An identity is an equation that is always true for all values of the variable. We use the symbol \(\equiv\) instead of \(=\) in an equality to demonstrate that it is an identity. We also use the terms left-hand side (LHS) and right-hand side (RHS).
An example of proving an identity
Prove that
\[ x^3 - y^3 \equiv (x-y)(x^2+xy+y^2).\]
Step 1: Consider one side of the expression.
Consider the RHS of the equation, \((x-y)(x^2+xy+y^2) \).
Step 2: Multiply the expressions inside the brackets.
\[ (x-y)(x^2+xy+y^2) \equiv x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3.\]
Step 3: Simplify the expression.
\[ x^3 + x^2y + xy^2 - x^2y - xy^2 - y^3 = x^3 - y^3 .\]
Step 4: The equation is now proven.
We see that the result is equivalent to the LHS of the equation. Therefore the identity
\[ x^3 - y^3 \equiv (x-y)(x^2+xy+y^2)\]
is proved as LHS \(\equiv\) RHS.
For example, \(3x + 6 = 3 (x + 2) \) is valid for all values of \(x\). This means that this is an identity, so we can write \(3x + 6 = 3 (x + 2) \).
Consider the equation \(x + 3 = 2x-5\).
This equation is valid for particular values of \(x\).
When \(x = 8\), then this equation is true. However, this equation doesn't hold for any other value of \(x\); therefore, it is not an identity.
Now let's consider how to prove an identity. When we solve an equation, we can manipulate terms from one side to the other. However, we cannot do this in the case of an identity. We must start from one side of the identity and then work towards the other side. Only then can we say that our identity is proved.
When first looking at an identity, it may be a good idea to substitute some values into the identity to check that the identity holds and that we understand how the identity works. However, this does not mean that the identity is proven - we can't check infinite possibilities of numbers, and one value might still not work. This means we need to prove unequivocally that the identity holds.
Let's demonstrate this point. Suppose we want to check whether the equation \(x^2 + 4x + 2 = 9x-4\) is an identity. To do this, we need to show that it works for all values of \(x\).
Let us check some values. Let's check \(x = 2\). On the left-hand side (LHS), we get \(2^2 + 4 (2) + 2 = 14\), and similarly, on the RHS, we also get \(14 = 9 (2) -4\). This means that we have not yet proven the identity, but equally, it's not disproved.
Now let's check \(x = 3\). On the LHS, we get \(3^2 + 4 (3) + 2 = 23\), and we get \(9 (3) -4 = 23\) on the RHS. So far, this is looking promising, as we have two values that hold.
Now, let's check \(x = 0\). On the left-hand side, we get \(2\), whereas, on the right-hand side, we get \(-4\), meaning that this doesn't work and thus isn't an identity. This demonstrates that just because we have a couple of cases that work, that doesn't mean we have a proof.
Proving an algebraic identity
With any identity, there are numerous ways to prove it. However, when we have an identity, we can apply a set of steps to prove it.
Step 1: Choose one side of the identity to work with. This should be the side that seems easier to work with.
Step 2: Try and manipulate this side. Algebraically, this will usually look like either a multiplication or factorization.
Step 3: Try to simplify this expression, and then if needed, manipulate further until your original expression is the same as the equivalent expression.
Step 4: The identity is proven.
Let's consider an algebraic identity to understand this better.
Prove that
\[ (x+y)(x-y) \equiv x^2 - y^2.\]
Step 1: Consider one side of the expression.
Consider the LHS of the equation, \((x+y)(x-y) \).
Step 2: Multiply the expressions inside the brackets.
\[ (x-y)(x+y) \equiv x^2 + xy - yx + y^2.\]
Step 3: Simplify the expression.
\[ x^2 + xy - yx + y^2 \equiv x^2 - y^2 .\]
Step 4: The equation is now proven.
We see that the result is equivalent to the LHS of the equation. Therefore the identity
\[ (x+y)(x-y) \equiv x^2 - y^2.\]
is proved as LHS \(\equiv\) RHS.
Let's look at another example.
Prove the identity \[ (x-y)^3 \equiv x^3 -3x^2y + 3xy^2 - y^3.\]
Step 1: Consider one side of the expression.
Consider the LHS of the equation, \((x-y)^3 \). We can write it as \( (x-y)(x-y)(x-y)\).
Step 2: Multiply the expressions inside the brackets.
\[ (x-y)^3 \equiv (x-y)(x-y)(x-y) \equiv x^3 - x^2y-2x^2y+2xy^2+xy^2-y^3.\]
Step 3: Simplify the expression.
\[ x^3 - x^2y-2x^2y+2xy^2+xy^2-y^3 \equiv x^3 -3x^2y + 3xy^2 - y^3 .\]
Step 4: The equation is now proven.
We see that the result is equivalent to the LHS of the equation. Therefore the identity
\[ (x-y)^3 \equiv x^3 -3x^2y + 3xy^2 - y^3.\]
is proved as LHS \(\equiv\) RHS.
Prove the identity \[ x^3 + y^3 + z^3 - 3xyz \equiv (x+y+x)(x^2+y^2+z^2 - xy-yz-xz).\]
Step 1: Consider one side of the expression.
Consider the RHS of the equation,
\[(x+y+x)(x^2+y^2+z^2 - xy-yz-xz).\]
Step 2: Multiply the expressions inside the brackets.
\[ \begin{align} (x+y+x)(x^2+y^2+z^2 - xy-yz-xz) &\equiv x^3 + xy^2 + xz^2 - x^2y -xyz \\ & \quad- x^2z+x^2y +y^3+ yz^2 -xy^2 \\ & \quad -y^2z - xyz + x^2z + x^2z + y^2z \\ & \quad + z^3 -xyz - yz^2 - xz^2 . \end{align}\]
Step 3: Simplify the expression.
\[ \begin{align} &x^3 + xy^2 + xz^2 - x^2y -xyz \\ & \quad- x^2z+x^2y +y^3+ yz^2 -xy^2 \\ & \quad -y^2z - xyz + x^2z + x^2z + y^2z \\ & \quad + z^3 -xyz - yz^2 - xz^2 \\ & \equiv x^2+y^2+z^2 - xy-yz-xz . \end{align} \]
Step 4: The equation is now proven.
We see that the result is equivalent to the LHS of the equation. Therefore the identity
\[ x^3 + y^3 + z^3 - 3xyz \equiv (x+y+x)(x^2+y^2+z^2 - xy-yz-xz).\]
is proved as LHS \(\equiv\) RHS.
Proving a trigonometric identity
As with any identity, we attempt to show that both sides of the identity are equivalent. We follow similar steps to those above.
Sometimes it is tricky to decide which side of an identity to start with. As a guide, the more complicated side is a good start. This means that there should be more possible steps to reduce it to the simpler side rather than adding terms to the more complex side.
If stuck, often a good principle is to convert every trigonometric function into a combination of sine and cosine functions. We will explore this topic further in trigonometric identities, but for now, let's look at an example to get our heads around the method.
Prove that
\[ \sec^2x - \csc^2x \equiv \tan^2x - cot^2x \]
given that
\[ \sin^2x + \cos^2x = 1\]
for all values of \(x\).
Let's start by manipulating the RHS. By definition,
\[ \tan x = \frac{\sin x}{\cos x} \mbox { and } \cot x = \frac{1}{\tan x}, \]
so
\[ \tan^2x - cot^2x \equiv \frac{ \sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\sin^2 x}.\]
Then by simplifying into one fraction, we get
\[ \frac{ \sin^2 x}{\cos^2 x} - \frac{\cos^2 x}{\sin^2 x} \equiv \frac{\sin^4 x - \cos^4 x}{\cos^2 x \sin^2 x}. \]
By the difference of two squares, we get
\[ \frac{\sin^4 x - \cos^4 x}{\cos^2 x \sin^2 x} \equiv \frac{(\sin^2 x - \cos^2 x)(\sin^2 x +\cos^2 x )}{\cos^2 x \sin^2 x} .\]
We can now use our given identity, to give us
\[ \frac{(\sin^2 x - \cos^2 x)(\sin^2 x +\cos^2 x )}{\cos^2 x \sin^2 x} \equiv \frac{(\sin^2 x - \cos^2 x)}{\cos^2 x \sin^2 x} .\]
Now we can split up the fraction to give
\[ \frac{(\sin^2 x - \cos^2 x)}{\cos^2 x \sin^2 x} \equiv \frac{\sin^2 x }{\cos^2 x \sin^2 x} - \frac{ \cos^2 x}{\cos^2 x \sin^2 x} \equiv \frac{1}{\cos^2 x } - \frac{1}{\sin^2 x } .\]
Now we can use the definition of secant and cosecant to get
\[\frac{1}{\cos^2 x } - \frac{1}{\sin^2 x } \equiv \tan^2x - cot^2x = \text{ LHS}.\]
Thus the identity is proven.
Proving an Identity - Key takeaways
An identity is an equation consisting of variables that is always true for all values of the variable.
To prove identity, you need to show that both sides (left-hand side and right-hand side) are the same by simplifying the expressions. You need to follow the logical steps to show that one side of the equation can be converted into the other side of the equation.
To prove identities, always start with the complex side of the equation as it is easier to eliminate or break down terms of a complex function to make it simple rather than finding terms to make a simple function complex.
The sign \(\equiv\) means equivalent and is shown in identities instead of the equals sign.
Explanations 
Exams 
Magazine 
