Proofs and Derivations

Study Sheet

Proofs and Derivations

Trig identities, the binomial theorem, and log rules — from scratch

The Trig Identities, Derived

Tip
Proof: sin⁡2θ+cos⁡2θ=1\sin^2\theta + \cos^2\theta = 1
(4/5, 3/5)cos θ = 4/5sin θ = 3/51
The example: sin⁡θ=3/5\sin\theta = 3/5 puts the point at (4/5,3/5)(4/5, 3/5), and (4/5)2+(3/5)2=1(4/5)^2 + (3/5)^2 = 1.

The unit-circle point (cos⁡θ,sin⁡θ)(\cos\theta, \sin\theta) lies at distance 11 from the origin, and the distance formula squares to cos⁡2θ+sin⁡2θ=1\cos^2\theta + \sin^2\theta = 1 — the Pythagorean theorem wearing trig notation. Dividing by cos⁡2θ\cos^2\theta yields 1+tan⁡2θ=sec⁡2θ1 + \tan^2\theta = \sec^2\theta. ■\blacksquare

In plain terms. It is the Pythagorean theorem on a radius-1 circle. Every "Pythagorean identity" is this one line, re-divided.

Example. sin⁡θ=35\sin\theta = \tfrac35 (Quadrant I) forces cos⁡θ=45\cos\theta = \tfrac45: no triangle needed.

Reminder — Distance and midpoint:  d=(x2−x1)2+(y2−y1)2,M=(x1+x22, y1+y22)d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}, \qquad M=\left(\frac{x_1+x_2}{2},\ \frac{y_1+y_2}{2}\right)
Reminder — The Pythagorean theorem:  a2+b2=c2a^2+b^2=c^2
Reminder — The Pythagorean identity:  sin⁡2θ+cos⁡2θ=1\sin^2\theta+\cos^2\theta=1
Tip
Derivation of the Law of Cosines

Place the triangle with CC at the origin and side aa along the xx-axis: B=(a,0)B = (a, 0) and A=(bcos⁡C,bsin⁡C)A = (b\cos C, b\sin C). The distance AB=cAB = c squares to c2=(bcos⁡C−a)2+(bsin⁡C)2=a2+b2−2abcos⁡Cc^2 = (b\cos C - a)^2 + (b\sin C)^2 = a^2 + b^2 - 2ab\cos C after applying sin⁡2+cos⁡2=1\sin^2 + \cos^2 = 1. ■\blacksquare

In plain terms. Drop the triangle onto coordinates and use the distance formula once — the identity mops up.

Example. Sides 5,75, 7 with a 60∘60^\circ angle between: c2=25+49−70⋅12=39c^2 = 25 + 49 - 70 \cdot \tfrac12 = 39.

Reminder — Law of Cosines:  c2=a2+b2−2abcos⁡Cc^{2}=a^{2}+b^{2}-2ab\cos C
Tip
Derivation of the Law of Sines

The area of a triangle is 12absin⁡C\tfrac12 ab\sin C — computed from ANY vertex. Equating the three versions, 12absin⁡C=12bcsin⁡A=12casin⁡B\tfrac12 ab \sin C = \tfrac12 bc \sin A = \tfrac12 ca \sin B, and dividing through by 12abc\tfrac12 abc gives sin⁡Aa=sin⁡Bb=sin⁡Cc\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}. ■\blacksquare

In plain terms. One area, three formulas for it — setting them equal is the whole proof.

Example. A 30∘30^\circ angle opposite a side of length 44 gives asin⁡A=41/2=8\dfrac{a}{\sin A} = \dfrac{4}{1/2} = 8, so in that triangle every side is 88 times the sine of the angle opposite it: a 90∘90^\circ angle, for instance, sits opposite a side of length 88.

Reminder — Law of Sines:  asin⁡A=bsin⁡B=csin⁡C\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}
Concept
Double angle from the addition formula

Set β=α\beta = \alpha in sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta to get sin⁡2α=2sin⁡αcos⁡α\sin 2\alpha = 2\sin\alpha\cos\alpha; the same move on cosine gives cos⁡2α=cos⁡2α−sin⁡2α\cos 2\alpha = \cos^2\alpha - \sin^2\alpha. Every double- and half-angle identity is the addition formula talking to itself. ■\blacksquare

In plain terms. Learn ONE formula (addition); the rest are its special cases.

Example. sin⁡2θ\sin 2\theta with sin⁡θ=35,cos⁡θ=45\sin\theta = \tfrac35, \cos\theta = \tfrac45: 2⋅35⋅45=24252 \cdot \tfrac35 \cdot \tfrac45 = \tfrac{24}{25}.

The Binomial Theorem, Proved

Tip
Proof by counting committees

Expanding (1+x)n(1 + x)^n means choosing, from each of the nn factors, either the 11 or the xx. A term xkx^k appears once for every way to pick WHICH kk factors contribute an xx — that is (nk)\binom{n}{k} ways. Hence (1+x)n=∑k(nk)xk(1+x)^n = \sum_k \binom{n}{k}x^k. ■\blacksquare

In plain terms. The coefficient counts choices, so it must be "n choose k" — the theorem is combinatorics, not algebra.

Example. The x2x^2 coefficient in (1+x)5(1+x)^5 is (52)=10\binom52 = 10.

Concept
Two identities that fall out immediately

Set x=1x = 1: ∑k(nk)=2n\sum_k \binom{n}{k} = 2^n (every subset picks in or out). Set x=−1x = -1: the alternating sum is 00 — so odd-sized and even-sized subsets are equally many. ■\blacksquare

In plain terms. Plugging numbers into a PROVED identity mass-produces new facts.

Example. (100)+(102)+⋯=2102=512\binom{10}{0} + \binom{10}{2} + \cdots = \tfrac{2^{10}}{2} = 512.

The Log Rules, Derived

Tip
Product and power rules from exponent laws

Let m=log⁡bxm = \log_b x, n=log⁡byn = \log_b y, so x=bmx = b^m, y=bny = b^n. Then xy=bm+nxy = b^{m+n} says exactly log⁡b(xy)=m+n\log_b(xy) = m + n, and xp=bmpx^p = b^{mp} says log⁡b(xp)=plog⁡bx\log_b(x^p) = p\log_b x. Logs ARE exponents, so exponent laws translate one-for-one. ■\blacksquare

In plain terms. Every log rule is an exponent rule read backwards.

Example. log⁡28+log⁡24=3+2=5=log⁡232\log_2 8 + \log_2 4 = 3 + 2 = 5 = \log_2 32.

Tip
Derivation of change of base

From x=blog⁡bxx = b^{\log_b x}, take log⁡c\log_c of both sides and pull the exponent down (power rule): log⁡cx=log⁡bx⋅log⁡cb\log_c x = \log_b x \cdot \log_c b. Divide: log⁡bx=log⁡cxlog⁡cb\log_b x = \dfrac{\log_c x}{\log_c b}. ■\blacksquare

In plain terms. One log in any base is two logs in the base your calculator likes.

Example. log⁡25=ln⁡5ln⁡2≈2.32\log_2 5 = \dfrac{\ln 5}{\ln 2} \approx 2.32.

Reminder — The differentiation rules:  (xn)′=nxn−1,(fg)′=f′g+fg′,(fg)′=f′g−fg′g2,(f(g(x)))′=f′(g(x))g′(x)(x^{n})'=nx^{n-1},\quad (fg)'=f'g+fg',\quad \left(\tfrac{f}{g}\right)'=\frac{f'g-fg'}{g^{2}},\quad \big(f(g(x))\big)'=f'(g(x))g'(x)

Two Derivations the AMC 12 Rewards

Tip
Derivation: the sum and product of roots for any degree

Vieta's relations follow from expanding the factored form.

Reminder — Vieta's formulas:  r1+r2=−ba,r1r2=car_1+r_2=-\frac{b}{a},\qquad r_1 r_2=\frac{c}{a}

Write anxn+⋯+a0=an(x−r1)⋯(x−rn)a_nx^n + \cdots + a_0 = a_n(x - r_1)\cdots(x - r_n) and expand the right side. The coefficient of xn−1x^{n-1} is −an∑ri-a_n\sum r_i and the constant term is an(−1)n∏ria_n(-1)^n\prod r_i; matching coefficients gives ∑ri=−an−1an\sum r_i = -\dfrac{a_{n-1}}{a_n} and ∏ri=(−1)na0an\prod r_i = (-1)^n\dfrac{a_0}{a_n}. ■\blacksquare Every symmetric function of the roots is likewise a coefficient ratio, which is why symmetric expressions never require finding the roots.

Tip
Derivation: the formula for cos⁡\cos of a sum, from rotation

cos⁡(α+β)=cos⁡αcos⁡β−sin⁡αsin⁡β\cos(\alpha+\beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta.

Multiply the unit complex numbers eiα=cos⁡α+isin⁡αe^{i\alpha} = \cos\alpha + i\sin\alpha and eiβ=cos⁡β+isin⁡βe^{i\beta} = \cos\beta + i\sin\beta. The product is (cos⁡αcos⁡β−sin⁡αsin⁡β)+i(sin⁡αcos⁡β+cos⁡αsin⁡β)\left(\cos\alpha\cos\beta - \sin\alpha\sin\beta\right) + i\left(\sin\alpha\cos\beta + \cos\alpha\sin\beta\right), and it also equals ei(α+β)=cos⁡(α+β)+isin⁡(α+β)e^{i(\alpha+\beta)} = \cos(\alpha+\beta) + i\sin(\alpha+\beta). Equating real and imaginary parts gives BOTH sum formulas at once. ■\blacksquare The same one-line argument regenerates the double- and triple-angle formulas, so there is nothing left to memorise.