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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.
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实践 (40)
先在纸上试一个。 将答案显示为检查; 校验的答案可以在求解器中打开, 每一步都可用 。
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Write the following product of cosines as a sum: \(2\ \cos (\frac{7x}{2})\ \cos \ \frac{3x}{2}.\)
发送答案
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: \(\cos (2\theta )\cos (4\theta ).\)
发送答案
\(\frac{1}{2}(\cos 6\theta +\cos 2\theta )\)
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Express the following product as a sum containing only sine or cosine and no products: \(\sin (4\theta )\cos (2\theta ).\)
发送答案
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: \(\sin (x+y)\cos (x-y).\)
发送答案
\(\frac{1}{2}(\sin 2x+\sin 2y)\)
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Write \(\cos (3\theta )\ \cos (5\theta )\) as a sum or difference.
发送答案
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 \(\cos \ \frac{11\pi }{12}\ \cos \ \frac{\pi }{12}.\)
发送答案
\(\frac{-2-\sqrt{3}}{4}\)
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Write the following difference of sines expression as a product: \(\sin (4\theta )-\sin (2\theta ).\)
发送答案
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: \(\sin (3\theta )+\sin (\theta ).\)
发送答案
\(2\sin (2\theta )\cos (\theta )\)
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Evaluate \(\cos (15^{\circ})-\cos (75^{\circ}).\) Check the answer with a graphing calculator.
发送答案
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:
\[\frac{\cos (4t)-\cos (2t)}{\sin (4t)+\sin (2t)}=-\tan \ t\]发送答案
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 \({\text{csc}}^{2}\theta -2=\frac{\cos (2\theta )}{{\sin }^{2}\theta }.\)
发送答案
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 \(\tan \ \theta \ \text{cot}\ \theta -{\cos }^{2}\theta ={\sin }^{2}\theta .\)
发送答案
\[\begin{array}{lll}\tan \ \theta \ \text{cot}\ \theta -{\cos }^{2}\theta & = & (\frac{\sin \ \theta }{\cos \ \theta })(\frac{\cos \ \theta }{\sin \ \theta })-{\cos }^{2}\theta \\ & = & 1-{\cos }^{2}\theta \\ & = & {\sin }^{2}\theta \end{array}\]
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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 \(\cos \ \alpha \ \sin \ \beta .\)
发送答案
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.
发送答案
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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\(16\ \sin (16x)\sin (11x)\)
发送答案
\(8(\cos (5x)-\cos (27x))\)
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\(20\ \cos (36t)\cos (6t)\)
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\(2\ \sin (5x)\cos (3x)\)
发送答案
\(\sin (2x)+\sin (8x)\)
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\(10\ \cos (5x)\sin (10x)\)
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\(\sin (-x)\sin (5x)\)
发送答案
\(\frac{1}{2}(\cos (6x)-\cos (4x))\)
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\(\sin (3x)\cos (5x)\)
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\(\cos (6t)+\cos (4t)\)
发送答案
\(2\ \cos (5t)\cos \ t\)
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\(\sin (3x)+\sin (7x)\)
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\(\cos (7x)+\cos (-7x)\)
发送答案
\(2\ \cos (7x)\)
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\(\sin (3x)-\sin (-3x)\)
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\(\cos (3x)+\cos (9x)\)
发送答案
\(2\ \cos (6x)\cos (3x)\)
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\(\sin \ h-\sin (3h)\)
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\(\cos (45^{\circ})\cos (15^{\circ})\)
发送答案
\(\frac{1}{4}(1+\sqrt{3})\)
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\(\cos (45^{\circ})\sin (15^{\circ})\)
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\(\sin (-345^{\circ})\sin (-15^{\circ})\)
发送答案
\(\frac{1}{4}(\sqrt{3}-2)\)
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\(\sin (195^{\circ})\cos (15^{\circ})\)
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\(\sin (-45^{\circ})\sin (-15^{\circ})\)
发送答案
\(\frac{1}{4}(\sqrt{3}-1)\)
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\(\cos (23^{\circ})\sin (17^{\circ})\)
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\(2\ \sin (100^{\circ})\sin (20^{\circ})\)
发送答案
\(\cos (80^{\circ})-\cos (120^{\circ})\)
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\(2\ \sin (-100^{\circ})\sin (-20^{\circ})\)
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\(\sin (213^{\circ})\cos (8^{\circ})\)
发送答案
\(\frac{1}{2}(\sin (221^{\circ})+\sin (205^{\circ}))\)
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\(2\ \cos (56^{\circ})\cos (47^{\circ})\)
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\(\sin (76^{\circ})+\sin (14^{\circ})\)
发送答案
\(\sqrt{2}\ \cos (31^{\circ})\)
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\(\cos (58^{\circ})-\cos (12^{\circ})\)
此处所用符号
输入完整定义、图片和其中每个字母的含义的符号。
如何: 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.
人们对人们提出的问题
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.
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