Proving a Trigonometric Identity

What a proof of an identity does

An identity claims that its left-hand side (LHS) and its right-hand side (RHS) are equal for every angle. Checking one angle cannot show that, because there are infinitely many angles to check. A proof shows it for all of them at once.

The method is to start from one side, usually the more complicated one, and rewrite it one step at a time until it becomes the other side. Each step uses algebra or an identity already known, so each line equals the line before it. The first line is the LHS and the last line is the RHS, so the LHS equals the RHS.

The identities to use are the ones already proved: ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1, and $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ. Two more names appear in proofs. The cotangent is the reciprocal of the tangent: $\cot \theta =1/\tan \theta =\cos \theta /\sin \theta$cot θ = 1 / tan θ = cos θ / sin θ. The secant is the reciprocal of the cosine: $\sec \theta =1/\cos \theta$sec θ = 1 / cos θ.

Why a proof never works on both sides

Suppose you start from the claim itself, written as an equation, and do the same thing to both sides until you reach something true. That shows only that the claim leads to something true, and a false claim can do that too.

Here is a false claim: $\cos \theta =\sqrt{1-{\sin }^2\theta }$cos θ = √(1 − sin²θ) for every angle. Square both sides and you get ${\cos }^2\theta =1-{\sin }^2\theta$cos²θ = 1 − sin²θ, which is true. But the claim is false at 120°: $\cos 120^{\circ }=-\frac{1}{2}$cos 120° = −1/2, while $\sqrt{1-{\sin }^{2}120^{\circ }}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$√(1 − sin²120°) = √(1 − 3/4) = 1/2. Squaring both sides hid the sign.

So never move a term across the equals sign, and never multiply or square both sides. Work on one side only, and let it arrive at the other.

y = cos x in gold and $y=\sqrt{1-{\sin }^{2}x}$y = √(1 − sin²x) in plain chalk, over one turn, with one square across for every 90°. From 0° to 90° and from 270° to 360° the two graphs are the same curve, so only the plain one shows there. Between 90° and 270° the cosine is negative and the plain graph is its reflection above the x-axis, so the claim is false there: at 120° the cosine is $-\frac{1}{2}$−1/2 and the plain graph is at $\frac{1}{2}$1/2.

A proof: $\tan \theta +\cot \theta =\frac{1}{\left(\sin \theta \cos \theta \right)}$tan θ + cot θ = 1/(sin θ cos θ)

Start from the LHS, $\tan \theta +\cot \theta$tan θ + cot θ, because it has more in it. Write everything in sines and cosines: $\tan \theta =\sin \theta /\cos \theta$tan θ = sin θ / cos θ and $\cot \theta =\cos \theta /\sin \theta$cot θ = cos θ / sin θ, so the LHS is $\frac{\left(\sin \theta \right)}{\left(\cos \theta \right)}+\frac{\left(\cos \theta \right)}{\left(\sin \theta \right)}$(sin θ)/(cos θ) + (cos θ)/(sin θ).

Put the two fractions over the common denominator $\sin \theta \cos \theta$sin θ cos θ. The first fraction becomes ${\sin }^2\theta$sin²θ over $\sin \theta \cos \theta$sin θ cos θ, and the second becomes ${\cos }^2\theta$cos²θ over $\sin \theta \cos \theta$sin θ cos θ. Together they make $\frac{\left({\sin }^2\theta +{\cos }^2\theta \right)}{\left(\sin \theta \cos \theta \right)}$(sin²θ + cos²θ)/(sin θ cos θ).

The numerator is ${\sin }^2\theta +{\cos }^2\theta$sin²θ + cos²θ, which is 1. So the LHS is $\frac{1}{\left(\sin \theta \cos \theta \right)}$1/(sin θ cos θ), which is the RHS, and the identity is proved.

Check at two angles. At 0.3 radians both sides come to 3.5421, to 4 decimal places, and at 1.1 radians both come to 2.4737. The checks do not prove anything, but a slip in the algebra would almost certainly show up in them.

A proof: $\left(1-\cos \theta \right)\left(1+\cos \theta \right)={\sin }^2\theta$(1 − cos θ)(1 + cos θ) = sin²θ

Start from the LHS and expand the brackets: $\left(1-\cos \theta \right)\left(1+\cos \theta \right)=1+\cos \theta -\cos \theta -{\cos }^2\theta =1-{\cos }^2\theta$(1 − cos θ)(1 + cos θ) = 1 + cos θ − cos θ − cos²θ = 1 − cos²θ. The two middle terms cancel, as they do in every difference of two squares, $\left(a-b\right)\left(a+b\right)=a^2-b^2$(a − b)(a + b) = a² − b².

Now use the identity. From ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1, subtracting ${\cos }^2\theta$cos²θ gives $1-{\cos }^2\theta ={\sin }^2\theta$1 − cos²θ = sin²θ. So the LHS is ${\sin }^2\theta$sin²θ, which is the RHS.

Check at 0.3 radians: 1 − cos 0.3 = 0.04466 and 1 + cos 0.3 = 1.95534, to 5 decimal places, and their product is 0.08733, which is ${\sin }^{2}0.3$sin²0.3. At 1.1 radians, 0.54640 × 1.45360 = 0.79425, which is ${\sin }^{2}1.1$sin²1.1.

Over one turn, with one square across for every 90°: the plain curve that starts at 0 is y = 1 − cos x, the plain curve that starts at 2 is y = 1 + cos x, and the gold curve is their product, (1 − cos x)(1 + cos x). The three dots mark 120°: the two factors there are $\frac{3}{2}$3/2 and ½, and the middle dot is their product, ¾, which is ${\sin }^{2}120^{\circ }={\left(\frac{\sqrt{3}}{2}\right)}^2$sin²120° = (√3/2)². The gold curve is the graph of $y={\sin }^{2}x$y = sin²x: 0 at 0°, 180° and 360°, and 1 at 90° and 270°.

Choosing the first move

Most proofs start with one of a few moves. Write tangents and reciprocals in sines and cosines. Put fractions over a common denominator. Expand brackets, or factor. Then look for ${\sin }^2\theta +{\cos }^2\theta$sin²θ + cos²θ, which can be replaced by 1, or for $1-{\cos }^2\theta$1 − cos²θ or $1-{\sin }^2\theta$1 − sin²θ, which can be replaced by a single square.

To prove $\tan \theta \cdot \cos \theta =\sin \theta$tan θ · cos θ = sin θ, the first move is to write $\tan \theta$tan θ as $\sin \theta /\cos \theta$sin θ / cos θ. Then the LHS is $\left(\frac{\sin \theta }{\cos \theta }\right)\times \cos \theta$(sin θ/cos θ) × cos θ, the $\cos \theta$cos θ cancels, and $\sin \theta$sin θ is left, which is the RHS.

A third example: prove ${\left(\sin \theta +\cos \theta \right)}^2=1+2\sin \theta \cos \theta$(sin θ + cos θ)² = 1 + 2 sin θ cos θ. Expand the LHS: ${\sin }^2\theta +2\sin \theta \cos \theta +{\cos }^2\theta$sin²θ + 2 sin θ cos θ + cos²θ. Group the squares: $\left({\sin }^2\theta +{\cos }^2\theta \right)+2\sin \theta \cos \theta =1+2\sin \theta \cos \theta$(sin²θ + cos²θ) + 2 sin θ cos θ = 1 + 2 sin θ cos θ, the RHS. At 0.3 radians both sides are 1.5646, and at 1.1 radians both are 1.8085.

Finishing with a Pythagorean identity

Prove ${\sec }^2\theta -{\tan }^2\theta =1$sec²θ − tan²θ = 1. In sines and cosines, the LHS is $\frac{1}{{\cos }^2\theta }-\frac{{\sin }^2\theta }{{\cos }^2\theta }$1/cos²θ − sin²θ/cos²θ. Both fractions have the denominator ${\cos }^2\theta$cos²θ, so the LHS is $\frac{\left(1-{\sin }^2\theta \right)}{{\cos }^2\theta }$(1 − sin²θ)/cos²θ. The numerator $1-{\sin }^2\theta$1 − sin²θ is ${\cos }^2\theta$cos²θ, so the LHS is $\frac{{\cos }^2\theta }{{\cos }^2\theta }=1$cos²θ/cos²θ = 1.

Dividing every term of ${\sin }^2\theta +{\cos }^2\theta =1$sin²θ + cos²θ = 1 by ${\cos }^2\theta$cos²θ gives $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ, and dividing by ${\sin }^2\theta$sin²θ gives $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$1 + cot²θ = cosec²θ, where $\mathrm{cosec}\theta =1/\sin \theta$cosec θ = 1 / sin θ. Once these are known, a line such as ${\sec }^2\theta -{\tan }^2\theta$sec²θ − tan²θ is finished in one step: replace ${\sec }^2\theta$sec²θ with $1+{\tan }^2\theta$1 + tan²θ, and $\left(1+{\tan }^2\theta \right)-{\tan }^2\theta =1$(1 + tan²θ) − tan²θ = 1. In the same way, $1+{\cot }^2\theta$1 + cot²θ is ${\mathrm{cosec}}^2\theta$cosec²θ.

Check at 1.1 radians: ${\sec }^{2}1.1=4.8603$sec²1.1 = 4.8603 and ${\tan }^{2}1.1=3.8603$tan²1.1 = 3.8603, and the difference is 1.

The usual mistakes

Working on both sides. Multiplying both sides by $\sin \theta$sin θ, or moving a term across the equals sign, treats the claim as already true. That is what the proof has to show.

Taking a check at one angle as a proof. At 45° many false claims hold. $\sin \theta =\cos \theta$sin θ = cos θ is true at 45° and false at 30°.

Canceling a term that is not a factor. In $\frac{\left(1-{\cos }^2\theta \right)}{\cos }\theta$(1 − cos²θ)/cos θ, the $\cos \theta$cos θ below cannot cancel with the ${\cos }^2\theta$cos²θ above, because the numerator is a difference. Factor or rewrite first.

Using the wrong Pythagorean form. A line holding ${\cot }^2\theta$cot²θ needs $1+{\cot }^2\theta ={\mathrm{cosec}}^2\theta$1 + cot²θ = cosec²θ; the form $1+{\tan }^2\theta ={\sec }^2\theta$1 + tan²θ = sec²θ pairs with ${\tan }^2\theta$tan²θ.

A lamp held aside by a rope

In the application below, an old handbook gives the pull in a rope as $W\left(\sec \theta -\cos \theta \right)/\sin \theta$W(sec θ − cos θ) / sin θ. The proof starts from the handbook's side only, writes the secant as $1/\cos \theta$1 / cos θ, and uses $1-{\cos }^2\theta ={\sin }^2\theta$1 − cos²θ = sin²θ to reach $W\tan \theta$W tan θ.