9.4 Sum-to-product and product-to-sum formulas (Page 2/6)

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\begin{array}{l} \underset{____________________________________________________}{\begin{array}{l} \begin{array}{l} \cos (α - β) = \cos α \cos β + \sin α \sin β \end{array} \\ - \cos (α + β) = - (\cos α \cos β - \sin α \sin β) \end{array}} \\ \cos (α - β) - \cos (α + β) = 2 \sin α \sin β \end{array}

Then, we divide by 2 to isolate the product of sines:

\sin α \sin β = \frac{1}{2} [\cos (α - β) - \cos (α + β)]

Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas.

A General Note

The product-to-sum formulas

The product-to-sum formulas are as follows:

\begin{matrix} \cos α \cos β & = & \frac{1}{2} [\cos (α - β) + \cos (α + β)] \end{matrix}

\begin{matrix} \sin α \cos β & = & \frac{1}{2} [\sin (α + β) + \sin (α - β)] \end{matrix}

\begin{matrix} \sin α \sin β & = & \frac{1}{2} [\cos (α - β) - \cos (α + β)] \end{matrix}

\begin{matrix} \cos α \sin β & = & \frac{1}{2} [\sin (α + β) - \sin (α - β)] \end{matrix}

Express the product as a sum or difference

Write $\cos (3 θ) \cos (5 θ)$ as a sum or difference.

We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.

\begin{matrix} \cos α \cos β & = & \frac{1}{2} [\cos (α - β) + \cos (α + β)] \\ \cos (3 θ) \cos (5 θ) & = & \frac{1}{2} [\cos (3 θ - 5 θ) + \cos (3 θ + 5 θ)] \\ = & \frac{1}{2} [\cos (2 θ) + \cos (8 θ)] & Use even-odd identity . \end{matrix}

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Try It

Use the product-to-sum formula to evaluate $\cos \frac{11 π}{12} \cos \frac{π}{12} .$

$\frac{- 2 - \sqrt{3}}{4}$

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Expressing sums as products

Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine . Let $\frac{u + v}{2} = α$ and $\frac{u - v}{2} = β .$

Then,

\begin{matrix} α + β & = & \frac{u + v}{2} + \frac{u - v}{2} \\ = & \frac{2 u}{2} \\ = & u \\ α - β & = & \frac{u + v}{2} - \frac{u - v}{2} \\ = & \frac{2 v}{2} \\ = & v \end{matrix}

Thus, replacing $α$ and $β$ in the product-to-sum formula with the substitute expressions, we have

\begin{matrix} \sin α \cos β & = & \frac{1}{2} [\sin (α + β) + \sin (α - β)] \\ \sin (\frac{u + v}{2}) \cos (\frac{u - v}{2}) & = & \frac{1}{2} [\sin u + \sin v] & Substitute for (α + β) and (α - β) \\ 2 \sin (\frac{u + v}{2}) \cos (\frac{u - v}{2}) & = & \sin u + \sin v \end{matrix}

The other sum-to-product identities are derived similarly.

A General Note

Sum-to-product formulas

The sum-to-product formulas are as follows:

\begin{matrix} \sin α + \sin β & = & 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \end{matrix}

\begin{matrix} \sin α - \sin β & = & 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2}) \end{matrix}

\begin{matrix} \cos α - \cos β & = & −2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2}) \end{matrix}

\begin{matrix} \cos α + \cos β & = & 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2}) \end{matrix}

Writing the difference of sines as a product

Write the following difference of sines expression as a product: $\sin (4 θ) - \sin (2 θ) .$

We begin by writing the formula for the difference of sines.

\sin α - \sin β = 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2})

Substitute the values into the formula, and simplify.

\begin{matrix} \sin (4 θ) - \sin (2 θ) & = & 2 \sin (\frac{4 θ - 2 θ}{2}) \cos (\frac{4 θ + 2 θ}{2}) \\ = & 2 \sin (\frac{2 θ}{2}) \cos (\frac{6 θ}{2}) \\ = & 2 \sin θ \cos (3 θ) \end{matrix}

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Use the sum-to-product formula to write the sum as a product: $\sin (3 θ) + \sin (θ) .$

$2 \sin (2 θ) \cos (θ)$

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Evaluating using the sum-to-product formula

Evaluate $\cos (15°) - \cos (75°) .$ Check the answer with a graphing calculator.

We begin by writing the formula for the difference of cosines.

\cos α - \cos β = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})

Then we substitute the given angles and simplify.

\begin{matrix} \cos (15°) - \cos (75°) & = & −2 \sin (\frac{15° + 75°}{2}) \sin (\frac{15° - 75°}{2}) \\ = & −2 \sin (45°) \sin (−30°) \\ = & −2 (\frac{\sqrt{2}}{2}) (- \frac{1}{2}) \\ = & \frac{\sqrt{2}}{2} \end{matrix}

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Proving an identity

Prove the identity:

\frac{\cos (4 t) - \cos (2 t)}{\sin (4 t) + \sin (2 t)} = - \tan t

We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.

\begin{matrix} \frac{\cos (4 t) - \cos (2 t)}{\sin (4 t) + \sin (2 t)} & = & \frac{- 2 \sin (\frac{4 t + 2 t}{2}) \sin (\frac{4 t - 2 t}{2})}{2 \sin (\frac{4 t + 2 t}{2}) \cos (\frac{4 t - 2 t}{2})} \\ = & \frac{- 2 \sin (3 t) \sin t}{2 \sin (3 t) \cos t} \\ = & \frac{- 2 \sin (3 t) \sin t}{2 \sin (3 t) \cos t} \\ = & - \frac{\sin t}{\cos t} \\ = & - \tan t \end{matrix}

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Questions & Answers

Why is b in the answer

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(Pcos∅+qsin∅)/(pcos∅-psin∅)

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Source: OpenStax, Algebra and trigonometry. OpenStax CNX. Nov 14, 2016 Download for free at https://legacy.cnx.org/content/col11758/1.6

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