Trigonometric identities: proving and simplifying expressions

A trigonometric identity is an equation that is true for every value of the variable for which both sides are defined. The most important identities at A level are

$$\boxed{\tan x=\frac{\sin x}{\cos x}}, \qquad \boxed{\sin^2x+\cos^2x=1},$$

together with the reciprocal definitions

$$\boxed{\sec x=\frac1{\cos x}}, \qquad \boxed{\cosec x=\frac1{\sin x}}, \qquad \boxed{\cot x=\frac1{\tan x}=\frac{\cos x}{\sin x}}.$$

They allow an unfamiliar expression to be rewritten using fewer trigonometric functions. This is the central technique in simplification, proof and many trigonometric equations.

Prerequisites

You should be able to:

• manipulate fractions and factorise algebraic expressions;
• solve basic equations without dividing by an expression that might be zero;
• use sine, cosine and tangent in degrees and radians;
• recognise the graphs and zeros of the trigonometric functions.

Review exact trigonometric values and trigonometric graphs if needed.

Identity or equation?

The statement

$$\sin^2x+\cos^2x=1$$

is an identity because it holds wherever sine and cosine are defined, which is for every real $x$. By contrast,

$$2\sin x=1$$

is an equation that is true only at particular values, such as $x=30^\circ$.

The symbol $\equiv$ is sometimes used to emphasise an identity:

$$\sin^2x+\cos^2x\equiv1.$$

In exam questions, “show that” or “prove the identity” means transform one side until it becomes the other. Checking several numerical values is useful for detecting errors, but it is not a proof.

Exponents and angles

$\sin^2x$ means $(\sin x)^2$. It does not mean $\sin(x^2)$. Similarly, $\cos^3x=(\cos x)^3$.

The quotient identity

From right angled triangle ratios,

$$\sin x=\frac{\text{opposite}}{\text{hypotenuse}}, \qquad \cos x=\frac{\text{adjacent}}{\text{hypotenuse}}.$$

Therefore

$$\frac{\sin x}{\cos x} =\frac{\text{opposite}}{\text{hypotenuse}} \div\frac{\text{adjacent}}{\text{hypotenuse}} =\frac{\text{opposite}}{\text{adjacent}} =\tan x.$$

Hence

$$\boxed{\tan x=\frac{\sin x}{\cos x}}.$$

It is valid where $\cos x\ne0$. At $x=90^\circ$, for example, both $\tan x$ and $\sin x/\cos x$ are undefined.

Worked example 1: rewrite using sine and cosine

Simplify

$$\frac{\tan x\cos x}{\sin x}.$$

Replace tangent by $\sin x/\cos x$:

$$\begin{aligned} \frac{\tan x\cos x}{\sin x} &=\frac{(\sin x/\cos x)\cos x}{\sin x}\\ &=\boxed{1}. \end{aligned}$$

The original expression is defined only when $\sin x\ne0$ and $\cos x\ne0$. The simplified value $1$ does not remove those restrictions.

Worked example 2: combine fractions

Show that

$$\frac{1+\tan x}{1-\tan x} =\frac{\cos x+\sin x}{\cos x-\sin x}.$$

Start with the more complicated side and replace tangent:

$$\begin{aligned} \frac{1+\tan x}{1-\tan x} &=\frac{1+\sin x/\cos x}{1-\sin x/\cos x}\\ &=\frac{(\cos x+\sin x)/\cos x} {(\cos x-\sin x)/\cos x}\\ &=\boxed{\frac{\cos x+\sin x}{\cos x-\sin x}}. \end{aligned}$$

Multiplying the numerator and denominator by $\cos x$ is often the cleanest way to remove fractions within fractions.

The fundamental Pythagorean identity

A point on the unit circle at angle $x$ has coordinates $(\cos x,\sin x)$. Since its distance from the origin is $1$, Pythagoras gives

$$(\cos x)^2+(\sin x)^2=1^2.$$

Thus

$$\boxed{\sin^2x+\cos^2x=1}.$$

This unit circle argument proves the result for angles in every quadrant, not just acute angles in a right angled triangle.

Three equivalent arrangements are worth recognising immediately:

$$\sin^2x=1-\cos^2x, \qquad \cos^2x=1-\sin^2x,$$

and, by difference of two squares,

$$\boxed{\cos^2x-\sin^2x=(\cos x-\sin x)(\cos x+\sin x)}.$$

Worked example 3: simplify a quadratic expression

Simplify

$$3\sin^2x+3\cos^2x-2.$$

Factor the common coefficient before using the identity:

$$\begin{aligned} 3\sin^2x+3\cos^2x-2 &=3(\sin^2x+\cos^2x)-2\\ &=3(1)-2\\ &=\boxed{1}. \end{aligned}$$

Worked example 4: factorise before cancelling

Simplify

$$\frac{1-\sin^2x}{1+\sin x}.$$

There are two useful routes. Using $1-\sin^2x=\cos^2x$ gives $\cos^2x/(1+\sin x)$, which is not yet simpler. Instead, factorise the difference of two squares:

$$\begin{aligned} \frac{1-\sin^2x}{1+\sin x} &=\frac{(1-\sin x)(1+\sin x)}{1+\sin x}\\ &=\boxed{1-\sin x}. \end{aligned}$$

The cancellation is valid where the original denominator is nonzero, so $\sin x\ne-1$.

Self check 1

Simplify:

1. $5\cos^2x+5\sin^2x-3$;
2. $\dfrac{\cos^2x}{1-\sin x}$;
3. $(\sec x-\tan x)(\sec x+\tan x)$.

Answers

1. $5(\cos^2x+\sin^2x)-3=5-3=\boxed{2}$.

2. Since $\cos^2x=1-\sin^2x$,

$$\frac{\cos^2x}{1-\sin x} =\frac{(1-\sin x)(1+\sin x)}{1-\sin x} =\boxed{1+\sin x},$$

where $\sin x\ne1$.

3. This is a difference of two squares:

$$(\sec x-\tan x)(\sec x+\tan x) =\sec^2x-\tan^2x =\boxed{1}.$$

Reciprocal functions and derived identities

The reciprocal functions are defined by

$$\sec x=\frac1{\cos x}, \qquad \cosec x=\frac1{\sin x}, \qquad \cot x=\frac1{\tan x}=\frac{\cos x}{\sin x}.$$

“Reciprocal” does not mean “inverse function”. For example,

$$\sec x=\frac1{\cos x},$$

whereas $\cos^{-1}x$ means the inverse cosine function. Learn more in reciprocal and inverse trigonometric functions.

Two further Pythagorean identities follow from $\sin^2x+\cos^2x=1$.

Divide every term by $\cos^2x$:

$$\frac{\sin^2x}{\cos^2x}+\frac{\cos^2x}{\cos^2x} =\frac1{\cos^2x},$$

so

$$\boxed{1+\tan^2x=\sec^2x}.$$

Divide instead by $\sin^2x$:

$$\frac{\sin^2x}{\sin^2x}+\frac{\cos^2x}{\sin^2x} =\frac1{\sin^2x},$$

so

$$\boxed{1+\cot^2x=\cosec^2x}.$$

Useful rearrangements include

$$\sec^2x-\tan^2x=1, \qquad \cosec^2x-\cot^2x=1.$$

Worked example 5: choose the matching identity

Simplify

$$\frac{\sec^2x-1}{\tan x}.$$

Because $\sec^2x-1=\tan^2x$,

$$\frac{\sec^2x-1}{\tan x} =\frac{\tan^2x}{\tan x} =\boxed{\tan x},$$

for values where the original expression is defined.

A reliable method for proving identities

Identity proofs are algebra, but the variables happen to be trigonometric functions. A reliable strategy is:

1. Start with one side, usually the more complicated side.
2. Rewrite reciprocal functions and tangent in terms of sine and cosine if no shorter route is visible.
3. Find a common denominator, factorise, or use a Pythagorean identity.
4. Work in a continuous chain until the target side appears.
5. Do not assume the result by manipulating both sides simultaneously.

Worked example 6: prove by using a common denominator

Prove that

$$\frac1{1-\sin x}+\frac1{1+\sin x}=2\sec^2x.$$

Begin with the left hand side:

$$\begin{aligned} \frac1{1-\sin x}+\frac1{1+\sin x} &=\frac{1+\sin x+1-\sin x}{(1-\sin x)(1+\sin x)}\\ &=\frac2{1-\sin^2x}\\ &=\frac2{\cos^2x}\\ &=2\sec^2x. \end{aligned}$$

This proves the identity wherever the original expressions are defined.

Worked example 7: convert everything to sine and cosine

Prove that

$$\frac{\sec x-\cos x}{\tan x}=\sin x.$$ $$\begin{aligned} \frac{\sec x-\cos x}{\tan x} &=\frac{1/\cos x-\cos x}{\sin x/\cos x}\\ &=\frac{(1-\cos^2x)/\cos x}{\sin x/\cos x}\\ &=\frac{\sin^2x}{\cos x}\times\frac{\cos x}{\sin x}\\ &=\sin x. \end{aligned}$$

The decisive step is replacing $1-\cos^2x$ by $\sin^2x$.

Worked example 8: use a conjugate

Show that

$$\frac1{\sec x+\tan x}=\sec x-\tan x.$$

Multiply by the conjugate $\sec x-\tan x$:

$$\begin{aligned} \frac1{\sec x+\tan x} &=\frac{\sec x-\tan x} {(\sec x+\tan x)(\sec x-\tan x)}\\ &=\frac{\sec x-\tan x}{\sec^2x-\tan^2x}\\ &=\boxed{\sec x-\tan x}. \end{aligned}$$

This is the trigonometric analogue of rationalising a denominator containing surds.

Work towards the target

If the target contains only $\sin x$, rewrite other functions in terms of sine and cosine and look for $1-\cos^2x=\sin^2x$. If it contains $\sec x$ and $\tan x$, preserve those functions and look for $\sec^2x-\tan^2x=1$.

Self check 2

Prove each identity.

1. $\dfrac{\tan x}{\sec x}=\sin x$
2. $\dfrac{1-\cos^2x}{\sin x}=\sin x$
3. $\dfrac{\cosec x-\sin x}{\cos x}=\cot x$

Answers

$$\frac{\tan x}{\sec x} =\frac{\sin x/\cos x}{1/\cos x} =\boxed{\sin x}.$$

$$\frac{1-\cos^2x}{\sin x} =\frac{\sin^2x}{\sin x} =\boxed{\sin x}.$$

$$\begin{aligned} \frac{\cosec x-\sin x}{\cos x} &=\frac{1/\sin x-\sin x}{\cos x}\\ &=\frac{1-\sin^2x}{\sin x\cos x}\\ &=\frac{\cos^2x}{\sin x\cos x}\\ &=\frac{\cos x}{\sin x}\\ &=\boxed{\cot x}. \end{aligned}$$

Using identities in equations

An equation involving both $\sin^2x$ and $\cos x$ can often be rewritten using only cosine. This turns it into a quadratic equation.

Worked example 9: reduce to one trigonometric function

Solve

$$2\sin^2x=3\cos x$$

for $0\le x<2\pi$.

Use $\sin^2x=1-\cos^2x$:

$$2(1-\cos^2x)=3\cos x.$$

Rearrange and factorise:

$$\begin{aligned} 2\cos^2x+3\cos x-2&=0\\ (2\cos x-1)(\cos x+2)&=0. \end{aligned}$$

Therefore

$$\cos x=\frac12 \qquad\text{or}\qquad \cos x=-2.$$

The second possibility is impossible because $-1\le\cos x\le1$. In the given interval, $\cos x=1/2$ at

$$\boxed{x=\frac\pi3,\ \frac{5\pi}3}.$$

Worked example 10: do not lose solutions by dividing

Solve

$$\sin x=\sin x\cos x$$

for $0^\circ\le x<360^\circ$.

Bring all terms to one side and factorise:

$$\sin x(1-\cos x)=0.$$

Hence

$$\sin x=0 \qquad\text{or}\qquad \cos x=1.$$

These give

$$x=0^\circ,180^\circ \qquad\text{or}\qquad x=0^\circ.$$

After removing the duplicate,

$$\boxed{x=0^\circ,180^\circ}.$$

Dividing the original equation by $\sin x$ would discard every solution for which $\sin x=0$. Factorising protects those solutions.

Continue with trigonometric equations for general solutions and more complicated intervals.

Common misconceptions

False distribution

In general,

$$\sin(A+B)\ne\sin A+\sin B.$$

For example, $\sin(30^\circ+60^\circ)=1$, but $\sin30^\circ+\sin60^\circ=(1+\sqrt3)/2$. Sums of angles require the compound angle formulae.

Squaring a sum incorrectly

$$(\sin x+\cos x)^2 =\sin^2x+2\sin x\cos x+\cos^2x =1+2\sin x\cos x,$$

not $1$. The middle term matters.

Taking square roots without both signs

From $\sin^2x=1/4$, it follows that

$$\sin x=\pm\frac12,$$

not only $1/2$. The interval then determines which angles are required.

Cancelling terms instead of factors

You may cancel common factors in a product, but not terms joined by addition. Thus

$$\frac{\sin x(1+\cos x)}{\sin x}=1+\cos x$$

where $\sin x\ne0$, but no cancellation is possible in

$$\frac{\sin x+\cos x}{\sin x}.$$

Ignoring domains

$\tan x$, $\sec x$ and any expression divided by $\cos x$ are undefined when $\cos x=0$. Likewise, $\cosec x$, $\cot x$ and expressions divided by $\sin x$ are undefined when $\sin x=0$. Algebraic simplification does not add excluded values back into the original expression.

Mixed self check

1. State whether $\sin x=\cos x$ is an identity or an equation.
2. Simplify $\dfrac{1+\tan^2x}{\sec x}$.
3. Prove $\dfrac{\cos x}{1-\sin x}=\sec x+\tan x$.
4. Solve $3\cos^2x+\sin^2x=2$ for $0\le x<2\pi$.

Answers

1. It is an equation. It holds only for particular angles, not for every $x$.

2. Since $1+\tan^2x=\sec^2x$,

$$\frac{1+\tan^2x}{\sec x} =\frac{\sec^2x}{\sec x} =\boxed{\sec x}.$$

3. Multiply numerator and denominator by $1+\sin x$:

$$\begin{aligned} \frac{\cos x}{1-\sin x} &=\frac{\cos x(1+\sin x)}{1-\sin^2x}\\ &=\frac{\cos x(1+\sin x)}{\cos^2x}\\ &=\frac1{\cos x}+\frac{\sin x}{\cos x}\\ &=\boxed{\sec x+\tan x}. \end{aligned}$$

4. Replace $\sin^2x$ by $1-\cos^2x$:

$$3\cos^2x+1-\cos^2x=2,$$

so

$$2\cos^2x=1 \quad\Longrightarrow\quad \cos x=\pm\frac1{\sqrt2}.$$

Therefore

$$\boxed{x=\frac\pi4,\ \frac{3\pi}4,\ \frac{5\pi}4,\ \frac{7\pi}4}.$$

What to remember

$$\boxed{\tan x=\frac{\sin x}{\cos x}}, \qquad \boxed{\sin^2x+\cos^2x=1},$$ $$\boxed{1+\tan^2x=\sec^2x}, \qquad \boxed{1+\cot^2x=\cosec^2x}.$$

• An identity is true for every value in its domain; an equation is true only for its solutions.
• In a proof, transform one side into the other using valid algebraic steps.
• Rewriting everything in sine and cosine is a dependable fallback, but a derived identity may give a shorter route.
• Factorise before dividing, or valid solutions may be lost.
• Keep the domain restrictions of the original expression.

Next, use these identities in the compound angle and double angle formulae, harmonic form and trigonometric proof and modelling.