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Sum-to-Product and Product-to-Sum Formulas
Express products as sums.
Expressing Products as Sums
We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity.
We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get:
\[\begin{array}{lllll}\underline{\begin{array}{lll}\cos \ \alpha \ \cos \ \beta +\sin \ \alpha \ \sin \ \beta & = & \cos (\alpha -\beta ) \\ +\ \cos \ \alpha \ \cos \ \beta -\sin \ \alpha \ \sin \ \beta & = & \cos (\alpha +\beta )\end{array}} \\ \begin{array}{lll}\ 2\ \cos \ \alpha \ \cos \ \beta & = & \cos (\alpha -\beta )+\cos (\alpha +\beta )\end{array}\end{array}\]Then, we divide by \(2\) to isolate the product of cosines:
\[\cos \ \alpha \ \cos \ \beta =\frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )]\]Example
Try it.
Write the following product of cosines as a sum: \(2\ \cos (\frac{7x}{2})\ \cos \ \frac{3x}{2}.\)
Solution
We begin by writing the formula for the product of cosines:
\[\cos \ \alpha \ \cos \ \beta =\frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )]\]We can then substitute the given angles into the formula and simplify.
\[\begin{array}{lll}2\ \cos (\frac{7x}{2})\cos (\frac{3x}{2}) & = & (2)(\frac{1}{2})[\cos (\frac{7x}{2}-\frac{3x}{2}))+\cos (\frac{7x}{2}+\frac{3x}{2})] \\ & = & [\cos (\frac{4x}{2})+\cos (\frac{10x}{2})] \\ & = & \cos \ 2x+\cos \ 5x\end{array}\]Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.
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 \(\frac{u+v}{2}=\alpha\) and \(\frac{u-v}{2}=\beta .\)
Then,
\[\begin{array}{lll}\alpha +\beta & = & \frac{u+v}{2}+\frac{u-v}{2} \\ & = & \frac{2u}{2} \\ & = & u \\ \alpha -\beta & = & \frac{u+v}{2}-\frac{u-v}{2} \\ & = & \frac{2v}{2} \\ & = & v\end{array}\]Thus, replacing \(\alpha\) and \(\beta\) in the product-to-sum formula with the substitute expressions, we have
\[\begin{array}{llll}\sin \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )] & \\ \sin (\frac{u+v}{2})\cos (\frac{u-v}{2}) & = & \frac{1}{2}[\sin \ u+\sin \ v] & \text{Substitute for}(\alpha +\beta )\text{ and }(\alpha -\beta ) \\ 2\ \sin (\frac{u+v}{2})\cos (\frac{u-v}{2}) & = & \sin \ u+\sin \ v & \end{array}\]The other sum-to-product identities are derived similarly.
Example
Try it.
Write the following difference of sines expression as a product: \(\sin (4\theta )-\sin (2\theta ).\)
Solution
We begin by writing the formula for the difference of sines.
\[\sin \ \alpha -\sin \ \beta =2\sin (\frac{\alpha -\beta }{2})\cos (\frac{\alpha +\beta }{2})\]Substitute the values into the formula, and simplify.
\[\begin{array}{lll}\sin (4\theta )-\sin (2\theta ) & = & 2\sin (\frac{4\theta -2\theta }{2})\ \cos (\frac{4\theta +2\theta }{2}) \\ & = & 2\sin (\frac{2\theta }{2})\ \cos (\frac{6\theta }{2}) \\ & = & 2\ \sin \ \theta \ \cos (3\theta )\end{array}\]Condensed: the full section is in OpenStax Algebra and Trigonometry 2e.
Key Equations
| Product-to-sum Formulas | \(\begin{array}{lll}\cos \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )] \\ \sin \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )] \\ \sin \ \alpha \ \sin \ \beta & = & \frac{1}{2}[\cos (\alpha -\beta )-\cos (\alpha +\beta )] \\ \cos \ \alpha \ \sin \ \beta & = & \frac{1}{2}[\sin (\alpha +\beta )-\sin (\alpha -\beta )]\end{array}\) |
| Sum-to-product Formulas | \(\begin{array}{lll}\sin \ \alpha +\sin \ \beta & = & 2\ \sin (\frac{\alpha +\beta }{2})\cos (\frac{\alpha -\beta }{2}) \\ \sin \ \alpha -\sin \ \beta & = & 2\ \sin (\frac{\alpha -\beta }{2})\cos (\frac{\alpha +\beta }{2}) \\ \cos \ \alpha -\cos \ \beta & = & -2\ \sin (\frac{\alpha +\beta }{2})\sin (\frac{\alpha -\beta }{2}) \\ \cos \ \alpha +\cos \ \beta & = & 2\ \cos (\frac{\alpha +\beta }{2})\cos (\frac{\alpha -\beta }{2})\end{array}\) |
Key Concepts
- From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine.
- We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines.
- We can also derive the sum-to-product identities from the product-to-sum identities using substitution.
- We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines.
- Trigonometric expressions are often simpler to evaluate using the formulas.
- The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side.
Expressing Products as Sums
We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity.
We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get:
\[\begin{array}{l}\underline{\begin{array}{l}\cos \ \alpha \ \cos \ \beta +\sin \ \alpha \ \sin \ \beta =\cos (\alpha -\beta ) \\ +\cos \ \alpha \ \cos \ \beta -\sin \ \alpha \ \sin \ \beta =\cos (\alpha +\beta )\end{array}} \\ 2\ \cos \ \alpha \ \cos \ \beta =\cos (\alpha -\beta )+\cos (\alpha +\beta )\end{array}\]Then, we divide by \(2\) to isolate the product of cosines:
\[\cos \ \alpha \ \cos \ \beta =\frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )]\]Example
Try it.
Write the following product of cosines as a sum: \(2\ \cos (\frac{7x}{2})\ \cos \ \frac{3x}{2}.\)
Solution
We begin by writing the formula for the product of cosines:
\[\cos \ \alpha \ \cos \ \beta =\frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )]\]We can then substitute the given angles into the formula and simplify.
\[\begin{array}{l}2\ \cos (\frac{7x}{2})\cos (\frac{3x}{2})=(2)(\frac{1}{2})[\cos (\frac{7x}{2}-\frac{3x}{2})+\cos (\frac{7x}{2}+\frac{3x}{2})] \\ =[\cos (\frac{4x}{2})+\cos (\frac{10x}{2})] \\ =\cos \ 2x+\cos \ 5x\end{array}\]Condensed: the full section is in OpenStax Precalculus 2e.
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 \(\frac{u+v}{2}=\alpha\) and \(\frac{u-v}{2}=\beta .\)
Then,
\[\begin{array}{l}\alpha +\beta =\frac{u+v}{2}+\frac{u-v}{2} \\ =\frac{2u}{2} \\ =u \\ \\ \alpha -\beta =\frac{u+v}{2}-\frac{u-v}{2} \\ =\frac{2v}{2} \\ =v\end{array}\]Thus, replacing \(\alpha\) and \(\beta\) in the product-to-sum formula with the substitute expressions, we have
\[\begin{array}{lll}\ \sin \ \alpha \ \cos \ \beta =\frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )] & & \\ \sin (\frac{u+v}{2})\cos (\frac{u-v}{2})=\frac{1}{2}[\sin \ u+\sin \ v] & & \text{Substitute for}(\alpha +\beta )\text{ and }(\alpha -\beta ) \\ 2\ \sin (\frac{u+v}{2})\cos (\frac{u-v}{2})=\sin \ u+\sin \ v & & \end{array}\]The other sum-to-product identities are derived similarly.
Example
Try it.
Write the following difference of sines expression as a product: \(\sin (4\theta )-\sin (2\theta ).\)
Solution
We begin by writing the formula for the difference of sines.
\[\sin \ \alpha -\sin \ \beta =2\sin (\frac{\alpha -\beta }{2})\cos (\frac{\alpha +\beta }{2})\]Substitute the values into the formula, and simplify.
\[\begin{array}{l}\sin (4\theta )-\sin (2\theta )=2\sin (\frac{4\theta -2\theta }{2})\ \cos (\frac{4\theta +2\theta }{2}) \\ =2\sin (\frac{2\theta }{2})\ \cos (\frac{6\theta }{2}) \\ =2\ \sin \ \theta \ \cos (3\theta )\end{array}\]Condensed: the full section is in OpenStax Precalculus 2e.
Key Equations
| Product-to-sum Formulas | \(\begin{array}{l} \\ \cos \ \alpha \ \cos \ \beta =\frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )] \\ \sin \ \alpha \ \cos \ \beta =\frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )] \\ \sin \ \alpha \ \sin \ \beta =\frac{1}{2}[\cos (\alpha -\beta )-\cos (\alpha +\beta )] \\ \cos \ \alpha \ \sin \ \beta =\frac{1}{2}[\sin (\alpha +\beta )-\sin (\alpha -\beta )]\end{array}\) |
| Sum-to-product Formulas | \(\begin{array}{l} \\ \sin \ \alpha +\sin \ \beta =2\ \sin (\frac{\alpha +\beta }{2})\cos (\frac{\alpha -\beta }{2}) \\ \sin \ \alpha -\sin \ \beta =2\ \sin (\frac{\alpha -\beta }{2})\cos (\frac{\alpha +\beta }{2}) \\ \cos \ \alpha -\cos \ \beta =-2\ \sin (\frac{\alpha +\beta }{2})\sin (\frac{\alpha -\beta }{2}) \\ \cos \ \alpha +\cos \ \beta =2\ \cos (\frac{\alpha +\beta }{2})\cos (\frac{\alpha -\beta }{2})\end{array}\) |
Key Concepts
- From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine.
- We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines.
- We can also derive the sum-to-product identities from the product-to-sum identities using substitution.
- We can use the sum-to-product formulas to rewrite sum or difference of sines, cosines, or products sine and cosine as products of sines and cosines.
- Trigonometric expressions are often simpler to evaluate using the formulas.
- The identities can be verified using other formulas or by converting the expressions to sines and cosines. To verify an identity, we choose the more complicated side of the equals sign and rewrite it until it is transformed into the other side.
이제 너 계산기는 이것을 해결하지 않지만, 그 조각은 계산 가능합니다. 아래의 하나를 시도하거나 자신의 것을 입력하십시오.
연습 (40)
먼저 종이에 각각을 시도해 보세요. 확인하기 위해 답을 드러내세요. 확인된 답은 각 단계의 솔버에서 열 수 있습니다.
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Write the following product of cosines as a sum: \(2\ \cos (\frac{7x}{2})\ \cos \ \frac{3x}{2}.\)
답을 드러내세요
We begin by writing the formula for the product of cosines:
\[\cos \ \alpha \ \cos \ \beta =\frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )]\]We can then substitute the given angles into the formula and simplify.
\[\begin{array}{lll}2\ \cos (\frac{7x}{2})\cos (\frac{3x}{2}) & = & (2)(\frac{1}{2})[\cos (\frac{7x}{2}-\frac{3x}{2}))+\cos (\frac{7x}{2}+\frac{3x}{2})] \\ & = & [\cos (\frac{4x}{2})+\cos (\frac{10x}{2})] \\ & = & \cos \ 2x+\cos \ 5x\end{array}\] -
Use the product-to-sum formula to write the product as a sum or difference: \(\cos (2\theta )\cos (4\theta ).\)
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\(\frac{1}{2}(\cos 6\theta +\cos 2\theta )\)
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Express the following product as a sum containing only sine or cosine and no products: \(\sin (4\theta )\cos (2\theta ).\)
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Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.
\[\begin{array}{lll}\sin \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )] \\ \sin (4\theta )\cos (2\theta ) & = & \frac{1}{2}[\sin (4\theta +2\theta )+\sin (4\theta -2\theta )] \\ & = & \frac{1}{2}[\sin (6\theta )+\sin (2\theta )]\end{array}\] -
Use the product-to-sum formula to write the product as a sum: \(\sin (x+y)\cos (x-y).\)
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\(\frac{1}{2}(\sin 2x+\sin 2y)\)
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Write \(\cos (3\theta )\ \cos (5\theta )\) as a sum or difference.
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We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.
\[\begin{array}{llll}\cos \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )] & \\ \cos (3\theta )\cos (5\theta ) & = & \frac{1}{2}[\cos (3\theta -5\theta )+\cos (3\theta +5\theta )] & \\ & = & \frac{1}{2}[\cos (2\theta )+\cos (8\theta )] & \text{Use even-odd identity}.\end{array}\] -
Use the product-to-sum formula to evaluate \(\cos \ \frac{11\pi }{12}\ \cos \ \frac{\pi }{12}.\)
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\(\frac{-2-\sqrt{3}}{4}\)
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Write the following difference of sines expression as a product: \(\sin (4\theta )-\sin (2\theta ).\)
답을 드러내세요
We begin by writing the formula for the difference of sines.
\[\sin \ \alpha -\sin \ \beta =2\sin (\frac{\alpha -\beta }{2})\cos (\frac{\alpha +\beta }{2})\]Substitute the values into the formula, and simplify.
\[\begin{array}{lll}\sin (4\theta )-\sin (2\theta ) & = & 2\sin (\frac{4\theta -2\theta }{2})\ \cos (\frac{4\theta +2\theta }{2}) \\ & = & 2\sin (\frac{2\theta }{2})\ \cos (\frac{6\theta }{2}) \\ & = & 2\ \sin \ \theta \ \cos (3\theta )\end{array}\] -
Use the sum-to-product formula to write the sum as a product: \(\sin (3\theta )+\sin (\theta ).\)
답을 드러내세요
\(2\sin (2\theta )\cos (\theta )\)
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Evaluate \(\cos (15^{\circ})-\cos (75^{\circ}).\) Check the answer with a graphing calculator.
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We begin by writing the formula for the difference of cosines.
\[\cos \ \alpha -\cos \ \beta =-2\ \sin (\frac{\alpha +\beta }{2})\ \sin (\frac{\alpha -\beta }{2})\]Then we substitute the given angles and simplify.
\[\begin{array}{lll}\cos (15^{\circ})-\cos (75^{\circ}) & = & -2\sin (\frac{15^{\circ}+75^{\circ}}{2})\ \sin (\frac{15^{\circ}-75^{\circ}}{2}) \\ & = & -2\sin (45^{\circ})\ \sin (-30^{\circ}) \\ & = & -2(\frac{\sqrt{2}}{2})(-\frac{1}{2}) \\ & = & \frac{\sqrt{2}}{2}\end{array}\] -
Prove the identity:
\[\frac{\cos (4t)-\cos (2t)}{\sin (4t)+\sin (2t)}=-\tan \ t\]답을 드러내세요
We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.
\[\begin{array}{lll}\frac{\cos (4t)-\cos (2t)}{\sin (4t)+\sin (2t)} & = & \frac{-2\ \sin (\frac{4t+2t}{2})\ \sin (\frac{4t-2t}{2})}{2\ \sin (\frac{4t+2t}{2})\ \cos (\frac{4t-2t}{2})} \\ & = & \frac{-2\ \sin (3t)\sin \ t}{2\ \sin (3t)\cos \ t} \\ & = & \frac{-2\sin (3t)\sin \ t}{2\sin (3t)\cos \ t} \\ & = & -\frac{\sin \ t}{\cos \ t} \\ & = & -\tan \ t\end{array}\] -
Verify the identity \({\text{csc}}^{2}\theta -2=\frac{\cos (2\theta )}{{\sin }^{2}\theta }.\)
답을 드러내세요
For verifying this equation, we are bringing together several of the identities. We will use the double-angle formula and the reciprocal identities. We will work with the right side of the equation and rewrite it until it matches the left side.
\[\begin{array}{lll}\frac{\cos (2\theta )}{{\sin }^{2}\theta } & = & \frac{1-2\ {\sin }^{2}\theta }{{\sin }^{2}\theta } \\ & = & \frac{1}{{\sin }^{2}\theta }-\frac{2\ {\sin }^{2}\theta }{{\sin }^{2}\theta } \\ & = & {\text{csc}}^{2}\theta -2\end{array}\] -
Verify the identity \(\tan \ \theta \ \text{cot}\ \theta -{\cos }^{2}\theta ={\sin }^{2}\theta .\)
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\[\begin{array}{lll}\tan \ \theta \ \text{cot}\ \theta -{\cos }^{2}\theta & = & (\frac{\sin \ \theta }{\cos \ \theta })(\frac{\cos \ \theta }{\sin \ \theta })-{\cos }^{2}\theta \\ & = & 1-{\cos }^{2}\theta \\ & = & {\sin }^{2}\theta \end{array}\]
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Starting with the product to sum formula \(\sin \ \alpha \ \cos \ \beta =\frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )],\) explain how to determine the formula for \(\cos \ \alpha \ \sin \ \beta .\)
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Substitute \(\ \alpha\) into cosine and \(\ \beta\) into sine and evaluate.
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Provide two different methods of calculating \(\cos (195^{\circ})\cos (105^{\circ}),\) one of which uses the product to sum. Which method is easier?
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Describe a situation where we would convert an equation from a sum to a product and give an example.
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Answers will vary. There are some equations that involve a sum of two trig expressions where when converted to a product are easier to solve. For example: \(\frac{\sin (3x)+\sin \ x}{\cos \ x}=1.\) When converting the numerator to a product the equation becomes: \(\frac{2\ \sin (2x)\cos \ x}{\cos \ x}=1\)
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Describe a situation where we would convert an equation from a product to a sum, and give an example.
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\(16\ \sin (16x)\sin (11x)\)
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\(8(\cos (5x)-\cos (27x))\)
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\(20\ \cos (36t)\cos (6t)\)
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\(2\ \sin (5x)\cos (3x)\)
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\(\sin (2x)+\sin (8x)\)
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\(10\ \cos (5x)\sin (10x)\)
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\(\sin (-x)\sin (5x)\)
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\(\frac{1}{2}(\cos (6x)-\cos (4x))\)
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\(\sin (3x)\cos (5x)\)
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\(\cos (6t)+\cos (4t)\)
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\(2\ \cos (5t)\cos \ t\)
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\(\sin (3x)+\sin (7x)\)
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\(\cos (7x)+\cos (-7x)\)
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\(2\ \cos (7x)\)
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\(\sin (3x)-\sin (-3x)\)
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\(\cos (3x)+\cos (9x)\)
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\(2\ \cos (6x)\cos (3x)\)
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\(\sin \ h-\sin (3h)\)
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\(\cos (45^{\circ})\cos (15^{\circ})\)
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\(\frac{1}{4}(1+\sqrt{3})\)
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\(\cos (45^{\circ})\sin (15^{\circ})\)
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\(\sin (-345^{\circ})\sin (-15^{\circ})\)
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\(\frac{1}{4}(\sqrt{3}-2)\)
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\(\sin (195^{\circ})\cos (15^{\circ})\)
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\(\sin (-45^{\circ})\sin (-15^{\circ})\)
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\(\frac{1}{4}(\sqrt{3}-1)\)
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\(\cos (23^{\circ})\sin (17^{\circ})\)
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\(2\ \sin (100^{\circ})\sin (20^{\circ})\)
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\(\cos (80^{\circ})-\cos (120^{\circ})\)
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\(2\ \sin (-100^{\circ})\sin (-20^{\circ})\)
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\(\sin (213^{\circ})\cos (8^{\circ})\)
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\(\frac{1}{2}(\sin (221^{\circ})+\sin (205^{\circ}))\)
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\(2\ \cos (56^{\circ})\cos (47^{\circ})\)
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\(\sin (76^{\circ})+\sin (14^{\circ})\)
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\(\sqrt{2}\ \cos (31^{\circ})\)
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\(\cos (58^{\circ})-\cos (12^{\circ})\)
여기서 사용된 기호
기호를 탭하면 전체 정의, 이미지 및 각 문자의 의미를 확인할 수 있습니다.
어떻게: Sum-to-Product and Product-to-Sum Formulas
- Express products as sums.
- Express sums as products.
- Write the formula for the product of cosines.
- Substitute the given angles into the formula.
- Simplify.
- From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine.
- We can use the product-to-sum formulas to rewrite products of sines, products of cosines, and products of sine and cosine as sums or differences of sines and cosines. See
- We can also derive the sum-to-product identities from the product-to-sum identities using substitution.
사람들이 묻는 질문
Why radians instead of degrees?
A radian is the angle whose arc equals the radius, so it is a pure ratio rather than an arbitrary 1/360 of a turn. With radians the derivative of sin x is exactly cos x; with degrees a factor of π/180 appears everywhere.
Why does sin x = 1/2 have infinitely many solutions?
Sine repeats every full turn, and within one turn it reaches 1/2 twice (at π/6 and 5π/6). Add any whole number of turns to either and it is still true.
How do I remember the exact values?
Two triangles: the 45-45-90 with sides 1, 1, √2 and the 30-60-90 with sides 1, √3, 2. Every value for 30°, 45° and 60° is a ratio of those sides; symmetry gives the rest of the circle.
이 페이지의 일부는 다음에서 변경되었습니다. OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). 여기서 압축하고 다시 설명; 오류는 우리의. 소스가 비상업적 사용을 위해 라이선스되어 있기 때문에, 이 레슨은 무료이며, 아무것도 판매하지 않으며, 동일한 라이선스로 공유됩니다.
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