  $\dfrac{\sqrt5+1}{4}$
- C
  $\dfrac{\sqrt5-1}{2}$
- D
  $\dfrac{\sqrt3-1}{2\sqrt2}$
cos 36°
$cos 3 6^{\circ} = \frac{5 + 1}{4}$
Q1MCQcos 36°
$\cos36^{\circ}$ equals:
- A
  $\dfrac{\sqrt5+1}{4}$
- B
  $\dfrac{\sqrt5-1}{4}$
- C
  $\dfrac{\sqrt5+1}{2}$
- D
  $\dfrac{\sqrt3+1}{2\sqrt2}$
LHS≤k≤RHS extremum trick
$if LHS \leq k \leq RHS, solve LHS = RHS = k$
Q1MCQExtremum trick
$\sin x+\csc x=2$ holds when:
- A
  $\sin x=1$ (LHS $\ge2$ with equality there)
- B
  $\sin x=-1$
- C
  $\sin x=0$
- D
  never
Reject undefined values
$discard θ making any term/denominator undefined$
Q1MCQReject undefined
While solving an equation with $\tan\theta$, we must reject:
- A
  $\theta=(2n+1)\dfrac{\pi}{2}$
- B
  $\theta=n\pi$
- C
  $\theta=2n\pi$
- D
  no value
Avoid squaring
$squaring may introduce extraneous roots — verify$
Q1MCQAvoid squaring
Squaring a trigonometric equation may:
- A
  introduce extraneous roots
- B
  lose all roots
- C
  never change solutions
- D
  make it linear
Don't cancel common factor
$cancelling f (θ) can lose solutions f (θ) = 0$
Q1MCQDon't cancel factor
Cancelling $\cos\theta$ from $\sin\theta\cos\theta=\cos\theta$ risks:
- A
  losing the solutions $\cos\theta=0$
- B
  nothing
- C
  gaining extraneous roots
- D
  changing the period
a cosθ+b sinθ as R form
$a cos θ + b sin θ = R cos (θ - ϕ), R = a^{2} + b^{2}$
Q1MCQR form
$a\cos\theta+b\sin\theta$ can be written as $R\cos(\theta-\phi)$ with $R=$:
- A
  $\sqrt{a^{2}+b^{2}}$
- B
  $a+b$
- C
  $a^{2}+b^{2}$
- D
  $\sqrt{ab}$

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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

More JEE Advanced Mathematics formula sheets

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