Inverse trigonometric functions

“A playground seesaw’s tilt angle = arcsin(rise / board length), so arcsin tells the exact angle from the height ratio.”

sin⁻¹x returns the unique angle θ whose sine equals x, so x must be in [–1, 1]. The principal-value range is restricted to –π/2 ≤ θ ≤ π/2, giving arcsin a single-valued output. In JEE keep the answer in radians unless degrees are explicitly required; mixing ranges costs marks.

Memory trick: Arcsin: ‘S’ for ‘Small’ range –π/2 to π/2.

Heights and distances

“Surveyors point a theodolite at a distant tower; the measured elevation angle lets them compute the tower’s height.”

In height-distance problems, draw the right triangle with the line of sight as the hypotenuse. Use tanθ = opposite/adjacent to relate the unknown height to the measured horizontal distance. A trap is using sin or cos instead of tan for elevation/depression angles, giving a wrong dimension.

Memory trick: Elevation: tanθ = height / base.

Polar form of a complex number

“A wind-vane’s length tells the speed (r) and its direction the angle θ; together they give the complex number in polar form.”

Any complex number can be written as z = r(cosθ + i sinθ) where r = |z| ≥0 and θ = arg(z) measured from the positive real axis. For multiplication, moduli multiply and arguments add – a quick exam shortcut.

Memory trick: r = radius, θ = direction

De Moivre's theorem

“Spinning a wheel n times adds up to n times the single turn; De Moivre does the same for angles in complex numbers.”

(cosθ + i sinθ)ⁿ = cos nθ + i sin nθ for any integer n. It lets you raise powers or extract roots by scaling the angle. Do not use it for non-integer exponents unless you first switch to the exponential form.

Memory trick: Power → multiply angle, raise length

Cube roots of unity

“Place three equally spaced points on a clock at 12, 4, 8 o’clock; those positions represent 1, ω, ω², the cube roots of 1.”

The solutions of z³=1 are 1, ω, ω² with ω = e^(2πi/3). They satisfy ω³=1, 1+ω+ω²=0 and lie on the unit circle at 0°, 120°, 240°. Use ω² = –1–ω to simplify expressions.

1+ω+ω²=0

Memory trick: 120° steps on unit circle sum to zero

Common root condition

“A common root is like a single key that can unlock two different locks – the same x-value fits both quadratic equations at zero.”

Two quadratics ax²+bx+c=0 and a' x²+b' x+c'=0 share a root iff their resultant, i.e. the determinant of the 4×4 Sylvester matrix, is zero. This condition is equivalent to the vanishing of the resultant polynomial in a,b,c,a',b',c'. A frequent trap is to use an incorrect simplified formula; always check the resultant or proportionality of coefficients.

Memory trick: Resultant = 0 → common root (R-zero means they’re linked).

Nature of a quadratic graph

“A roller-coaster hill (a>0) creates a valley at its lowest point, while a dip (a<0) creates a peak at its highest point.”

A parabola y=ax²+bx+c opens upward when a>0, giving a minimum at the vertex; it opens downward when a<0, giving a maximum. The vertex y-value is −Δ/(4a). The usual error is to reverse the sign of a when deciding max/min.

Memory trick: a>0 → smile (min), a<0 → frown (max).

Sum of first n natural numbers

“Pairing 1 with n, 2 with n-1, … each gives the same sum n+1, giving the total quickly.”

1+2+…+n = n(n+1)/2. The trap is forgetting the division by 2 or writing n(n-1)/2, which is the sum of first (n-1) numbers, not n.

Σn = n(n+1)/2

Memory trick: Sum = n·(first+last)/2

Sum of squares

“Stacking square tiles to form a stepped pyramid uses the sum-of-squares formula.”

1²+2²+…+n² = n(n+1)(2n+1)/6. Students often mix up the denominator with 3 or 12; the correct factor is exactly 6.

n(n+1)(2n+1)/6

Memory trick: Squares sum = n(n+1)(2n+1)/6

Sum of cubes

“Counting every rectangle on an n×n chessboard gives 1³+2³+…+n³, which equals (1+2+…+n)².”

The sum of the first n cubes equals the square of the sum of the first n natural numbers: 1³+2³+…+n³ = [n(n+1)/2]². It follows from combinatorial rectangle counting or algebraic pairing. A frequent mistake is writing n(n+1)/2 without squaring the whole term.

[n(n+1)/2]²

Memory trick: Cubes sum = (sum of 1 to n)² → (n(n+1)/2)².

Harmonic progression

“If you run the same 5 km each day and your time each day grows by 2 min, the daily speeds form a harmonic progression.”

A sequence is a harmonic progression when the reciprocals of its terms form an arithmetic progression. Hence aₙ = 1/(A+(n-1)d). The trap is assuming the HP itself has a constant difference; only its reciprocals do.

Memory trick: HP ⇔ 1/(AP): aₙ = 1/(A+(n-1)d).

Arithmetic mean between two numbers

“Finding the midpoint of two houses on a road gives the arithmetic mean of their distances.”

The arithmetic mean of a and b is (a+b)/2, the exact midpoint on the number line. It balances the two numbers equally. Trap: forgetting parentheses leads to a+b/2, which is wrong.

(a+b)/2

Memory trick: Midpoint = (a+b)/2

Middle term of a binomial

“A palindrome’s centre letters are the middle terms of a binomial expansion.”

If n is even, there is one middle term at r = n/2. If n is odd, there are two middle terms at r = (n−1)/2 and r = (n+1)/2. Students often swap the even-odd cases.

Memory trick: Even n → r=n/2; odd n → r=(n±1)/2.

Sum of binomial coefficients

“Flipping a fair coin n times gives 2ⁿ outcomes, exactly the sum of all nCr.”

Σ_(r=0)ⁿ nCr = (1+1)ⁿ = 2ⁿ. The exam trick is to set x=1 in (1+x)ⁿ. Forgetting the r=0 term loses the +1.

ΣnCr = 2ⁿ

Memory trick: Set x=1 in (1+x)ⁿ → sum = 2ⁿ.

Section formula

“When you cut a rope so the left piece is m parts and the right piece n parts, the cut point’s coordinates follow the same rule.”

For A(x1,y1) and B(x2,y2), if P divides AB such that AP:PB = m:n, then P = ((n·x1 + m·x2)/(m+n), (n·y1 + m·y2)/(m+n)). The coefficient of each endpoint is the opposite segment length. A common trap is swapping m and n, which yields the wrong point.

Memory trick: n with first point, m with second – ‘n-first, m-second’.

Distance of a point from a line

“A laser beam hits a wall at the shortest possible path, which is the perpendicular distance given by the formula.”

The perpendicular distance from (x₀,y₀) to ax+by+c=0 is d = |ax₀+by₀+c| / √(a²+b²). Absolute value removes sign; the denominator comes from the line’s normal vector length. Forgetting the absolute value gives a signed distance, not the required length.

d=|ax₀+by₀+c|/√(a²+b²)

Memory trick: |ax₀+by₀+c| over √(a²+b²).

Angle between two lines

“The angle where two intersecting roads meet is found from their slopes using tanθ = |(m₁−m₂)/(1+m₁m₂)|.”

For lines with slopes m₁ and m₂, tan θ = |(m₁−m₂)/(1+m₁m₂)| gives the acute angle between them. The denominator must be 1 + m₁m₂, not a sum of slopes. Using the wrong sign or omitting the absolute value yields the wrong angle.

Memory trick: tanθ = |Δm|/(1+m₁m₂).

Tangent to a circle

“A pencil touching a coin at one point draws the tangent, which is perpendicular to the radius through that point.”

A tangent touches a circle at exactly one point and the radius drawn to that point is perpendicular to the tangent line. This follows from the shortest-distance property of a radius. In problems, any line through the point is not automatically a tangent; the ⟂ condition must be checked.

Memory trick: Tangent ⟂ radius at point of contact

Ellipse equation

“A planet’s orbit around the Sun traces an ellipse described by x²/a² + y²/b² = 1.”

The equation x²/a² + y²/b² = 1 represents an ellipse centered at the origin with semi-major axis a along the x-axis and semi-minor axis b along the y-axis. Any point (x,y) on it satisfies that the sum of distances to the two foci equals 2a. Remember: a and b are semi-axes, not full lengths; if a<b the major axis is vertical.

x²/a²+y²/b²=1

Memory trick: a = half the longest diameter, b = half the shortest.

Hyperbola equation

“The shape of a cooling tower’s cross-section matches the hyperbola x²/a²−y²/b²=1, opening left-right.”

For the hyperbola x²/a²−y²/b²=1 the transverse axis lies along the x-direction (the term with +). The difference of distances to the foci is constant =2a. A sign mistake (using + for both terms) turns the equation into an ellipse, a frequent pitfall.

x²/a²−y²/b²=1

Memory trick: Hyperbola: one plus, one minus → opens along plus sign

Latus rectum

“The latus-rectum is like the width of a satellite dish at the focus, measured perpendicular to the dish’s axis.”

In any conic the latus rectum is the chord through the focus and perpendicular to the major (or transverse) axis. For a parabola y²=4ax its length is 4a; for ellipse/hyperbola it is 2b²/a. Remember: it must be perpendicular—do not mix with any focal chord.

Memory trick: L = line through focus ⟂ axis

Derivative of sine and cosine

“A pendulum’s angle swings, and its velocity follows cosine, while the acceleration follows –sine.”

d/dx(sin x)=cos x and d/dx(cos x)=−sin x. The derivative of sine leads the function by 90°, cosine leads with a minus sign. Mistake: mixing up the sign for cosine’s derivative.

Memory trick: sin→cos, cos→−sin.

Derivative of e^x and ln x

“A bank account with continuous compounding grows at a rate equal to its balance—just like eˣ.”

d/dx(eˣ)=eˣ; the exponential is its own derivative. For ln x, derivative is 1/x, reflecting the inverse relationship with eˣ. Trap: applying ln rule at x≤0 where it’s undefined.

Memory trick: eˣ self-derivative; ln' = 1/x.

Quotient rule

“Dividing the fuel flow (u) by the wheel radius (v) while both change, the bike’s speed follows (u'v-uv')/v².”

The quotient rule gives the derivative of u/v as (u'v-uv')/v². It comes from writing u·v⁻¹ and applying the product rule. Common trap: forgetting to square v or reversing the sign of the second term.

(u/v)'=(u'v−uv')/v²

Memory trick: Low·dHigh − High·dLow over Low².