Double-Angle and Half-Angle Formulas
2倍角の公式
Overview
The double-angle formulas and half-angle formulas are important identities derived directly from the addition theorems. They appear frequently in integration and in solving trigonometric equations.
Prerequisite: Addition Theorems
The following addition theorems are used (see Chapter 4):
$$\sin(\alpha + \beta) = \sin\alpha\cos\beta + \cos\alpha\sin\beta$$ $$\cos(\alpha + \beta) = \cos\alpha\cos\beta - \sin\alpha\sin\beta$$Double-Angle Formulas
Setting $\alpha = \beta = \theta$ in the addition theorems yields the double-angle formulas.
Proof of sin 2θ
Step 1: Substitute α = β = θ into the addition theorem
$$\sin 2\theta = \sin(\theta + \theta) = \sin\theta\cos\theta + \cos\theta\sin\theta$$Step 2: Simplify
$$\sin 2\theta = 2\sin\theta\cos\theta$$Figure 5.1: Relationship between sin 2θ and sin θ, cos θ (example with θ = 40°).
The y-coordinate of the point at angle 2θ on the unit circle equals sin 2θ, shown here as a vertical segment.
Double-Angle Formula for Sine
$$\sin 2\theta = 2\sin\theta\cos\theta \quad \blacksquare$$Proof of cos 2θ
Step 1: Substitute α = β = θ into the addition theorem
$$\cos 2\theta = \cos(\theta + \theta) = \cos\theta\cos\theta - \sin\theta\sin\theta$$Step 2: Simplify
$$\cos 2\theta = \cos^2\theta - \sin^2\theta$$This expression can be rewritten using $\sin^2\theta + \cos^2\theta = 1$.
Variant 1: Eliminate sin²
Substituting $\sin^2\theta = 1 - \cos^2\theta$:
$$\cos 2\theta = \cos^2\theta - (1 - \cos^2\theta) = 2\cos^2\theta - 1$$Variant 2: Eliminate cos²
Substituting $\cos^2\theta = 1 - \sin^2\theta$:
$$\cos 2\theta = (1 - \sin^2\theta) - \sin^2\theta = 1 - 2\sin^2\theta$$Figure 5.2: Three equivalent representations of $\cos 2\theta$
Double-Angle Formula for Cosine (Three Forms)
\begin{align} \cos 2\theta &= \cos^2\theta - \sin^2\theta \\ \cos 2\theta &= 2\cos^2\theta - 1 \label{eq:cos2theta-form2} \\ \cos 2\theta &= 1 - 2\sin^2\theta \label{eq:cos2theta-form3} \quad \blacksquare \end{align}Proof of tan 2θ
Step 1: Substitute α = β = θ into the addition theorem for tangent
$$\tan 2\theta = \tan(\theta + \theta) = \frac{\tan\theta + \tan\theta}{1 - \tan\theta \cdot \tan\theta}$$Step 2: Simplify
$$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$Double-Angle Formula for Tangent
$$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta} \quad \blacksquare$$Half-Angle Formulas
Rearranging the double-angle formulas yields the half-angle formulas.
Proof of cos²(θ/2)
Step 1: Start from the double-angle formula
Replacing $\theta$ by $\dfrac{\theta}{2}$ in equation \eqref{eq:cos2theta-form2}, the left-hand side becomes $\cos\!\left(2 \cdot \dfrac{\theta}{2}\right) = \cos\theta$:
$$\cos\theta = 2\cos^2\frac{\theta}{2} - 1$$Step 2: Solve for cos²(θ/2)
$$2\cos^2\frac{\theta}{2} = 1 + \cos\theta$$ $$\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}$$Proof of sin²(θ/2)
Step 1: Start from the double-angle formula
Similarly, replacing $\theta$ by $\dfrac{\theta}{2}$ in equation \eqref{eq:cos2theta-form3}:
$$\cos\theta = 1 - 2\sin^2\frac{\theta}{2}$$Step 2: Solve for sin²(θ/2)
$$2\sin^2\frac{\theta}{2} = 1 - \cos\theta$$ $$\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$$Half-Angle Formulas
$$\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$$ $$\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}$$Proof of tan(θ/2)
Method 1: From the ratio of sine and cosine
$$\tan^2\frac{\theta}{2} = \frac{\sin^2\frac{\theta}{2}}{\cos^2\frac{\theta}{2}} = \frac{\dfrac{1-\cos\theta}{2}}{\dfrac{1+\cos\theta}{2}} = \frac{1-\cos\theta}{1+\cos\theta}$$Method 2: An alternative form
Multiplying numerator and denominator by $(1-\cos\theta)$:
$$\tan^2\frac{\theta}{2} = \frac{(1-\cos\theta)^2}{(1+\cos\theta)(1-\cos\theta)} = \frac{(1-\cos\theta)^2}{1-\cos^2\theta} = \frac{(1-\cos\theta)^2}{\sin^2\theta}$$Taking the square root of both sides would normally introduce a $\pm$ sign, but the sign is determined automatically. Since $1-\cos\theta$ is always non-negative, the sign of the denominator $\sin\theta$ determines the sign of the entire right-hand side. This always matches the sign of $\tan\dfrac{\theta}{2}$, so no $\pm$ is needed. Therefore:
$$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta}$$Method 3: Yet another form
Similarly, multiplying numerator and denominator by $(1+\cos\theta)$:
$$\tan^2\frac{\theta}{2} = \frac{(1-\cos\theta)(1+\cos\theta)}{(1+\cos\theta)^2} = \frac{1-\cos^2\theta}{(1+\cos\theta)^2} = \frac{\sin^2\theta}{(1+\cos\theta)^2}$$Taking the square root:
$$\tan\frac{\theta}{2} = \frac{\sin\theta}{1+\cos\theta}$$(Since $1+\cos\theta$ is always non-negative, the sign is automatically correct in this form as well.)
Half-Angle Formulas for Tangent (Three Forms)
$$\tan^2\frac{\theta}{2} = \frac{1-\cos\theta}{1+\cos\theta}$$ $$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta} \quad \blacksquare$$Triple-Angle Formulas
By combining the double-angle formulas with the addition theorems, one can derive the triple-angle formulas.
Proof of sin 3θ
Step 1: Decompose 3θ = 2θ + θ
$$\sin 3\theta = \sin(2\theta + \theta)$$Step 2: Apply the addition theorem
$$= \sin 2\theta \cos\theta + \cos 2\theta \sin\theta$$Step 3: Substitute the double-angle formulas
Using $\sin 2\theta = 2\sin\theta\cos\theta$ and $\cos 2\theta = 1 - 2\sin^2\theta$ (choosing this form to express everything in terms of $\sin\theta$):
$$= 2\sin\theta\cos\theta \cdot \cos\theta + (1 - 2\sin^2\theta) \cdot \sin\theta$$ $$= 2\sin\theta\cos^2\theta + \sin\theta - 2\sin^3\theta$$Step 4: Substitute cos²θ = 1 - sin²θ and simplify
$$= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta$$ $$= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta$$ $$= 3\sin\theta - 4\sin^3\theta$$Proof of cos 3θ
Step 1: Apply the addition theorem
$$\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta$$Step 2: Substitute the double-angle formulas
Using $\cos 2\theta = 2\cos^2\theta - 1$ (choosing this form to express everything in terms of $\cos\theta$) and $\sin 2\theta = 2\sin\theta\cos\theta$:
$$= (2\cos^2\theta - 1)\cos\theta - 2\sin\theta\cos\theta \cdot \sin\theta$$ $$= 2\cos^3\theta - \cos\theta - 2\sin^2\theta\cos\theta$$Step 3: Substitute sin²θ = 1 - cos²θ and simplify
$$= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta$$ $$= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta$$ $$= 4\cos^3\theta - 3\cos\theta$$Triple-Angle Formulas
$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$ $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta \quad \blacksquare$$Tips for Mastering the Double-Angle Formulas
sin 2θ = 2 sin θ cos θ
"Two-sin-cos"
The coefficient 2 comes first, followed by one factor each of sin and cos. Simply recite "two, sin, cos" — there are only three elements, making this the easiest formula to memorize.
cos 2θ — Choosing among the three forms
Which form to use?
cos 2θ has three equivalent expressions; choose based on what you need to eliminate.
- To eliminate sin → $\cos 2\theta = 2\cos^2\theta - 1$ (cosine only)
- To eliminate cos → $\cos 2\theta = 1 - 2\sin^2\theta$ (sine only)
- Either may remain → $\cos 2\theta = \cos^2\theta - \sin^2\theta$ (basic form)
Read in reverse, these become the half-angle formulas: $\cos^2\theta = \dfrac{1 + \cos 2\theta}{2}$, $\sin^2\theta = \dfrac{1 - \cos 2\theta}{2}$.
tan 2θ = 2 tan θ / (1 − tan²θ)
Derive it in three seconds from the addition theorem
Simply set $\alpha = \beta = \theta$ in the addition formula $\tan(\alpha+\beta) = \dfrac{\tan\alpha + \tan\beta}{1 - \tan\alpha\tan\beta}$. If you know the addition theorem, you can derive this formula instantly. Practicing derivation is more practical than rote memorization.
Worked Examples
Example 1: Finding sin 2θ and cos 2θ
Problem: Given $\sin\theta = \dfrac{3}{5}$ with $0 < \theta < \dfrac{\pi}{2}$, find $\sin 2\theta$ and $\cos 2\theta$.
Step 1: Find cos θ
From $\sin^2\theta + \cos^2\theta = 1$, we get $\cos^2\theta = 1 - \left(\dfrac{3}{5}\right)^2 = 1 - \dfrac{9}{25} = \dfrac{16}{25}$.
Since $0 < \theta < \dfrac{\pi}{2}$, we have $\cos\theta > 0$, so $\cos\theta = \dfrac{4}{5}$.
Step 2: Apply the double-angle formulas
$$\sin 2\theta = 2\sin\theta\cos\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = \frac{24}{25}$$ $$\cos 2\theta = \cos^2\theta - \sin^2\theta = \frac{16}{25} - \frac{9}{25} = \frac{7}{25}$$Answer
$$\sin 2\theta = \frac{24}{25}, \quad \cos 2\theta = \frac{7}{25}$$Example 2: Trigonometric Equation
Problem: Solve $\cos 2\theta + 3\cos\theta + 2 = 0$ for $0 \leq \theta < 2\pi$.
Step 1: Express cos 2θ in terms of cos θ only
Using $\cos 2\theta = 2\cos^2\theta - 1$ (choosing the form that eliminates sin):
$$(2\cos^2\theta - 1) + 3\cos\theta + 2 = 0$$ $$2\cos^2\theta + 3\cos\theta + 1 = 0$$Step 2: Solve the quadratic in cos θ
Setting $t = \cos\theta$, we have $2t^2 + 3t + 1 = 0$. Factoring:
$$(2t + 1)(t + 1) = 0$$$t = -\dfrac{1}{2}$ or $t = -1$.
Step 3: Find θ
$\cos\theta = -\dfrac{1}{2}$ → $\theta = \dfrac{2\pi}{3},\; \dfrac{4\pi}{3}$
$\cos\theta = -1$ → $\theta = \pi$
Answer
$$\theta = \frac{2\pi}{3},\; \pi,\; \frac{4\pi}{3}$$Example 3: Applying the Half-Angle Formula
Problem: Find the value of $\sin^2 15°$ using the half-angle formula.
Step 1: Apply the half-angle formula
Since $15° = \dfrac{30°}{2}$, we use $\sin^2\dfrac{\theta}{2} = \dfrac{1 - \cos\theta}{2}$ with $\theta = 30°$:
$$\sin^2 15° = \frac{1 - \cos 30°}{2} = \frac{1 - \dfrac{\sqrt{3}}{2}}{2}$$Step 2: Simplify
$$= \frac{\dfrac{2 - \sqrt{3}}{2}}{2} = \frac{2 - \sqrt{3}}{4}$$Answer
$$\sin^2 15° = \frac{2 - \sqrt{3}}{4}$$Practice Problems
Problem 1
Given $\cos\theta = \dfrac{1}{3}$ with $0 < \theta < \dfrac{\pi}{2}$, find $\sin 2\theta$.
Hint
Find $\sin\theta$ from $\sin^2\theta + \cos^2\theta = 1$, then substitute into $\sin 2\theta = 2\sin\theta\cos\theta$.
Solution
$\sin^2\theta = 1 - \dfrac{1}{9} = \dfrac{8}{9}$. Since $0 < \theta < \dfrac{\pi}{2}$, $\sin\theta = \dfrac{2\sqrt{2}}{3}$.
$$\sin 2\theta = 2 \cdot \frac{2\sqrt{2}}{3} \cdot \frac{1}{3} = \frac{4\sqrt{2}}{9}$$Problem 2
Solve the equation $\cos 2\theta - \cos\theta = 0$ for $0 \leq \theta < 2\pi$.
Hint
Substitute $\cos 2\theta = 2\cos^2\theta - 1$ to obtain a quadratic equation in $\cos\theta$.
Solution
$2\cos^2\theta - 1 - \cos\theta = 0$ → $2\cos^2\theta - \cos\theta - 1 = 0$.
$(2\cos\theta + 1)(\cos\theta - 1) = 0$, giving $\cos\theta = -\dfrac{1}{2}$ or $\cos\theta = 1$.
$$\theta = 0,\; \frac{2\pi}{3},\; \frac{4\pi}{3}$$Problem 3
Find the value of $\cos^2 75°$ using the half-angle formula.
Hint
Note that $75° = \dfrac{150°}{2}$ and substitute $\theta = 150°$ into $\cos^2\dfrac{\theta}{2} = \dfrac{1 + \cos\theta}{2}$.
Solution
Problem 4
Given $\tan\theta = 2$, find $\tan 2\theta$.
Hint
Substitute directly into $\tan 2\theta = \dfrac{2\tan\theta}{1 - \tan^2\theta}$.
Solution
Problem 5
Given $\sin\theta + \cos\theta = \dfrac{1}{2}$, find $\sin 2\theta$.
Hint
Square both sides to obtain $\sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta$. Use $\sin^2\theta + \cos^2\theta = 1$ and $\sin 2\theta = 2\sin\theta\cos\theta$.
Solution
$(\sin\theta + \cos\theta)^2 = \sin^2\theta + 2\sin\theta\cos\theta + \cos^2\theta = 1 + \sin 2\theta$
$\left(\dfrac{1}{2}\right)^2 = 1 + \sin 2\theta$, hence
$$\sin 2\theta = \frac{1}{4} - 1 = -\frac{3}{4}$$Applications
Power Reduction in Integration
Integrals involving $\sin^2 x$ or $\cos^2 x$ are routinely handled by using the half-angle formulas (i.e., rearranged forms of the double-angle formula for cosine) to reduce the power:
$$\int \sin^2 x\, dx = \int \frac{1 - \cos 2x}{2}\, dx = \frac{x}{2} - \frac{\sin 2x}{4} + C$$Trigonometric Equations and Inequalities
By converting $\cos 2\theta$ into a quadratic expression in $\cos\theta$, trigonometric equations can be reduced to algebraic equations (see Example 2).
Summary
Double-Angle Formulas
$$\sin 2\theta = 2\sin\theta\cos\theta$$ $$\cos 2\theta = \cos^2\theta - \sin^2\theta = 2\cos^2\theta - 1 = 1 - 2\sin^2\theta$$ $$\tan 2\theta = \frac{2\tan\theta}{1 - \tan^2\theta}$$Half-Angle Formulas
$$\sin^2\frac{\theta}{2} = \frac{1 - \cos\theta}{2}$$ $$\cos^2\frac{\theta}{2} = \frac{1 + \cos\theta}{2}$$ $$\tan\frac{\theta}{2} = \frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$$Triple-Angle Formulas
$$\sin 3\theta = 3\sin\theta - 4\sin^3\theta$$ $$\cos 3\theta = 4\cos^3\theta - 3\cos\theta$$Column: Why Is There No "One-Third-Angle Formula"?
The half-angle formulas were obtained by "solving the double-angle formulas in reverse." One might then expect that solving the triple-angle formulas in reverse would yield "one-third-angle formulas." The answer, however, is that in general no closed-form expression exists.
Why the Half-Angle Formula Works
Recall that the derivation of the half-angle formula starts from
$$\cos\theta = 2\cos^2\frac{\theta}{2} - 1$$This is a quadratic equation in $t = \cos\dfrac{\theta}{2}$, namely $2t^2 - 1 = \cos\theta$, which is easily solved by the quadratic formula: $t = \pm\sqrt{\dfrac{1+\cos\theta}{2}}$.
Why the One-Third-Angle Case Fails
Reading the triple-angle formula $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ in reverse, $t = \cos\dfrac{\theta}{3}$ satisfies the cubic equation:
$$4t^3 - 3t = \cos\theta$$Cardano's formula (16th century) does exist for cubic equations, but when all three roots are real — as is often the case here — the formula necessarily involves cube roots of complex numbers (casus irreducibilis). The roots cannot be expressed using only real-valued arithmetic operations and radicals.
Connection to the angle trisection problem — The impossibility of trisecting an arbitrary angle with straightedge and compass alone, one of the three famous ancient Greek construction problems, corresponds precisely to the fact that this cubic equation cannot be solved using only quadratic radicals. Bisecting an angle (the half-angle case) reduces to a quadratic equation and is therefore constructible; that is why the half-angle formula has a neat closed form.
In other words, the existence of the half-angle formula and the non-existence of a one-third-angle formula is not a coincidence but reflects the essential difference between the degree of the underlying equation being 2 versus 3.