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NDA Mathematics · Formula sheet

Trigonometric Equations formulas

9 formulas and 12 common traps for NDA Mathematics Trigonometric Equations, grouped by subtopic.

Full notes with worked examples

General Solutions & Counting Solutions

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The General-Solution Formulas

General solutions

sin⁡θ=sin⁡α: θ=nπ+(−1)nα;cos⁡θ=cos⁡α: θ=2nπ±α;tan⁡θ=tan⁡α: θ=nπ+α;sin⁡θ=0: θ=nπ;cos⁡θ=0: θ=(2n+1)π2;sin⁡2θ=sin⁡2α: θ=nπ±α\sin\theta=\sin\alpha:\ \theta=n\pi+(-1)^n\alpha; \quad \cos\theta=\cos\alpha:\ \theta=2n\pi\pm\alpha; \quad \tan\theta=\tan\alpha:\ \theta=n\pi+\alpha; \quad \sin\theta=0:\ \theta=n\pi; \quad \cos\theta=0:\ \theta=(2n+1)\tfrac{\pi}{2}; \quad \sin^2\theta=\sin^2\alpha:\ \theta=n\pi\pm\alpha

Reducing an Equation to Standard Form

Co-function reduction

cos⁡θ=sin⁡ ⁣(π2−θ),sin⁡θ=cos⁡ ⁣(π2−θ)\cos\theta = \sin\!\left(\tfrac{\pi}{2} - \theta\right), \quad \sin\theta = \cos\!\left(\tfrac{\pi}{2} - \theta\right)

Counting Solutions in an Interval

Count = solutions of the reduced equation in range

cot⁡2x cot⁡3x=1 ⇒ cos⁡5x=0 ⇒ 5x=(2n+1)π2\cot 2x\,\cot 3x = 1 \ \Rightarrow\ \cos 5x = 0 \ \Rightarrow\ 5x = (2n+1)\tfrac{\pi}{2}

Range & Existence Conditions

Existence bound

asin⁡x=b  solvable  ⟺  ∣b∣≤∣a∣a\sin x = b \ \text{ solvable} \iff |b| \le |a|

Common traps

sin and cos use DIFFERENT general forms

sin⁡θ=sin⁡α\sin\theta=\sin\alpha uses nπ+(−1)nαn\pi+(-1)^n\alpha; cos⁡θ=cos⁡α\cos\theta=\cos\alpha uses 2nπ±α2n\pi\pm\alpha. Swapping them is the most common error and changes which solutions you count.

The (−1)n(-1)^n belongs to SINE only

The alternating factor (−1)n(-1)^n appears ONLY in the sine general solution θ=nπ+(−1)nα\theta=n\pi+(-1)^n\alpha. Cosine uses 2nπ±α2n\pi\pm\alpha and tangent uses nπ+αn\pi+\alpha — neither carries a (−1)n(-1)^n. Writing cos⁡θ=cos⁡α⇒θ=nπ+(−1)nα\cos\theta=\cos\alpha\Rightarrow\theta=n\pi+(-1)^n\alpha is wrong.

The ±\pm is the whole point of the cosine form

cos⁡θ=cos⁡α\cos\theta=\cos\alpha gives BOTH branches 2nπ+α2n\pi+\alpha and 2nπ−α2n\pi-\alpha, so the formula is 2nπ±α2n\pi\pm\alpha. Dropping the ±\pm and writing only 2nπ+α2n\pi+\alpha loses half the solutions.

tan and sin share nπn\pi but differ in the add-on

tan⁡θ=tan⁡α\tan\theta=\tan\alpha gives θ=nπ+α\theta=n\pi+\alpha — note it is nπn\pi (period π\pi), NOT 2nπ2n\pi. Tangent repeats every π\pi, so its general solution steps by π\pi, unlike cosine's 2nπ2n\pi.

Squaring can add false roots — collapse to a half-angle instead when you can

1+cos⁡x=3sin⁡x1 + \cos x = \sqrt3\sin x squared gives a quadratic whose roots include cos⁡x=−1\cos x = -1 — which makes the original csc⁡x+cot⁡x\csc x + \cot x undefined. Substitute every root back before counting. Better: recognise 1+cos⁡xsin⁡x=cot⁡x2\dfrac{1+\cos x}{\sin x} = \cot\tfrac{x}{2} and never square at all.

Dividing by cos⁡x\cos x (or sin⁡x\sin x) can LOSE roots

Going from sin⁡x=cos⁡x\sin x = \cos x to tan⁡x=1\tan x = 1 by dividing by cos⁡x\cos x silently assumes cos⁡x≠0\cos x \ne 0. That is fine here, but in general dividing by a trig factor throws away every solution where that factor is zero — factor instead of divide.

sin⁡x=2\sin x = 2 has NO solution

After reducing, always check the range: sin⁡x\sin x and cos⁡x\cos x live in [−1,1][-1,1]. An equation that forces sin⁡x=2\sin x = 2 (or cos⁡x=−3\cos x = -3) has no real solution at all — don't write a general solution for it.

Scale the interval when the angle is multiplied

For sin⁡2x=k\sin 2x = k on 0≤x<2π0 \le x < 2\pi, substitute u=2xu = 2x so uu runs over [0,4π)[0, 4\pi) — TWICE the length. Counting solutions in the original xx-interval instead of the stretched uu-interval halves the count.

Discard solutions where the equation is undefined

When a reduced equation like cos⁡5x=0\cos 5x = 0 came from cot⁡2x cot⁡3x=1\cot 2x\,\cot 3x = 1, drop any value of xx where cot⁡2x\cot 2x or cot⁡3x\cot 3x blows up — those are not genuine solutions of the original, even though they satisfy the reduced form.

Count integers INCLUSIVELY across the range

For 3cos⁡x=k3\cos x = k the bound is k∈[−3,3]k \in [-3,3], which contains 77 integers −3,…,3-3,\dots,3 — not 66. Endpoints ±3\pm 3 are achievable (at cos⁡x=±1\cos x = \pm 1), so include them; forgetting one or both endpoints is the classic off-by-one.

Specific Forms — Vieta, Products & Logarithms

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Trig Values as Roots of a Quadratic (Vieta)

Vieta + tan-sum

tan⁡(α+β)=tan⁡α+tan⁡β1−tan⁡αtan⁡β=−b/a1−c/a\tan(\alpha+\beta) = \dfrac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta} = \dfrac{-b/a}{1-c/a}

Product & Sum-to-Product Forms

The product identity

(1+tan⁡A)(1+tan⁡B)=2  ⟺  A+B=π4(1+\tan A)(1+\tan B) = 2 \iff A + B = \tfrac{\pi}{4}

Logarithmic & Special Trig Equations

Reciprocal-log trick

t+1t=2  ⟺  t=1t + \tfrac{1}{t} = 2 \iff t = 1

Common traps

Vieta's product is c/ac/a, not −c/a-c/a

For ax2+bx+c=0ax^2+bx+c=0 the sum of roots is −ba-\tfrac{b}{a} (sign flips) but the product is +ca+\tfrac{c}{a} (no sign flip). Putting a minus on the product, or forgetting it on the sum, breaks the tan⁡(α+β)\tan(\alpha+\beta) computation.

A log base must be positive and ≠1\ne 1

In log⁡cos⁡xsin⁡x\log_{\cos x}\sin x the base cos⁡x\cos x must satisfy cos⁡x>0\cos x > 0 and cos⁡x≠1\cos x \ne 1, and the argument needs sin⁡x>0\sin x > 0. After solving tan⁡x=1\tan x = 1, keep only roots in the first quadrant — x=π4x = \tfrac{\pi}{4} — and reject any where the base/argument condition fails.

Simultaneous & Combined Trigonometric Systems

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Solving Two Equations Together

Common solution = intersection

{θ:eqn 1}∩{θ:eqn 2}\{\theta : \text{eqn 1}\} \cap \{\theta : \text{eqn 2}\}

Reducing a Combined System

The s-substitution

s=sin⁡x+cos⁡x ⇒ sin⁡xcos⁡x=s2−12,sin⁡2x=s2−1s = \sin x + \cos x \ \Rightarrow\ \sin x\cos x = \tfrac{s^2-1}{2},\quad \sin 2x = s^2 - 1

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