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Trigonometric Identities

Two small identities do a surprising amount of heavy lifting. Learn how \(\tan x = \tfrac{\sin x}{\cos x}\) and \(\sin^2 x + \cos^2 x = 1\) fit together and you can simplify messy expressions and prove new results with confidence.

MYP 5Extended MathsAdvanced Geometry & TrigonometryCriteria A · C~10 min read

Written by Fayek · MYP student

What an identity is

An identity is an equation that is true for every value of the variable, not just for a special few. That is different from an equation you solve, where you hunt for the particular values that work.

Identity
A statement that two expressions are always equal. We often write it with \(\equiv\) to stress it holds for all \(x\), for example \(\tan x \equiv \dfrac{\sin x}{\cos x}\).

Everything in this note is built from just two identities, so it is worth committing them to memory before anything else.

The quotient identity

The first links all three ratios together:

\[ \tan x = \frac{\sin x}{\cos x} \]

This is why any \(\tan\) can be swapped for a sine over a cosine, which often lets terms cancel. It also explains why \(\tan x\) is undefined wherever \(\cos x = 0\): the denominator would be zero.

The Pythagorean identity

The second comes straight from the unit circle, where a point has coordinates \((\cos x, \sin x)\) and lies on a circle of radius 1. Pythagoras on that point gives:

\[ \sin^2 x + \cos^2 x = 1 \]

The real power is in its rearrangements, because they let you replace one squared ratio with the other:

\[ \sin^2 x = 1 - \cos^2 x \qquad\qquad \cos^2 x = 1 - \sin^2 x \]

Spot the swap

Any time you see \(1 - \cos^2 x\) or \(1 - \sin^2 x\) in an expression, replace it in one step with \(\sin^2 x\) or \(\cos^2 x\). That single move unlocks most simplifying questions.

Simplifying and proving

To prove an identity, start with one side (usually the messier one) and rewrite it step by step until it matches the other. Never move terms across the \(=\) sign as if you were solving, that assumes the very thing you are trying to show.

Worked example

Prove that \(\dfrac{\sin x}{\cos x} + \dfrac{\cos x}{\sin x} = \dfrac{1}{\sin x\cos x}\).

1
Start with the left side and put the two fractions over a common denominator \(\sin x\cos x\): \(\dfrac{\sin x \cdot \sin x}{\sin x\cos x} + \dfrac{\cos x \cdot \cos x}{\sin x\cos x}\).
2
Combine the numerators: \(\dfrac{\sin^2 x + \cos^2 x}{\sin x\cos x}\).
3
Use \(\sin^2 x + \cos^2 x = 1\) on the top: \(\dfrac{1}{\sin x\cos x}\), which is the right side.
Left side \(=\) right side, so the identity is proved.

Where this is assessed

Choosing the right identity to apply is Criterion A, and a proof laid out as a clear chain of equal steps with a stated conclusion is Criterion C. Work on one side only until it matches the other.

Check yourself

1. Simplify \(\cos x\,\tan x\). +

Replace \(\tan x\) with \(\dfrac{\sin x}{\cos x}\): \(\cos x \times \dfrac{\sin x}{\cos x}\). The \(\cos x\) cancels, leaving \(\mathbf{\sin x}\).

2. Given \(\sin x = \tfrac{3}{5}\) and \(x\) is acute, find \(\cos x\). +

Use \(\cos^2 x = 1 - \sin^2 x = 1 - \tfrac{9}{25} = \tfrac{16}{25}\). Then \(\cos x = \sqrt{\tfrac{16}{25}} = \mathbf{\tfrac{4}{5}}\), taking the positive root because \(x\) is acute.

3. Simplify \(1 - \sin^2 x\). +

Rearranging the Pythagorean identity gives \(1 - \sin^2 x = \mathbf{\cos^2 x}\).

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