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Section 3.3 Double-Angle and Half-Angle Formulas

Suppose we want to accurately position the Hawaiian Star Compass on the Unit Circle. In FigureΒ 1.1.4, the house for Manu is located halfway between Hikina and Κ»Δ€kau, resulting in an angle of \(45^{\circ}\text{.}\) By applying right triangle trigonometry, we can determine the exact coordinates of Manu as \(\left(\cos(45^{\circ}), \sin(45^{\circ})\right) = \left(\frac{1}{\sqrt{2}} , \frac{1}{\sqrt{2}} \right)\text{.}\) However, as we move to the house of Κ»Δ€ina, located halfway between Manu and Hikina, we encounter a problem. The angle for Κ»Δ€ina is \(22.5^{\circ}\text{,}\) which is not explicitly listed in TableΒ 1.5.18. Therefore, we must resort to a calculator for numerical approximations.
In this section, we will learn about the double and half-angle formulas for trigonometry. These formulas allow us to determine exact trigonometric function values for angles that are double or half of the common angles. This will enable us to use our existing knowledge of trigonometric functions at \(45^{\circ}\) and apply the half-angle formulas to obtain exact values at \(22.5^{\circ}\text{.}\)

Subsection 3.3.1 Double-Angle Formulas

Recall the addition formula for sine:
\begin{equation} \sin (\alpha+\beta)=\sin \alpha \cos \beta + \cos \alpha \sin \beta\text{.}\tag{3.3.1} \end{equation}
Consider the case when the two angles are equal. We will call this angle \(\theta\text{,}\) so let \(\alpha=\theta\) and \(\beta=\theta\text{.}\) Then EqΒ (3.3.1) becomes
\begin{align*} \sin(\theta+\theta)\amp = \sin \theta \cos \theta + \cos \theta \sin \theta\\ \sin (2\theta)\amp = 2\sin \theta \cos \theta\text{.} \end{align*}
Thus we obtain a formula for sine of twice the angle \(\theta\text{.}\)

Definition 3.3.1. Double-Angle Formulas.

\begin{align*} \sin 2\theta \amp = 2\sin \theta \cos \theta\\ \cos 2\theta \amp = \cos^2 \theta - \sin^2 \theta\\ \cos 2\theta \amp = 1-2\sin^2 \theta\\ \cos 2\theta \amp = 2\cos^2 \theta-1\\ \tan 2\theta \amp = \frac{2\tan \theta}{1-\tan^2 \theta} \end{align*}
The proofs for the Double-Angle Formulas for Cosine and Tangent are left as exercises (ExerciseΒ 3.3.4.6-ExerciseΒ 3.3.4.9).

Remark 3.3.2.

Notice that there are three variations of the double-angle formula for cosine. All three equations give the correct answer, however, one version may be more convenient depending on the given information. For example, if we are given the value of \(\sin\theta\text{,}\) it may be easier to select the version that solely involves \(\sin\theta\) and does not include \(\cos\theta\text{.}\)

Example 3.3.3.

Given \(\sin\theta=-\frac{5}{13}\) and \(\theta\) lies in Quadrant III, find the exact value of:

(a)

\(\sin(2\theta)\)
Solution.
By the double-angle formula, we have \(\sin(2\theta)=2\sin\theta\cos\theta\text{.}\) We are given the value of \(\sin\theta\text{,}\) but we do not have \(\cos\theta\text{.}\) To find \(\cos\theta\text{,}\) we will draw the triangle formed from \(\sin\theta=-\frac{5}{13}\) where \(\theta\) lies in Quadrant III.
Using the Pythagorean Theorem, we can solve the triangle:
\begin{align*} x^2+(-5)^2\amp = 13^2\\ x^2+25\amp = 169\\ x^2\amp = 144\\ x\amp = 12\text{.} \end{align*}
Thus we have
\begin{equation*} \cos\theta=\frac{\mbox{adjacent}}{\mbox{hypotenuse}}=-\frac{12}{13} \end{equation*}
and
\begin{equation*} \tan\theta=\frac{\mbox{opposite}}{\mbox{adjacent}}=\frac{5}{12}\text{.} \end{equation*}
With this new information, we can use the double-angle formula to find \(\sin(2\theta)\text{:}\)
\begin{equation*} \sin(2\theta)=2\sin\theta\cos\theta=2\left(-\frac{5}{13}\right)\left(-\frac{12}{13}\right)=\frac{120}{169}\text{.} \end{equation*}

(b)

\(\cos(2\theta)\)
Solution.
To compute \(\cos2\theta\text{,}\) notice there are three different formulas: \(\cos2\theta=\cos^2\theta-\sin^2\theta\text{,}\) \(\cos2\theta=1-2\sin^2\theta\text{,}\) or \(\cos2\theta=2\cos^2\theta-1\text{.}\) Using any of the three equations will give us the correct answer. However, given that we know \(\sin\theta=-\frac{5}{13}\text{,}\) it may be easier to use \(\cos2\theta=1-2\sin^2\theta\text{,}\) since the other two equations require us to know \(\cos\theta\text{.}\)
Without having to draw the triangle, we get
\begin{align*} \cos2\theta \amp= 1-2\sin^2\theta \\ \amp = 1-2\left(-\frac{5}{13}\right)^2 \\ \amp = 1-2\left(\frac{25}{169}\right) \\ \amp = \frac{169}{169}-\frac{50}{169} \\ \amp = \frac{119}{169} \text{.} \end{align*}

(c)

\(\tan(2\theta)\)
Solution.
Using the double-angle formula for tangent, we get
\begin{align*} \tan2\theta \amp =\frac{2\tan\theta}{1-\tan^2\theta} \\ \amp = \frac{2\left(\frac{5}{12}\right)}{1-\left(\frac{5}{12}\right)^2} \\ \amp = \frac{\frac{10}{12}}{1-\frac{25}{144}} \\ \amp = \frac{\frac{10}{12}}{\frac{119}{144}} \\ \amp = \frac{10}{12}\cdot\frac{144}{119}\\ \amp = \frac{120}{119} \text{.} \end{align*}

Example 3.3.4.

Write \(\sin(3\theta)\) in terms of \(\sin\theta\text{.}\)

Solution.

\begin{align*} \sin(3\theta)\amp =\sin(2\theta+\theta)\\ \amp =\sin(2\theta)\cos\theta+\cos(2\theta)\sin\theta\amp \amp \mbox{addition formula}\\ \amp =(2\sin\theta\cos\theta)\cos\theta+(\cos^2\theta-\sin^2\theta)\sin\theta\amp \amp \mbox{double-angle formula}\\ \amp =2\sin\theta\cos^2\theta+\sin\theta\cos^2\theta-\sin^3\theta\\ \amp =3\sin\theta\cos^2\theta-\sin^3\theta\\ \amp =3\sin\theta(1-\sin^2\theta)-\sin^3\theta\amp \amp \mbox{Pythagorean Identity}\\ \amp =3\sin\theta-3\sin^3\theta-\sin^3\theta\\ \amp =3\sin\theta-4\sin^3\theta\text{.} \end{align*}

Subsection 3.3.2 Reducing Powers Formulas

The double-angle formula for cosine expresses a trigonometric function in terms of the square of another trigonometric function. By rearranging the terms, we can derive formulas for reducing the powers of sine, cosine, and tangent in expressions with even powers to terms involving only cosine. These formulas are particularly useful in calculus.

Definition 3.3.5. Formulas for Reducing Powers.

\begin{equation*} \sin^2\theta=\frac{1-\cos(2\theta)}{2} \end{equation*}
\begin{equation*} \cos^2\theta=\frac{1+\cos(2\theta)}{2} \end{equation*}
\begin{equation*} \tan^2\theta=\frac{1-\cos(2\theta)}{1+\cos(2\theta)} \end{equation*}

Proof.

To prove the first formula, solve for \(\sin^2\theta\) in the double-angle formula: \(\cos2\theta=1-2\sin^2\theta\text{.}\) The second formula is obtained similarly by solving for \(\cos^2\theta\) in the formula \(\cos2\theta=2\cos^2\theta-1\text{.}\) The first two formulas can be used to obtain the third formula:
\begin{equation*} \tan^2\theta=\frac{\sin^2\theta}{\cos^2\theta}=\frac{\dfrac{1-\cos(2\theta)}{2}}{\dfrac{1+\cos(2\theta)}{2}}=\frac{1-\cos(2\theta)}{1+\cos(2\theta)}\text{.} \end{equation*}

Example 3.3.6.

Write \(\sin^4\theta\) as an expression that does not involve powers of sine or cosine greater than 1.

Solution.

We will use the Reducing Powers Formula twice.
\begin{align*} \sin^4\theta \amp = (\sin^2\theta)^2\\ \amp =\left(\frac{1-\cos(2\theta)}{2}\right)^2\amp \amp \mbox{reducing powers}\\ \amp =\frac{1}{4}\left(1-2\cos(2\theta)+\cos^2(2\theta)\right)\\ \amp =\frac{1}{4}\left(1-2\cos(2\theta)+\frac{1+\cos(4\theta)}{2}\right)\amp \amp \mbox{reducing powers}\\ \amp =\frac{1}{4}\left(1-2\cos(2\theta)+\frac{1}{2}+\frac{\cos(4\theta)}{2}\right)\\ \amp =\frac{1}{4}\left(\frac{3}{2}-\frac{4}{2}\cos(2\theta)+\frac{1}{2}\cos(4\theta)\right)\\ \amp =\frac{1}{8}\left(3-4\cos(2\theta)+\cos(4\theta)\right)\text{.} \end{align*}

Subsection 3.3.3 Half-Angle Formulas

Another set of useful formulas are the half-angle formulas.

Definition 3.3.7. Half-Angle Formulas.

\begin{align*} \sin\frac{\theta}{2}\amp =\pm\sqrt{\frac{1-\cos\theta}{2}},\amp \cos\frac{\theta}{2}\amp =\pm\sqrt{\frac{1+\cos\theta}{2}},\amp \tan\frac{\theta}{2}\amp =\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}} \end{align*}
The choice of the + or - sign depends on the Quadrant in which \(\theta/2\) lies.

Proof.

We take the square root of both sides of the Formulas for Reducing Powers (DefinitionΒ 3.3.5) and halve the angle (\(\theta\) becomes \(\frac{\theta}{2}\) and \(2\theta\) becomes \(\theta\)) to arrive at our formulas.

Example 3.3.8. Locating Κ»Δ€ina.

We are now ready to revisit the problem posed at the start of this section where we were asked to determine the exact coordinates of the house Κ»Δ€ina on the Unit Circle.

Solution.

We know the coordinates are at
\begin{equation*} (\cos22.5^{\circ}, \sin22.5^{\circ})\text{.} \end{equation*}
To determine the exact value of \(\cos 22.5^{\circ}\text{,}\) we use the half-angle formula along with the known value \(\cos 45^{\circ} =\frac{1}{\sqrt{2}} =\frac{\sqrt{2}}{2}\text{:}\)
\begin{align*} \cos22.5^{\circ} \amp = \cos\left(\frac{45}{2}\right)^{\circ} \\ \amp = \sqrt{\frac{1+\cos45^{\circ}}{2}} \\ \amp = \sqrt{\frac{1+\frac{\sqrt{2}}{2}}{2}}\\ \amp = \sqrt{\frac{\frac{2}{2}+\frac{\sqrt{2}}{2}}{4}} \\ \amp = \sqrt{\frac{2+\sqrt{2}}{4}} \\ \amp = \frac{1}{2}\sqrt{2+\sqrt{2}}\text{.} \end{align*}
Since the half-angle formula has \(\pm\text{,}\) we check the quadrant. In this case, our angle is \(22.5^{\circ}\text{,}\) which is in Quadrant I. Therefore, we choose the positive value.
Finding the exact value of \(\sin22.5^{\circ}\) is left for ExerciseΒ 3.3.4.1.

Example 3.3.9.

Given \(\sin\theta=-\frac{5}{13}\) and \(\theta\) lies in Quadrant III, find the exact value of:

(a)

\(\sin\frac{\theta}{2}\)
Solution.
Notice the Half-Angle Formulas all require us to know \(\cos\theta\text{.}\) Since the given information describes the same triangle as in ExampleΒ 3.3.3, we refer to that problem to get \(\cos\theta=-\frac{12}{13}\text{.}\)
Next, since \(\theta\) is in Quadrant III, \(180^{\circ}\lt \theta\lt 270^{\circ}\text{,}\) dividing by 2 gives us \(\frac{180^{\circ}}{2}\lt \frac{\theta}{2}\lt \frac{270^{\circ}}{2}\) or \(90^{\circ}\lt \frac{\theta}{2}\lt 135^{\circ}\text{.}\) Therefore, we conclude that \(\frac{\theta}{2}\) lies in Quadrant II.
To calculate \(\sin\frac{\theta}{2}\text{,}\) we first note that because \(\frac{\theta}{2}\) lies in Quadrant II, \(\sin\frac{\theta}{2}>0\) so we will choose the positive (+) sign in the Half-Angle Formula:
\begin{align*} \sin\frac{\theta}{2}\amp = \sqrt{\frac{1-\cos\theta}{2}}\\ \amp = \sqrt{\frac{1-\left(-\frac{12}{13}\right)}{2}}\\ \amp = \sqrt{\frac{1+\frac{12}{13}}{2}}\\ \amp = \sqrt{\frac{\frac{13}{13}+\frac{12}{13}}{2}}\\ \amp = \sqrt{\frac{\frac{25}{13}}{2}}\\ \amp = \sqrt{\frac{25}{26}}\text{.} \end{align*}

(b)

\(\cos\frac{\theta}{2}\)
Solution.
Since \(\frac{\theta}{2}\) is in Quadrant II, we know that \(\cos\frac{\theta}{2}\lt 0\) so we will choose the negative (-) sign in the Half-Angle Formula:
\begin{align*} \cos\frac{\theta}{2}\amp = -\sqrt{\frac{1+\cos\theta}{2}}\\ \amp = -\sqrt{\frac{1+\left(-\frac{12}{13}\right)}{2}}\\ \amp = -\sqrt{\frac{1-\frac{12}{13}}{2}}\\ \amp = -\sqrt{\frac{\frac{13}{13}-\frac{12}{13}}{2}}\\ \amp = -\sqrt{\frac{\frac{1}{13}}{2}}\\ \amp = -\sqrt{\frac{1}{26}}\text{.} \end{align*}

(c)

\(\tan\frac{\theta}{2}\)
Solution.
Since \(\frac{\theta}{2}\) is in Quadrant II, we know that \(\tan\frac{\theta}{2}\lt 0\) so we will choose the negative (-) sign in the Half-Angle Formula:
\begin{align*} \tan\frac{\theta}{2}\amp = -\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}\\ \amp = -\sqrt{\frac{1-\left(-\frac{12}{13}\right)}{1+\left(-\frac{12}{13}\right)}}\\ \amp = -\sqrt{\frac{1+\frac{12}{13}}{1-\frac{12}{13}}}\\ \amp = -\sqrt{\frac{\frac{13}{13}+\frac{12}{13}}{\frac{13}{13}-\frac{12}{13}}}\\ \amp = -\sqrt{\frac{\frac{25}{13}}{\frac{1}{13}}}\\ \amp = -\sqrt{\frac{25}{13}\cdot\frac{13}{1}}\\ \amp = -\sqrt{25}\\ \amp = -5\text{.} \end{align*}
We can derive another formula for \(\tan\frac{\theta}{2}\) that does not involve \(\pm\text{.}\)

Definition 3.3.10. Half-Angle Formulas for Tangent.

\begin{equation*} \tan\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{1+\cos\theta}}=\frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta} \end{equation*}

Proof.

We begin by first multiplying both sides of the Sine formula for Reducing Powers by 2 and halving the angle:
\begin{equation*} 1-\cos\theta=2\sin^2\left(\frac{\theta}{2}\right) \end{equation*}
and applying the double-angle formula to:
\begin{equation*} \sin\theta=\sin\left(2\cdot\frac{\theta}{2}\right)=2\sin\frac{\theta}{2}\cos\frac{\theta}{2}\text{.} \end{equation*}
By dividing the two preceding results we get
\begin{equation*} \frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}=\tan\frac{\theta}{2}\text{.} \end{equation*}
Thus
\begin{equation*} \tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}\text{.} \end{equation*}
Similarly, it can be shown that
\begin{equation*} \tan\frac{\theta}{2}=\frac{\sin\theta}{1+\cos\theta}\text{.} \end{equation*}

Example 3.3.11.

Calculate \(\tan\frac{\theta}{2}\) from ExampleΒ 3.3.9 using the above formula.

Solution.

\begin{equation*} \tan\frac{\theta}{2}=\frac{1-\cos\theta}{\sin\theta}=\frac{1-\left(-\frac{12}{13}\right)}{-\frac{5}{13}}=\frac{\frac{25}{13}}{-\frac{5}{13}}=\frac{25}{13}\cdot\left(-\frac{13}{5}\right)=-5\text{.} \end{equation*}
Note: We obtained the same result for \(\tan\frac{\theta}{2}\) as we did in ExampleΒ 3.3.9. In this example, we did not have to determine whether \(\tan\frac{\theta}{2}\) was positive or negative, however, we need to know the values of both \(\sin\theta\) and \(\cos\theta\text{.}\)

Exercises 3.3.4 Exercises

Exercise Group.

The house Κ»Δ€ina is located at \(22.5^{\circ}=\frac{45^{\circ}}{2}\) and the house Nā Leo is located at \(67.5^{\circ}=\frac{135^{\circ}}{2}\text{.}\) Use the half-angle formulas to evaluate the exact value of the given expression at each of these houses.

6.

Use the Addition Formula, \(\cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta\text{,}\) to prove the double angle formula for cosine:
\begin{equation*} \cos2\theta=\cos^2\theta-\sin^2\theta\text{.} \end{equation*}

7.

Use the Pythagorean Identity (\(\sin^2\theta+\cos^2\theta=1\)) and the result from ExerciseΒ 3.3.4.6 to prove
\begin{equation*} \cos2\theta=1-2\sin^2\theta\text{.} \end{equation*}

8.

Use the Pythagorean Identity (\(\sin^2\theta+\cos^2\theta=1\)) and the result from ExerciseΒ 3.3.4.6 to prove
\begin{equation*} \cos2\theta=2\cos^2\theta-1\text{.} \end{equation*}

9.

Use the addition formula, \(\tan(\alpha+\beta)=\frac{\tan\alpha+\tan\beta}{1-\tan\alpha\tan\beta}\text{,}\) to prove the double angle formula for tangent:
\begin{equation*} \tan2\theta=\frac{2\tan\theta}{1-\tan^2\theta}\text{.} \end{equation*}

Exercise Group.

Use the figure below to find the exact values for each of the following exercises.

Exercise Group.

Use the figure below to find the exact values for each of the following exercises.

Exercise Group.

Find the exact value of each expression given \(\cos\theta=-\frac{3}{5}\) and \(\theta\) is in Quadrant III.

Exercise Group.

Find the exact value of each expression given \(\sin\theta=-\frac{8}{17}\) and \(270^{\circ}\lt\theta\lt360^{\circ}\text{.}\)

Exercise Group.

Find the exact value of each expression given \(\cos\theta=\frac{2}{3}\) and \(\frac{3\pi}{2}\lt\theta\lt2\pi\text{.}\)

Exercise Group.

Use the Half-Angle Formula to find the exact value of each of the following

Exercise Group.

Write each of the following as expressions that do not involve powers of sine or cosine greater than 1.

Exercise Group.

Use half angle to find the exact deviation for the indicated angle, \(\theta\text{.}\)

Exercise Group.

Verify the identity

69.

\(\cos(2\theta)=\frac{\cot\theta-\tan\theta}{\cot\theta+\tan\theta}\)

75.

\(\csc^2\left(\frac{\theta}{2}\right)=\frac{2}{1-\cos\theta}\)

76.

\(\sec^2\left(\frac{\theta}{2}\right)=\frac{2}{1+\cos\theta}\)
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