1.05p - Proving trigonometric identities

1.05p - Proving trigonometric identities

A trigonometric identity can hide a very simple expression inside a complicated fraction. Learn to uncover it with a general proof, choose identities that fit the target, and keep track of the inputs your algebra permits. You will also repair arguments that reach a plausible answer by an invalid step.

Planning a proof and its domain

Can two expressions agree wherever you can compare them, yet have different domains? At x=π/4x=\pi/4, both (1−cos⁡2x)/sin⁡2x(1-\cos2x)/\sin2x and tan⁡x\tan x equal 11. At x=0x=0, tangent exists but the fraction does not. A successful proof explains both the agreement and that missing input.

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. A fraction requires a nonzero denominator. Compound-angle identities give the double-angle formulas.

Begin by asking where the two expressions exist. Then work from the more complicated side towards the simpler one, or transform each independently to a common expression. You may plan backwards on scrap paper; your finished argument must start from known facts.

Trigonometric identity

An equality true for every input in the common domain of its two expressions. Checking a few angles can disprove a false claim, but cannot prove an identity for all allowed angles.

The numerator often tells you which double-angle form will help.

Double-angle forms for a quotient

For real xx,

1−cos⁡2x=2sin⁡2x,1+cos⁡2x=2cos⁡2x,sin⁡2x=2sin⁡xcos⁡x1-\cos2x=2\sin^2x,\qquad 1+\cos2x=2\cos^2x,\qquad \sin2x=2\sin x\cos x

These follow from the supplied compound-angle identities; the double-angle forms themselves are not printed there.

Not in the formula sheet: learn it.

Keep a denominator condition beside the calculation so cancellation cannot quietly enlarge the domain.

Worked example 1: a quotient with a missing input

Question. Prove 1−cos⁡2xsin⁡2x=tan⁡x\dfrac{1-\cos2x}{\sin2x}=\tan x, stating the common domain. Angles are in radians.

Plan. Replace numerator and denominator by double-angle expressions, then cancel a nonzero sine factor.

The common domain requires sin⁡2x≠0\sin2x\ne0, equivalently x≠kπ/2x\ne k\pi/2 for integer kk. Both sin⁡x\sin x and cos⁡x\cos x are then nonzero. On this domain,

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

Why this step? The chosen cosine form creates exactly the sine factor in the denominator.

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

Check. At x=π/4x=\pi/4, both sides are 11. At x=0x=0, the starting fraction is undefined; cancellation has not supplied a value there.

Change just the numerator's minus sign to a plus. The useful squared function changes with it.

Worked example 2: changing the numerator's sign

Question. Prove 1+cos⁡2xsin⁡2x=cot⁡x\dfrac{1+\cos2x}{\sin2x}=\cot x, stating the common domain.

Plan. Use the cosine-square double-angle form and cancel a nonzero cosine factor.

Again sin⁡2x≠0\sin2x\ne0 excludes x=kπ/2x=k\pi/2. On this domain,

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

Why this step? The plus sign makes a cosine square, so cosine is the factor to cancel.

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

Check. At x=π/4x=\pi/4, both sides equal 11. At x=π/2x=\pi/2, cotangent is zero but the starting fraction is undefined.

A complete proof is a chain of justified general statements, not a final expression followed by “as required”.

Watch out

Wrong: cancel a factor and then claim the equality for every input of the simplified expression. Right: retain the starting common domain, including inputs excluded by a cancelled factor.

Now change only the scale of the angle in the second example.

Rational expressions and conjugate factors

A conjugate factor turns a difference involving sine or cosine into a square that cancels cleanly.

Difference of squares in trigonometry

For cos⁡x≠0\cos x\ne0,

(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

This follows by dividing sin⁡2x+cos⁡2x=1\sin^2x+\cos^2x=1 by cos⁡2x\cos^2x.

Not in the formula sheet: learn it.

Here is a surprising consequence: wherever they exist, sec⁡x+tan⁡x\sec x+\tan x and sec⁡x−tan⁡x\sec x-\tan x are reciprocals. They may look unrelated, yet their product is always 11. A conjugate proof explains the cancellation rather than relying on a numerical coincidence.

Worked example 3: a conjugate removes a sine term

Question. Prove sec⁡x−tan⁡x=cos⁡x1+sin⁡x\sec x-\tan x=\dfrac{\cos x}{1+\sin x} on its common domain.

Plan. Write the left side as one fraction, then multiply by a conjugate to produce 1−sin⁡2x1-\sin^2x.

The common domain is cos⁡x≠0\cos x\ne0. This also ensures sin⁡x≠−1\sin x\ne-1. Thus

sec⁡x−tan⁡x=1−sin⁡xcos⁡xreciprocal definitions=(1−sin⁡x)(1+sin⁡x)cos⁡x(1+sin⁡x)1+sin⁡x≠0=cos⁡2xcos⁡x(1+sin⁡x)unit-circle identity=cos⁡x1+sin⁡xcos⁡x≠0\begin{aligned} \sec x-\tan x &=\frac{1-\sin x}{\cos x} && \text{reciprocal definitions}\\ &=\frac{(1-\sin x)(1+\sin x)}{\cos x(1+\sin x)} && 1+\sin x\ne0\\ &=\frac{\cos^2x}{\cos x(1+\sin x)} && \text{unit-circle identity}\\ &=\frac{\cos x}{1+\sin x} && \cos x\ne0 \end{aligned}

Why this step? Multiplying by 1+sin⁡x1+\sin x makes a difference of squares, not a longer unrelated expansion.

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

Check. At x=0x=0, both sides equal 11. Multiplying the result by sec⁡x+tan⁡x\sec x+\tan x gives the reciprocal product above.

Change only the sign between secant and tangent; its conjugate changes sign too.

Worked example 4: the other conjugate

Question. Prove sec⁡x+tan⁡x=cos⁡x1−sin⁡x\sec x+\tan x=\dfrac{\cos x}{1-\sin x} on its common domain.

Plan. Combine the left side and multiply by 1−sin⁡x1-\sin x.

On cos⁡x≠0\cos x\ne0, also 1−sin⁡x≠01-\sin x\ne0, so

sec⁡x+tan⁡x=1+sin⁡xcos⁡xcombine fractions=(1+sin⁡x)(1−sin⁡x)cos⁡x(1−sin⁡x)multiply by 1=cos⁡2xcos⁡x(1−sin⁡x)1−sin⁡2x=cos⁡2x=cos⁡x1−sin⁡xnonzero cancellation\begin{aligned} \sec x+\tan x &=\frac{1+\sin x}{\cos x} && \text{combine fractions}\\ &=\frac{(1+\sin x)(1-\sin x)}{\cos x(1-\sin x)} && \text{multiply by }1\\ &=\frac{\cos^2x}{\cos x(1-\sin x)} && 1-\sin^2x=\cos^2x\\ &=\frac{\cos x}{1-\sin x} && \text{nonzero cancellation} \end{aligned}

Why this step? The denominator in the target tells you which conjugate to introduce.

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

Check. At x=π/6x=\pi/6, the left side is 2/3+1/3=32/\sqrt3+1/\sqrt3=\sqrt3, and the right side is (3/2)/(1/2)=3(\sqrt3/2)/(1/2)=\sqrt3.

Interchanging sine and cosine gives the partner result cosec⁡x−cot⁡x=sin⁡x/(1+cos⁡x)\operatorname{cosec}x-\cot x=\sin x/(1+\cos x): combine the left side as (1−cos⁡x)/sin⁡x(1-\cos x)/\sin x, multiply by 1+cos⁡x1+\cos x, and use 1−cos⁡2x=sin⁡2x1-\cos^2x=\sin^2x. Its common domain is sin⁡x≠0\sin x\ne0.

Problem-solving

Choose the shortest structure you can justify. A product of conjugates suggests a difference of squares; a single fraction with 1±sin⁡x1\pm\sin x suggests introducing its conjugate. Do not quote the desired identity as a step in its own proof.

Use the other cosine conjugate in this fresh proof.

Continue this lesson.

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