Tutorials › Calculus I › Deriving the Double-Angle Identities

Trigonometric Function Toolkit · Tutorial 140 of 1495

Deriving the Double-Angle Identities

Learn how setting the two angles equal in the angle addition formulas produces the double-angle identities and their domain conditions.

Beginner 9 min read

What You'll Learn

  • Derive the sine and cosine double-angle identities from the angle addition formulas.
  • Obtain three equivalent expressions for the cosine of twice an angle.
  • Derive the tangent double-angle identity by taking a quotient.
  • Identify when the tangent double-angle formula is defined.
  • Apply and check the identities using exact values.

Why Double-Angle Identities Follow from Angle Addition

An angle addition formula describes the sine or cosine of a sum of two angles. A double-angle expression is the special case in which those two angles are equal: \(x+x=2x\). This simple substitution produces useful identities for \(\sin(2x)\), \(\cos(2x)\), and, when the relevant denominators are nonzero, \(\tan(2x)\).

The sine and cosine identities are valid for every real \(x\). The tangent identity needs more care: tangent is a quotient, so it is defined only when its cosine denominator is nonzero. Keeping track of those restrictions is part of deriving the identity correctly.

Starting point: For all real angles \(a\) and \(b\), the angle addition formulas are \( \sin(a+b)=\sin a\cos b+\cos a\sin b \) and \( \cos(a+b)=\cos a\cos b-\sin a\sin b \).

Deriving the Sine Double-Angle Identity

Set \(a=x\) and \(b=x\) in the sine addition formula. Each sine and cosine on the right is then evaluated at \(x\):

$$ \begin{aligned} \sin(2x) &=\sin(x+x)\\ &=\sin x\cos x+\cos x\sin x\\ &=2\sin x\cos x. \end{aligned} $$

The last equality uses the commutativity of multiplication: \(\cos x\sin x=\sin x\cos x\). There are no divisions in this derivation, so the result holds for every real \(x\).

Identity: For every real \(x\), $$ \sin(2x)=2\sin x\cos x. $$

Deriving the Cosine Double-Angle Identities

Now set \(a=b=x\) in the cosine addition formula. This gives the first form directly:

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

Two other forms follow from the Pythagorean identity \(\sin^2 x+\cos^2 x=1\). To obtain a form involving only \(\cos x\), replace \(\sin^2 x\) with \(1-\cos^2 x\):

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

Alternatively, replace \(\cos^2 x\) with \(1-\sin^2 x\):

$$ \begin{aligned} \cos(2x) &=\cos^2 x-\sin^2 x\\ &=(1-\sin^2 x)-\sin^2 x\\ &=1-2\sin^2 x. \end{aligned} $$
Identities: For every real \(x\), $$ \cos(2x)=\cos^2 x-\sin^2 x =2\cos^2 x-1 =1-2\sin^2 x. $$ The equalities follow from the cosine addition formula and the Pythagorean identity.

These are not three separate rules about different angles. They are equal expressions for the same quantity. The first comes directly from angle addition; the other two are algebraic consequences of the Pythagorean identity.

Deriving the Tangent Double-Angle Identity

Tangent is defined by \(\tan u=\frac{\sin u}{\cos u}\) when \(\cos u\ne 0\). Therefore \(\tan(2x)\) is defined only if \(\cos(2x)\ne 0\). To express it in terms of \(\tan x\), we also need \(\tan x\) to be defined, which requires \(\cos x\ne 0\).

Under those conditions, use the sine and cosine double-angle identities and divide:

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

The division by \(\cos^2 x\) is valid because \(\cos x\ne 0\). The final denominator must also be nonzero: \(1-\tan^2 x\ne 0\). With \(\cos x\ne 0\), the cosine identity gives \(\cos(2x)=\cos^2 x(1-\tan^2 x)\), so this last condition is exactly what ensures \(\cos(2x)\ne 0\).

Identity and conditions: The formula $$ \tan(2x)=\frac{2\tan x}{1-\tan^2 x} $$ applies when \(\cos x\ne 0\) and \(1-\tan^2 x\ne 0\). Equivalently, both \(\tan x\) and \(\tan(2x)\) must be defined.

A Complete Derivation in Steps

The derivations can be organized as a short sequence. The domain check belongs specifically to the tangent step because that step involves division.

1
Set the angles equal.
Use \(a=b=x\) in the sine and cosine addition formulas, so \(a+b=2x\).
2
Simplify the sine expression.
The two terms \(\sin x\cos x\) and \(\cos x\sin x\) are equal, giving \(\sin(2x)=2\sin x\cos x\).
3
Simplify the cosine expression.
The cosine addition formula gives \(\cos(2x)=\cos^2 x-\sin^2 x\); the Pythagorean identity yields the other two forms.
4
Take a quotient only where it is defined.
Write \(\tan(2x)=\sin(2x)/\cos(2x)\), then divide by \(\cos^2 x\). This requires \(\cos x\ne 0\) and \(1-\tan^2 x\ne 0\).

Worked Examples

Worked Example: Evaluate the Double Angle at \(x=\frac{\pi}{6}\)

Use \(\sin(\pi/6)=1/2\) and \(\cos(\pi/6)=\sqrt{3}/2\). First apply the sine identity:

$$ \sin\left(2\cdot\frac{\pi}{6}\right) =2\sin\left(\frac{\pi}{6}\right)\cos\left(\frac{\pi}{6}\right) =2\cdot\frac12\cdot\frac{\sqrt3}{2} =\frac{\sqrt3}{2}. $$

For cosine, use the difference of squares:

$$ \cos\left(2\cdot\frac{\pi}{6}\right) =\left(\frac{\sqrt3}{2}\right)^2-\left(\frac12\right)^2 =\frac34-\frac14 =\frac12. $$

Since \(\tan(\pi/6)=\sqrt3/3\), the tangent formula gives

$$ \tan\left(2\cdot\frac{\pi}{6}\right) =\frac{2(\sqrt3/3)}{1-(\sqrt3/3)^2} =\frac{2\sqrt3/3}{1-1/3} =\frac{2\sqrt3/3}{2/3} =\sqrt3. $$

Check: \(2(\pi/6)=\pi/3\), where the exact values are \(\sin(\pi/3)=\sqrt3/2\), \(\cos(\pi/3)=1/2\), and \(\tan(\pi/3)=\sqrt3\). They agree with all three results.

Worked Example: Use Given Sine and Cosine Values

Suppose \(\sin x=5/13\) and \(\cos x=12/13\). These values are consistent with the Pythagorean identity because \( (5/13)^2+(12/13)^2=25/169+144/169=1 \). Calculate the three double-angle values.

$$ \begin{aligned} \sin(2x) &=2\sin x\cos x\\ &=2\cdot\frac5{13}\cdot\frac{12}{13}\\ &=\frac{120}{169}, \end{aligned} \qquad \begin{aligned} \cos(2x) &=\cos^2 x-\sin^2 x\\ &=\left(\frac{12}{13}\right)^2-\left(\frac5{13}\right)^2\\ &=\frac{144}{169}-\frac{25}{169}\\ &=\frac{119}{169}. \end{aligned} $$

For tangent, first calculate \(\tan x=(5/13)/(12/13)=5/12\). Then

$$ \begin{aligned} \tan(2x) &=\frac{2(5/12)}{1-(5/12)^2}\\ &=\frac{5/6}{1-25/144}\\ &=\frac{5/6}{119/144}\\ &=\frac56\cdot\frac{144}{119}\\ &=\frac{120}{119}. \end{aligned} $$

Check the tangent result using the sine and cosine values just found: \(\sin(2x)/\cos(2x)=(120/169)/(119/169)=120/119\). The quotient agrees.

Worked Example: Recognize When the Tangent Formula Cannot Be Used

Take \(x=\pi/4\). Here \(\sin x=\cos x=\sqrt2/2\), so the sine and cosine identities are still available:

$$ \sin(2x)=2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt2}{2}=1, \qquad \cos(2x)=\left(\frac{\sqrt2}{2}\right)^2-\left(\frac{\sqrt2}{2}\right)^2=0. $$

But \(\tan(2x)=\tan(\pi/2)\) is undefined because its cosine denominator is zero. The tangent double-angle expression also signals the problem: \(\tan(\pi/4)=1\), so \(1-\tan^2(\pi/4)=1-1=0\). The formula would require division by zero and must not be used here.

Check: \(2x=\pi/2\), and \(\cos(\pi/2)=0\), confirming directly that \(\tan(2x)\) is undefined. The sine and cosine identities remain valid because neither requires division.

A Common Mistake: Ignoring the Denominator

A frequent error is to use the tangent identity whenever \(\tan x\) exists, without checking the denominator \(1-\tan^2 x\). For example, at \(x=\pi/4\), \(\tan x=1\) exists, but the proposed calculation

$$ \frac{2\tan(\pi/4)}{1-\tan^2(\pi/4)} =\frac{2}{1-1} =\frac{2}{0} $$

is not a real number. It does not mean that \(\tan(\pi/2)\) has some value obtained by this calculation. Rather, both the right-hand expression and \(\tan(\pi/2)\) are undefined. The correction is to check the denominator before applying the formula and to use the sine and cosine identities when those quantities are what is needed.

Why These Identities Matter

The identities connect trigonometric values at \(x\) to values at \(2x\). They let a calculation involving a doubled angle be rewritten using the sine, cosine, or tangent of the original angle. The derivation also illustrates a general mathematical habit: algebraic rearrangement is safe only after checking that every division is permitted.

For sine and cosine, setting equal angles in the addition formulas introduces no restrictions. For tangent, the quotient definition creates domain conditions that cannot be dropped. Remembering this distinction makes the identities both useful and precise.

Check Your Understanding

Try each question using the identities and conditions developed above.

  1. Starting from the sine addition formula, show why \(\sin(2x)=2\sin x\cos x\).
  2. Use the Pythagorean identity to obtain \(1-2\sin^2 x\) as a form of \(\cos(2x)\).
  3. For which values of \(x\) is the expression \(\frac{2\tan x}{1-\tan^2 x}\) defined?
  4. Given \(\sin x=3/5\) and \(\cos x=4/5\), calculate \(\sin(2x)\) and \(\cos(2x)\).
  5. Explain why the tangent identity cannot be evaluated at \(x=\pi/4\), even though \(\tan(\pi/4)\) exists.

Test yourself: Calculus I final exam practice

Ready to check what you know? Work a full-length practice final, then open the complete worked solutions.

Want all 1,114 practice problems?

This tutorial covers one topic. Calc Pass covers the full Calculus AB & BC curriculum (the core of Calculus I and II): full review, every practice question, and step-by-step solutions to each one.

QR code to download the Calc Pass app

Get the Calc Pass App

Scan to download and practice on the go, with instant scoring.

Calc Pass Volume 1: 1,114 Practice Questions book cover

Vol. 1: 1,114 Questions

The complete practice question bank, organized by AB & BC topic.

View on Amazon
Calc Pass Volume 2: Solutions Manual book cover

Vol. 2: Solutions Manual

Fully worked solutions to every question in Volume 1.

View on Amazon