One Identity, Three Useful Forms
The cosine double-angle identity can be written in three equivalent ways. Each form gives the same value for \(\cos(2x)\), but a particular form may make a given expression much easier to simplify. The key is to look at the terms already present: if an expression contains \(\cos^2 x\), use the form involving cosine squared; if it contains \(\sin^2 x\), use the form involving sine squared; if it contains both, the difference form may fit directly.
Throughout, \(x\) is a real angle, measured in radians as is standard in calculus. The identities below hold for every real \(x\). Restrictions arise only when an identity is used inside a quotient: the denominator of the original expression must remain nonzero.
These are not three different identities with different conditions. They are three ways to express the same quantity. The first form uses both squared functions, while the second and third each use only one of them.
Why the Forms Are Equivalent
The angle-addition formula for cosine gives \(\cos(a+b)=\cos a\cos b-\sin a\sin b\). Taking \(a=b=x\) gives the first form. The other two follow from the Pythagorean identity \(\sin^2 x+\cos^2 x=1\). Here are the steps explicitly.
Set both angles equal to \(x\): $$ \cos(2x)=\cos(x+x)=\cos x\cos x-\sin x\sin x =\cos^2 x-\sin^2 x. $$
Since \(\sin^2 x=1-\cos^2 x\), $$ \cos^2 x-\sin^2 x =\cos^2 x-(1-\cos^2 x) =2\cos^2 x-1. $$
Since \(\cos^2 x=1-\sin^2 x\), $$ \cos^2 x-\sin^2 x =(1-\sin^2 x)-\sin^2 x =1-2\sin^2 x. $$
This derivation also explains why the formulas work: each transformation uses an equality that holds for every real angle. No division is involved, so these identities do not introduce restrictions on \(x\).
How to Choose a Form
Before substituting, inspect the expression you are simplifying. Choose the form that allows terms to combine, cancel, or match a denominator. The best form is not always the one that looks shortest by itself; it is the one that fits the whole expression.
| What the expression contains | Useful form to consider |
|---|---|
| A term involving \(\cos^2 x\), together with a constant | \(\cos(2x)=2\cos^2 x-1\) |
| A term involving \(\sin^2 x\), together with a constant | \(\cos(2x)=1-2\sin^2 x\) |
| Both \(\cos^2 x\) and \(\sin^2 x\) | \(\cos(2x)=\cos^2 x-\sin^2 x\) |
This table is a guide, not a rule that forces one choice. You may use any form and still obtain a correct result, but a less suitable choice can create extra algebra instead of reducing it.
Worked Examples
Worked Example: Simplifying an Expression with Sine Squared
Simplify \[ \frac{1-\cos(2x)}{2\sin^2 x}. \] The denominator contains \(\sin^2 x\), so choose \(\cos(2x)=1-2\sin^2 x\). The original expression is defined only when \(2\sin^2 x\ne 0\), which is equivalent to \(\sin x\ne 0\).
Substitute the selected form in the numerator: $$ \frac{1-\cos(2x)}{2\sin^2 x} =\frac{1-(1-2\sin^2 x)}{2\sin^2 x}. $$ Simplify the numerator, keeping the parentheses: $$ \frac{1-1+2\sin^2 x}{2\sin^2 x} =\frac{2\sin^2 x}{2\sin^2 x} =1, \qquad \sin x\ne 0. $$ Check by replacing \(\cos(2x)\) in the original expression again: \(1-(1-2\sin^2 x)=2\sin^2 x\), so the numerator equals the denominator wherever the denominator is nonzero. Thus the result is \(1\) on the original domain, not at inputs where \(\sin x=0\).
Worked Example: Simplifying an Expression with Cosine Squared
Simplify \[ \frac{1+\cos(2x)}{2\cos^2 x}. \] Since the denominator contains \(\cos^2 x\), use \(\cos(2x)=2\cos^2 x-1\). The original quotient requires \(\cos x\ne 0\).
Substitute and simplify: $$ \frac{1+\cos(2x)}{2\cos^2 x} =\frac{1+(2\cos^2 x-1)}{2\cos^2 x} =\frac{1+2\cos^2 x-1}{2\cos^2 x} =\frac{2\cos^2 x}{2\cos^2 x} =1, \qquad \cos x\ne 0. $$ Check the numerator directly: \(1+(2\cos^2 x-1)=2\cos^2 x\), exactly the denominator. The quotient is therefore \(1\) at every input in its original domain. At inputs where \(\cos x=0\), the original expression is undefined, so the simplification does not make it defined there.
Worked Example: Simplifying a Quotient with Both Squares
Simplify \[ \frac{1+\cos(2x)}{1-\cos(2x)}. \] Both the numerator and denominator can be rewritten using the sine-squared and cosine-squared forms. The denominator of the original quotient must be nonzero.
Use \(\cos(2x)=2\cos^2 x-1\) in the numerator and \(\cos(2x)=1-2\sin^2 x\) in the denominator: $$ 1+\cos(2x)=1+(2\cos^2 x-1)=2\cos^2 x, $$ $$ 1-\cos(2x)=1-(1-2\sin^2 x)=2\sin^2 x. $$ Therefore, $$ \frac{1+\cos(2x)}{1-\cos(2x)} =\frac{2\cos^2 x}{2\sin^2 x} =\frac{\cos^2 x}{\sin^2 x} =\left(\frac{\cos x}{\sin x}\right)^2, \qquad \sin x\ne 0. $$ Here the final quotient is \(\cot^2 x\), using \(\cot x=\cos x/\sin x\). Check the domain: the original denominator is \(1-\cos(2x)=2\sin^2 x\), which is nonzero exactly when \(\sin x\ne 0\). On that domain, substituting the rewritten numerator and denominator back gives the same quotient, so the simplification is valid.
Conditions and Domain Awareness
The three double-angle forms are valid for all real \(x\), because they follow from the angle-addition formula and the Pythagorean identity. A quotient formed from them has an additional condition: its denominator cannot equal zero. When simplifying a quotient, keep the original domain in mind even if the final expression appears to be defined at more inputs.
For example, in the first worked example the simplified expression is the constant \(1\), which is defined for every real number. But the original quotient has denominator \(2\sin^2 x\), which is zero whenever \(\sin x=0\). The simplification gives an equal value wherever the original expression exists; it does not repair the original expression at excluded inputs.
It is also useful to distinguish an identity from a single numerical check. The identities state equalities for every real \(x\). Substituting one value can check an example, but it does not establish the identity for all inputs. The derivation above establishes the general equalities.
Common Mistake: Losing a Minus Sign
A frequent error occurs when replacing \(\cos^2 x\) by \(1-\sin^2 x\) in the difference form. The minus sign in \(\cos^2 x-\sin^2 x\) applies to the entire substituted quantity only where parentheses show it clearly.
Incorrect calculation: $$ \cos^2 x-\sin^2 x =(1-\sin^2 x)-\sin^2 x \mathrel{\color{red}{=}}1-\sin^2 x. $$ The last step incorrectly drops one \(-\sin^2 x\) term.
Correction: Combine both sine-squared terms: $$ (1-\sin^2 x)-\sin^2 x =1-\sin^2 x-\sin^2 x =1-2\sin^2 x. $$ The corrected result is the sine-squared form of the double-angle identity. A useful check is to compare the expression before and after substitution: the term being subtracted must still be accounted for.
A Reliable Selection Process
When simplifying an expression involving \(\cos(2x)\), use a deliberate sequence rather than choosing a formula by habit.
Identify where \(\cos(2x)\) occurs and note the other terms around it.
Use the cosine-squared form when it pairs with \(\cos^2 x\), the sine-squared form when it pairs with \(\sin^2 x\), or the difference form when both squares are present.
Parentheses make signs clear, especially when the double-angle term is preceded by a minus sign.
Combine like terms, cancel only common nonzero factors, and retain any restrictions from the original denominator.
Why the Choice Matters
The three forms let you adapt an identity to the structure of a problem. In the first example, the sine-squared form turns the numerator into the denominator. In the second, the cosine-squared form does the same. In the third, separate choices for the numerator and denominator expose a quotient of squares. In each case, the identity does not change the value; it changes how easily the expression’s structure can be seen.
This is a general algebraic habit: equivalent forms are tools, and the useful form depends on the task. A good simplification preserves equality, makes the next step more transparent, and respects the domain of the original expression.
Check Your Understanding
For each question, show the identity form you chose and keep track of any restrictions on the original expression.
- Write \(\cos(2x)\) in terms of \(\cos^2 x\), then derive that form from \(\cos^2 x-\sin^2 x\).
- Simplify \(\dfrac{1-\cos(2x)}{\sin^2 x}\). For which real inputs is the original quotient defined?
- Simplify \(\dfrac{1+\cos(2x)}{1-\cos(2x)}\) and state the restriction on \(x\).
- Explain why \(\cos(2x)=\cos^2 x-\sin^2 x\) has no restrictions on real \(x\), while a quotient using this identity may have restrictions.
- Find and correct the error in the calculation \(\cos^2 x-\sin^2 x=(1-\sin^2 x)-\sin^2 x=1-\sin^2 x\).
