What Is the Cos2x Formula?
The cos2x formula is the double-angle identity for cosine. It expresses $\cos$ of $2x$ — twice an angle — in terms of $\cos x$ and $\sin x$ alone.
Three equivalent forms (every form is exactly equal — pick the one that fits your problem):
$$\cos 2x = \cos^2 x - \sin^2 x \tag{Form 1}$$
$$\cos 2x = 2\cos^2 x - 1 \tag{Form 2}$$
$$\cos 2x = 1 - 2\sin^2 x \tag{Form 3}$$
There is also a fourth form in terms of $\tan x$:
$$\cos 2x = \frac{1 - \tan^2 x}{1 + \tan^2 x} \tag{Form 4}$$
How Do You Prove Cos2x = Cos²x − Sin²x?
Start from the sum formula for cosine:
$$\cos(A + B) = \cos A \cos B - \sin A \sin B$$
Set $A = B = x$:
$$\cos(x + x) = \cos x \cos x - \sin x \sin x$$
$$\cos 2x = \cos^2 x - \sin^2 x$$
That's Form 1, derived in two lines.
How Do You Get the Other Two Forms?
Use the Pythagorean identity $\sin^2 x + \cos^2 x = 1$.
Form 2: Cos2x in Terms of Cos Only
Replace $\sin^2 x$ with $1 - \cos^2 x$:
$$\cos 2x = \cos^2 x - (1 - \cos^2 x) = 2\cos^2 x - 1$$
Form 3: Cos2x in Terms of Sin Only
Replace $\cos^2 x$ with $1 - \sin^2 x$:
$$\cos 2x = (1 - \sin^2 x) - \sin^2 x = 1 - 2\sin^2 x$$
Form 4: Cos2x in Terms of Tan
Divide Form 1's numerator and denominator (1 in the denominator) by $\cos^2 x$:
$$\cos 2x = \frac{\cos^2 x - \sin^2 x}{\cos^2 x + \sin^2 x} = \frac{1 - \tan^2 x}{1 + \tan^2 x}$$
Is Cos2x the Same as Cos²x?
No. $\cos 2x$ and $\cos^2 x$ look similar on the page but are entirely different functions. Confusing them is the single most common notation mistake in trigonometry — and the source of countless lost exam marks.
Notation | Meaning | Worked at $x = 30°$ |
|---|---|---|
$\cos 2x$ | Cosine of the angle $2x$ | $\cos 60° = \tfrac{1}{2}$ |
$\cos^2 x$ | Cosine of $x$, squared: $(\cos x)^2$ | $(\cos 30°)^2 = \left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{3}{4}$ |
Two different values from the same $x = 30°$.
How to tell them apart at a glance.
The "2" sits before the variable in $\cos 2x$ → "double angle." It's part of the argument.
The "2" sits between cos and the variable in $\cos^2 x$ → "cosine squared." It's an exponent applied to the output.
Worked relationship. The cos2x identity in Form 2 connects the two:
$$\cos 2x = 2\cos^2 x - 1 \quad \Leftrightarrow \quad \cos^2 x = \frac{1 + \cos 2x}{2}$$
That last rearrangement — $\cos^2 x = \tfrac{1}{2}(1 + \cos 2x)$ — is the power-reduction formula used in calculus to integrate $\cos^2 x$. Similarly:
$$\sin^2 x = \frac{1 - \cos 2x}{2}$$
So while $\cos 2x \neq \cos^2 x$, the two are linked by a clean algebraic identity that's essential for integration. The notation distinction is the entry point; the power-reduction identity is the payoff.
When Do You Use Each Form?
Form | Use When |
|---|---|
$\cos^2 x - \sin^2 x$ | Both $\sin x$ and $\cos x$ are known |
$2\cos^2 x - 1$ | Only $\cos x$ is known |
$1 - 2\sin^2 x$ | Only $\sin x$ is known |
$\frac{1 - \tan^2 x}{1 + \tan^2 x}$ | Only $\tan x$ is known |
Picking the right form saves a step.
Worked Examples
Example 1: Direct Computation
Compute $\cos 60°$ using the cos2x identity with $x = 30°$.
Use Form 2 (only $\cos 30°$ needed):
$$\cos 60° = 2 \cos^2 30° - 1 = 2 \cdot \tfrac{3}{4} - 1 = \tfrac{3}{2} - 1 = \tfrac{1}{2}$$
Check against the standard value: $\cos 60° = \tfrac{1}{2}$ ✓.
Example 2: When Sin is Given
If $\sin x = \tfrac{3}{5}$, find $\cos 2x$.
Use Form 3 (only $\sin x$ needed):
$$\cos 2x = 1 - 2 \sin^2 x = 1 - 2 \cdot \tfrac{9}{25} = 1 - \tfrac{18}{25} = \tfrac{7}{25}$$
Example 3: Solving a Trigonometric Equation
Solve $\cos 2x + \cos x = 0$ for $x \in [0, 2\pi)$.
Substitute Form 2:
$$2\cos^2 x - 1 + \cos x = 0$$
Let $u = \cos x$:
$$2u^2 + u - 1 = 0$$
Factor: $(2u - 1)(u + 1) = 0$, so $u = \tfrac{1}{2}$ or $u = -1$.
$\cos x = \tfrac{1}{2}$ gives $x = \tfrac{\pi}{3}$ or $x = \tfrac{5\pi}{3}$. $\cos x = -1$ gives $x = \pi$.
Solutions: $x = \tfrac{\pi}{3}, \pi, \tfrac{5\pi}{3}$.
Why Does Cos2x Matter?
"The double-angle formula is the gateway to power-reduction in calculus." — adapted from any standard calculus text.
The cos2x identity isn't just an academic identity. It's the key to several practical computations:
Integration in calculus. $\int \cos^2 x , dx$ is hard to integrate directly. Using $\cos^2 x = \frac{1 + \cos 2x}{2}$ (rearranged from Form 2) makes it easy: $\int \frac{1 + \cos 2x}{2} dx = \frac{x}{2} + \frac{\sin 2x}{4} + C$.
Audio synthesis — AM radio. Amplitude modulation creates side bands at $f_{\text{carrier}} \pm f_{\text{audio}}$. The math behind this uses the product-to-sum formulas, which come from cos2x.
Wave interference. When two waves with similar frequencies interfere, the resulting amplitude varies according to cos((f₁ − f₂)t/2) — directly the cos of a double angle.
Physics — power in AC circuits. Instantaneous power $P(t) = V_0 I_0 \cos^2(\omega t + \phi)$ uses Form 2 to find average power.
Signal processing. The discrete cosine transform — at the heart of JPEG and MP3 — uses cosine identities heavily.
The cos2x identity is named in Leonhard Euler's 1748 Introductio in Analysin Infinitorum, though it had been used by Indian mathematicians like Bhaskara II centuries earlier in their trigonometric work.
Learn more: Trigonometry Formulas — Full List
A Worked Example
Find $\cos 2x$ given $\sin x = \tfrac{4}{5}$ and $x$ is in Q2.
The intuitive (wrong) approach. A student in a hurry uses Form 1 without finding $\cos x$ first:
$$\cos 2x \stackrel{?}{=} \cos^2 x - \sin^2 x = ? - \tfrac{16}{25}$$
But they don't know $\cos x$, so they can't complete the calculation — and they pick the wrong form.
Why it fails. Form 1 requires both $\sin x$ and $\cos x$. If only one is given, Form 2 or Form 3 saves the work.
The correct method. Since only $\sin x$ is given, use Form 3:
$$\cos 2x = 1 - 2 \sin^2 x = 1 - 2 \cdot \tfrac{16}{25} = 1 - \tfrac{32}{25} = -\tfrac{7}{25}$$
Check. $\cos 2x = -\tfrac{7}{25}$. (Negative because $2x$ is in Q3 or Q4 when $x$ is in Q2.)
At Bhanzu, our trainers teach the form-selection rule — pick the form that matches what you have — before deriving the identities. The student who learns this rule first never gets stuck choosing.
What Are the Most Common Mistakes With Cos2x?
Mistake 1: Treating $\cos 2x$ as $2 \cos x$
Where it slips in: Linear-distribution intuition applied to a non-linear function.
Don't do this: $\cos 2x = 2 \cos x$.
The correct way: $\cos 2x = \cos^2 x - \sin^2 x$ (or equivalent forms). The cosine of a doubled angle is not double the cosine. Check with $x = 30°$: $2 \cos 30° = \sqrt{3} \approx 1.73$, but $\cos 60° = 0.5$. Different.
Mistake 2: Forgetting Form 2 has a minus, not a plus
Where it slips in: Recalling $\cos 2x = 2 \cos^2 x \pm 1$ and guessing the sign.
Don't do this: $\cos 2x = 2 \cos^2 x + 1$.
The correct way: $\cos 2x = 2 \cos^2 x - 1$. The minus sign matters. Memory aid: at $x = 90°$, $\cos 2 \cdot 90° = \cos 180° = -1$, and Form 2 gives $2 \cdot 0 - 1 = -1$ ✓; plus would give $+1$, wrong.
Mistake 3: Picking the slowest form for the problem
Where it slips in: Using Form 1 when only $\sin x$ is given.
Don't do this: Use Form 1 always, regardless of what's given.
The correct way: Match the form to the data. Form 1 if both known; Form 2 if only cos; Form 3 if only sin; Form 4 if only tan. Pick wisely; save work.
The Mathematicians Who Shaped the Cos2x Identity
Leonhard Euler (1707–1783, Switzerland) — Standardised the modern statement of the double-angle identities in his 1748 Introductio in Analysin Infinitorum.
Bhaskara II (1114–1185, India) — Used double-angle relationships in his trigonometric work for astronomy in Siddhanta-Shiromani.
Hipparchus of Nicaea (c. 190–c. 120 BCE, Greece) — His chord-table calculations implicitly contained the double-angle relationships, though without modern notation.
A Practical Next Step
Try these three before moving on to triple-angle formulas.
Use $\cos 2x = 2\cos^2 x - 1$ to find $\cos 90°$ given $\cos 45° = \tfrac{1}{\sqrt{2}}$.
If $\sin x = \tfrac{12}{13}$ and $x$ is in Q1, find $\cos 2x$.
Solve $\cos 2x = \tfrac{1}{2}$ for $x \in [0, \pi]$.
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