Mt20 - Proving trigonometric identities

Mt20 - Proving trigonometric identities

Build a general argument that explains why two trigonometric expressions agree. Learn to choose a productive starting side, handle fractions and shifted angles, and read a proposed proof critically. The algebra must preserve the domain as well as the answer.

Planning a proof and its domain

Does simplifying a fraction tell you where the fraction exists? Compare sin⁡2x/(1+cos⁡2x)\sin2x/(1+\cos2x) with tan⁡x\tan x. At x=0x=0, both are zero; at x=π/2x=\pi/2, neither exists. Now compare (1−cos⁡2x)/sin⁡2x(1-\cos2x)/\sin2x with the same tangent: at zero, only tangent exists. Similar-looking identities can have different exclusions.

You already know

sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1, tan⁡x=sin⁡x/cos⁡x\tan x=\sin x/\cos x and cot⁡x=cos⁡x/sin⁡x\cot x=\cos x/\sin x. Double-angle identities follow by using equal angles in compound-angle formulas. Division by zero is never allowed.

A useful proof begins with the side containing more structure and ends with the other side. Alternatively, transform both sides independently to one common expression. You can explore from the target while planning, but the written proof must not assume what it is supposed to establish.

Common domain

The inputs for which both expressions are defined. An identity states equality for every input in this common domain. Agreement at several selected angles is evidence for a conjecture, not a proof of it.

Choose the double-angle version that exposes a factor already present in the fraction.

Double-angle substitutions

For every real xx,

sin⁡2x=2sin⁡xcos⁡x,1+cos⁡2x=2cos⁡2x,1−cos⁡2x=2sin⁡2x\sin2x=2\sin x\cos x,\qquad 1+\cos2x=2\cos^2x,\qquad 1-\cos2x=2\sin^2x

They are derived from the compound-angle formulas printed in the sheet, rather than printed as separate results.

Not in the formula sheet: learn it.

The following pair differs in one denominator sign. That small change determines which inputs are excluded.

Worked example 1: matching a cosine square

Question. Prove sin⁡2x1+cos⁡2x=tan⁡x\dfrac{\sin2x}{1+\cos2x}=\tan x, stating its common domain in radians.

Plan. Express the denominator as a cosine square and cancel a nonzero cosine.

The denominator 1+cos⁡2x=2cos⁡2x1+\cos2x=2\cos^2x is nonzero exactly when cos⁡x≠0\cos x\ne0. This is also tangent's domain. Therefore

sin⁡2x1+cos⁡2x=2sin⁡xcos⁡x2cos⁡2xdouble-angle substitutions=sin⁡xcos⁡xcos⁡x≠0=tan⁡xdefinition\begin{aligned} \frac{\sin2x}{1+\cos2x} &=\frac{2\sin x\cos x}{2\cos^2x} && \text{double-angle substitutions}\\ &=\frac{\sin x}{\cos x} && \cos x\ne0\\ &=\tan x && \text{definition} \end{aligned}

Why this step? Keeping one cosine in the denominator produces exactly the tangent definition.

Answer. The identity holds for x≠π/2+kπx\ne\pi/2+k\pi, k∈Zk\in\mathbb Z.

Check. At zero both sides are zero; at π/4\pi/4 both equal 11.

Change just the denominator's plus to minus. Its square becomes a sine square.

Worked example 2: matching a sine square

Question. Prove sin⁡2x1−cos⁡2x=cot⁡x\dfrac{\sin2x}{1-\cos2x}=\cot x, stating its common domain.

Plan. Use 1−cos⁡2x=2sin⁡2x1-\cos2x=2\sin^2x and cancel a nonzero sine.

Both sides exist exactly when sin⁡x≠0\sin x\ne0. On that domain,

sin⁡2x1−cos⁡2x=2sin⁡xcos⁡x2sin⁡2xdouble-angle substitutions=cos⁡xsin⁡xsin⁡x≠0=cot⁡xdefinition\begin{aligned} \frac{\sin2x}{1-\cos2x} &=\frac{2\sin x\cos x}{2\sin^2x} && \text{double-angle substitutions}\\ &=\frac{\cos x}{\sin x} && \sin x\ne0\\ &=\cot x && \text{definition} \end{aligned}

Why this step? The changed sign determines the cancelled factor and the resulting reciprocal family.

Answer. The common domain is x≠kπx\ne k\pi, k∈Zk\in\mathbb Z.

Check. At π/2\pi/2, the fraction and cotangent are both zero; this input is allowed here.

Compare the quotient (1−cos⁡2x)/sin⁡2x(1-\cos2x)/\sin2x. Its numerator becomes 2sin⁡2x2\sin^2x, so it also simplifies to tangent, but only on sin⁡2x≠0\sin2x\ne0. Likewise (1+cos⁡2x)/sin⁡2x=cot⁡x(1+\cos2x)/\sin2x=\cot x retains this smaller domain. Cancelling does not fill the missing inputs.

Watch out

Wrong: simplify to tangent, then give tangent's domain without checking the starting fraction. Right: find the common domain first and retain every original exclusion.

Change the input scale in the second example and determine the exclusions again.

Rational expressions and conjugate factors

Conjugate denominators often turn an awkward sum of fractions into a familiar squared reciprocal.

The factors 1−sin⁡x1-\sin x and 1+sin⁡x1+\sin x multiply to 1−sin⁡2x=cos⁡2x1-\sin^2x=\cos^2x. Similarly, (sec⁡x+tan⁡x)(sec⁡x−tan⁡x)=sec⁡2x−tan⁡2x=1(\sec x+\tan x)(\sec x-\tan x)=\sec^2x-\tan^2x=1, wherever cos⁡x≠0\cos x\ne0. That is a neat surprise: adding or subtracting tangent from secant produces a reciprocal pair.

Reciprocal Pythagorean identities

sec⁡2x−tan⁡2x=1(cos⁡x≠0)\sec^2x-\tan^2x=1\qquad(\cos x\ne0) cosec⁡2x−cot⁡2x=1(sin⁡x≠0)\operatorname{cosec}^2x-\cot^2x=1\qquad(\sin x\ne0)

Divide the unit-circle identity by the appropriate nonzero square to derive them.

Not in the formula sheet: learn it.

A common denominator makes the cancellation transparent; do not cross out terms across a plus sign.

Worked example 3: adding conjugate fractions

Question. Prove 11−sin⁡x+11+sin⁡x=2sec⁡2x\dfrac1{1-\sin x}+\dfrac1{1+\sin x}=2\sec^2x, stating the common domain.

Plan. Combine the fractions and replace the difference of squares by cos⁡2x\cos^2x.

Both denominators are nonzero exactly when sin⁡x≠±1\sin x\ne\pm1, or cos⁡x≠0\cos x\ne0. Thus

11−sin⁡x+11+sin⁡x=(1+sin⁡x)+(1−sin⁡x)1−sin⁡2xcommon denominator=2cos⁡2xcancel numerator terms=2sec⁡2xreciprocal definition\begin{aligned} \frac1{1-\sin x}+\frac1{1+\sin x} &=\frac{(1+\sin x)+(1-\sin x)}{1-\sin^2x} && \text{common denominator}\\ &=\frac2{\cos^2x} && \text{cancel numerator terms}\\ &=2\sec^2x && \text{reciprocal definition} \end{aligned}

Why this step? The conjugate denominator product becomes one square, and addition removes the sine terms upstairs.

Answer. The identity holds on cos⁡x≠0\cos x\ne0.

Check. At x=π/6x=\pi/6, the left side is 2+2/3=8/32+2/3=8/3; the right side is 2(4/3)=8/32(4/3)=8/3.

Change only the operation between the two fractions from addition to subtraction.

Worked example 4: subtracting conjugate fractions

Question. Prove 11−sin⁡x−11+sin⁡x=2sec⁡xtan⁡x\dfrac1{1-\sin x}-\dfrac1{1+\sin x}=2\sec x\tan x on its common domain.

Plan. Use the same denominator but subtract the entire second numerator.

On cos⁡x≠0\cos x\ne0,

11−sin⁡x−11+sin⁡x=(1+sin⁡x)−(1−sin⁡x)1−sin⁡2xcommon denominator=2sin⁡xcos⁡2xsimplify=2(1cos⁡x)(sin⁡xcos⁡x)split the factors=2sec⁡xtan⁡xdefinitions\begin{aligned} \frac1{1-\sin x}-\frac1{1+\sin x} &=\frac{(1+\sin x)-(1-\sin x)}{1-\sin^2x} && \text{common denominator}\\ &=\frac{2\sin x}{\cos^2x} && \text{simplify}\\ &=2\left(\frac1{\cos x}\right)\left(\frac{\sin x}{\cos x}\right) && \text{split the factors}\\ &=2\sec x\tan x && \text{definitions} \end{aligned}

Why this step? The changed operation preserves the sine terms rather than cancelling them; the denominator still supplies two cosine factors.

Answer. The identity is proved on cos⁡x≠0\cos x\ne0.

Check. At π/6\pi/6, both sides equal 4/34/3.

The same conjugate mechanism also proves the useful single-fraction identity

sec⁡x−tan⁡x=1−sin⁡xcos⁡x=1−sin⁡2xcos⁡x(1+sin⁡x)=cos⁡x1+sin⁡x\begin{aligned} \sec x-\tan x &=\frac{1-\sin x}{\cos x} =\frac{1-\sin^2x}{\cos x(1+\sin x)} =\frac{\cos x}{1+\sin x} \end{aligned}

on cos⁡x≠0\cos x\ne0. Interchange sine and cosine to obtain cosec⁡x−cot⁡x=sin⁡x/(1+cos⁡x)\operatorname{cosec}x-\cot x=\sin x/(1+\cos x) on sin⁡x≠0\sin x\ne0. In each case the introduced conjugate is nonzero on the stated domain.

Watch out

Wrong: (1+sin⁡x)−(1−sin⁡x)=0(1+\sin x)-(1-\sin x)=0. Right: distribute the minus to both terms to get 2sin⁡x2\sin x. Brackets protect the very sign the proof depends on.

Change the trigonometric family while retaining the subtraction structure.

Continue this lesson.

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