7.3 Double-angle, half-angle, and reduction formulas (Page 2/8)

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Given $\sin α = \frac{5}{8},$ with $θ$ in quadrant I, find $\cos (2 α) .$

$\cos (2 α) = \frac{7}{32}$

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Using the double-angle formula for cosine without exact values

Use the double-angle formula for cosine to write $\cos (6 x)$ in terms of $\cos (3 x) .$

\begin{array}{l} \cos (6 x) = \cos (3 x + 3 x) \\ = \cos 3 x \cos 3 x - \sin 3 x \sin 3 x \\ = \cos^{2} 3 x - \sin^{2} 3 x \end{array}

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Using double-angle formulas to verify identities

Establishing identities using the double-angle formulas is performed using the same steps we used to derive the sum and difference formulas. Choose the more complicated side of the equation and rewrite it until it matches the other side.

Using the double-angle formulas to establish an identity

Establish the following identity using double-angle formulas:

1 + \sin (2 θ) = {(\sin θ + \cos θ)}^{2}

We will work on the right side of the equal sign and rewrite the expression until it matches the left side.

\begin{array}{l} {(\sin θ + \cos θ)}^{2} = \sin^{2} θ + 2 \sin θ \cos θ + \cos^{2} θ \\ = (\sin^{2} θ + \cos^{2} θ) + 2 \sin θ \cos θ \\ = 1 + 2 \sin θ \cos θ \\ = 1 + \sin (2 θ) \end{array}

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Establish the identity: $\cos^{4} θ - \sin^{4} θ = \cos (2 θ) .$

$\cos^{4} θ - \sin^{4} θ = (\cos^{2} θ + \sin^{2} θ) (\cos^{2} θ - \sin^{2} θ) = \cos (2 θ)$

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Verifying a double-angle identity for tangent

Verify the identity:

\tan (2 θ) = \frac{2}{\cot θ - \tan θ}

In this case, we will work with the left side of the equation and simplify or rewrite until it equals the right side of the equation.

\begin{array}{l} \tan (2 θ) = \frac{2 \tan θ}{1 - \tan^{2} θ} & Double-angle formula \\ = \frac{2 \tan θ (\frac{1}{\tan θ})}{(1 - \tan^{2} θ) (\frac{1}{\tan θ})} \begin{matrix}  \end{matrix} & Multiply by a term that results in desired numerator . \\ = \frac{2}{\frac{1}{\tan θ} - \frac{\tan^{2} θ}{\tan θ}} \\ = \frac{2}{\cot θ - \tan θ} & Use reciprocal identity for \frac{1}{\tan θ} . \end{array}

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Try It

Verify the identity: $\cos (2 θ) \cos θ = \cos^{3} θ - \cos θ \sin^{2} θ .$

$\cos (2 θ) \cos θ = (\cos^{2} θ - \sin^{2} θ) \cos θ = \cos^{3} θ - \cos θ \sin^{2} θ$

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Use reduction formulas to simplify an expression

The double-angle formulas can be used to derive the reduction formulas , which are formulas we can use to reduce the power of a given expression involving even powers of sine or cosine. They allow us to rewrite the even powers of sine or cosine in terms of the first power of cosine. These formulas are especially important in higher-level math courses, calculus in particular. Also called the power-reducing formulas, three identities are included and are easily derived from the double-angle formulas.

We can use two of the three double-angle formulas for cosine to derive the reduction formulas for sine and cosine. Let’s begin with $\cos (2 θ) = 1 - 2 \sin^{2} θ .$ Solve for $\sin^{2} θ :$

\begin{array}{l} \cos (2 θ) = 1 - 2 \sin^{2} θ \\ 2 \sin^{2} θ = 1 - \cos (2 θ) \\ \sin^{2} θ = \frac{1 - \cos (2 θ)}{2} \end{array}

Next, we use the formula $\cos (2 θ) = 2 \cos^{2} θ - 1.$ Solve for $\cos^{2} θ :$

\begin{array}{l} \cos (2 θ) = 2 \cos^{2} θ - 1 \\ 1 + \cos (2 θ) = 2 \cos^{2} θ \\ \frac{1 + \cos (2 θ)}{2} = \cos^{2} θ \end{array}

The last reduction formula is derived by writing tangent in terms of sine and cosine:

\begin{array}{l} \tan^{2} θ = \frac{\sin^{2} θ}{\cos^{2} θ} \\ = \frac{\frac{1 - \cos (2 θ)}{2}}{\frac{1 + \cos (2 θ)}{2}} & Substitute the reduction formulas. \\ = (\frac{1 - \cos (2 θ)}{2}) (\frac{2}{1 + \cos (2 θ)}) \begin{matrix}  \end{matrix} \\ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)} \end{array}

A General Note

Reduction formulas

The reduction formulas are summarized as follows:

\sin^{2} θ = \frac{1 - \cos (2 θ)}{2}

\cos^{2} θ = \frac{1 + \cos (2 θ)}{2}

\tan^{2} θ = \frac{1 - \cos (2 θ)}{1 + \cos (2 θ)}

Writing an equivalent expression not containing powers greater than 1

Write an equivalent expression for $\cos^{4} x$ that does not involve any powers of sine or cosine greater than 1.

We will apply the reduction formula for cosine twice.

\begin{array}{l} \cos^{4} x = {(\cos^{2} x)}^{2} \\ = {(\frac{1 + \cos (2 x)}{2})}^{2} & {Substitute reduction formula for cos}^{2} x . \\ = \frac{1}{4} (1 + 2 \cos (2 x) + \cos^{2} (2 x)) \\ = \frac{1}{4} + \frac{1}{2} \cos (2 x) + \frac{1}{4} (\frac{1 + \cos 2 (2 x)}{2}) \begin{matrix}  \end{matrix} & {Substitute reduction formula for cos}^{2} x . \\ = \frac{1}{4} + \frac{1}{2} \cos (2 x) + \frac{1}{8} + \frac{1}{8} \cos (4 x) \\ = \frac{3}{8} + \frac{1}{2} \cos (2 x) + \frac{1}{8} \cos (4 x) \end{array}

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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