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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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Exercices (40)

Essayez chacun d'abord sur papier. Révélez la réponse pour vérifier; les vérifiés peuvent être ouverts dans le solveur pour chaque étape.

  1. Write the following product of cosines as a sum: \(2\ \cos (\frac{7x}{2})\ \cos \ \frac{3x}{2}.\)

    Révèle la réponse

    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.

    \[\begin{array}{lll}2\ \cos (\frac{7x}{2})\cos (\frac{3x}{2}) & = & (2)(\frac{1}{2})[\cos (\frac{7x}{2}-\frac{3x}{2}))+\cos (\frac{7x}{2}+\frac{3x}{2})] \\ & = & [\cos (\frac{4x}{2})+\cos (\frac{10x}{2})] \\ & = & \cos \ 2x+\cos \ 5x\end{array}\]

  2. Use the product-to-sum formula to write the product as a sum or difference: \(\cos (2\theta )\cos (4\theta ).\)

    Révèle la réponse

    \(\frac{1}{2}(\cos 6\theta +\cos 2\theta )\)

  3. Express the following product as a sum containing only sine or cosine and no products: \(\sin (4\theta )\cos (2\theta ).\)

    Révèle la réponse

    Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.

    \[\begin{array}{lll}\sin \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha -\beta )] \\ \sin (4\theta )\cos (2\theta ) & = & \frac{1}{2}[\sin (4\theta +2\theta )+\sin (4\theta -2\theta )] \\ & = & \frac{1}{2}[\sin (6\theta )+\sin (2\theta )]\end{array}\]

  4. Use the product-to-sum formula to write the product as a sum: \(\sin (x+y)\cos (x-y).\)

    Révèle la réponse

    \(\frac{1}{2}(\sin 2x+\sin 2y)\)

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

    Révèle la réponse

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

    \[\begin{array}{llll}\cos \ \alpha \ \cos \ \beta & = & \frac{1}{2}[\cos (\alpha -\beta )+\cos (\alpha +\beta )] & \\ \cos (3\theta )\cos (5\theta ) & = & \frac{1}{2}[\cos (3\theta -5\theta )+\cos (3\theta +5\theta )] & \\ & = & \frac{1}{2}[\cos (2\theta )+\cos (8\theta )] & \text{Use even-odd identity}.\end{array}\]

  6. Use the product-to-sum formula to evaluate \(\cos \ \frac{11\pi }{12}\ \cos \ \frac{\pi }{12}.\)

    Révèle la réponse

    \(\frac{-2-\sqrt{3}}{4}\)

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

    Révèle la réponse

    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.

    \[\begin{array}{lll}\sin (4\theta )-\sin (2\theta ) & = & 2\sin (\frac{4\theta -2\theta }{2})\ \cos (\frac{4\theta +2\theta }{2}) \\ & = & 2\sin (\frac{2\theta }{2})\ \cos (\frac{6\theta }{2}) \\ & = & 2\ \sin \ \theta \ \cos (3\theta )\end{array}\]

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

    Révèle la réponse

    \(2\sin (2\theta )\cos (\theta )\)

  9. Evaluate \(\cos (15^{\circ})-\cos (75^{\circ}).\) Check the answer with a graphing calculator.

    Révèle la réponse

    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.

    \[\begin{array}{lll}\cos (15^{\circ})-\cos (75^{\circ}) & = & -2\sin (\frac{15^{\circ}+75^{\circ}}{2})\ \sin (\frac{15^{\circ}-75^{\circ}}{2}) \\ & = & -2\sin (45^{\circ})\ \sin (-30^{\circ}) \\ & = & -2(\frac{\sqrt{2}}{2})(-\frac{1}{2}) \\ & = & \frac{\sqrt{2}}{2}\end{array}\]

  10. Prove the identity:

    \[\frac{\cos (4t)-\cos (2t)}{\sin (4t)+\sin (2t)}=-\tan \ t\]
    Révèle la réponse

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

    \[\begin{array}{lll}\frac{\cos (4t)-\cos (2t)}{\sin (4t)+\sin (2t)} & = & \frac{-2\ \sin (\frac{4t+2t}{2})\ \sin (\frac{4t-2t}{2})}{2\ \sin (\frac{4t+2t}{2})\ \cos (\frac{4t-2t}{2})} \\ & = & \frac{-2\ \sin (3t)\sin \ t}{2\ \sin (3t)\cos \ t} \\ & = & \frac{-2\sin (3t)\sin \ t}{2\sin (3t)\cos \ t} \\ & = & -\frac{\sin \ t}{\cos \ t} \\ & = & -\tan \ t\end{array}\]

  11. Verify the identity \({\text{csc}}^{2}\theta -2=\frac{\cos (2\theta )}{{\sin }^{2}\theta }.\)

    Révèle la réponse

    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.

    \[\begin{array}{lll}\frac{\cos (2\theta )}{{\sin }^{2}\theta } & = & \frac{1-2\ {\sin }^{2}\theta }{{\sin }^{2}\theta } \\ & = & \frac{1}{{\sin }^{2}\theta }-\frac{2\ {\sin }^{2}\theta }{{\sin }^{2}\theta } \\ & = & {\text{csc}}^{2}\theta -2\end{array}\]

  12. Verify the identity \(\tan \ \theta \ \text{cot}\ \theta -{\cos }^{2}\theta ={\sin }^{2}\theta .\)

    Révèle la réponse

    \[\begin{array}{lll}\tan \ \theta \ \text{cot}\ \theta -{\cos }^{2}\theta & = & (\frac{\sin \ \theta }{\cos \ \theta })(\frac{\cos \ \theta }{\sin \ \theta })-{\cos }^{2}\theta \\ & = & 1-{\cos }^{2}\theta \\ & = & {\sin }^{2}\theta \end{array}\]

  13. 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 \(\cos \ \alpha \ \sin \ \beta .\)

    Révèle la réponse

    Substitute \(\ \alpha\) into cosine and \(\ \beta\) into sine and evaluate.

  14. Provide two different methods of calculating \(\cos (195^{\circ})\cos (105^{\circ}),\) one of which uses the product to sum. Which method is easier?

  15. Describe a situation where we would convert an equation from a sum to a product and give an example.

    Révèle la réponse

    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: \(\frac{2\ \sin (2x)\cos \ x}{\cos \ x}=1\)

  16. Describe a situation where we would convert an equation from a product to a sum, and give an example.

  17. \(16\ \sin (16x)\sin (11x)\)

    Révèle la réponse

    \(8(\cos (5x)-\cos (27x))\)

  18. \(20\ \cos (36t)\cos (6t)\)

  19. \(2\ \sin (5x)\cos (3x)\)

    Révèle la réponse

    \(\sin (2x)+\sin (8x)\)

  20. \(10\ \cos (5x)\sin (10x)\)

  21. \(\sin (-x)\sin (5x)\)

    Révèle la réponse

    \(\frac{1}{2}(\cos (6x)-\cos (4x))\)

  22. \(\sin (3x)\cos (5x)\)

  23. \(\cos (6t)+\cos (4t)\)

    Révèle la réponse

    \(2\ \cos (5t)\cos \ t\)

  24. \(\sin (3x)+\sin (7x)\)

  25. \(\cos (7x)+\cos (-7x)\)

    Révèle la réponse

    \(2\ \cos (7x)\)

  26. \(\sin (3x)-\sin (-3x)\)

  27. \(\cos (3x)+\cos (9x)\)

    Révèle la réponse

    \(2\ \cos (6x)\cos (3x)\)

  28. \(\sin \ h-\sin (3h)\)

  29. \(\cos (45^{\circ})\cos (15^{\circ})\)

    Révèle la réponse

    \(\frac{1}{4}(1+\sqrt{3})\)

  30. \(\cos (45^{\circ})\sin (15^{\circ})\)

  31. \(\sin (-345^{\circ})\sin (-15^{\circ})\)

    Révèle la réponse

    \(\frac{1}{4}(\sqrt{3}-2)\)

  32. \(\sin (195^{\circ})\cos (15^{\circ})\)

  33. \(\sin (-45^{\circ})\sin (-15^{\circ})\)

    Révèle la réponse

    \(\frac{1}{4}(\sqrt{3}-1)\)

  34. \(\cos (23^{\circ})\sin (17^{\circ})\)

  35. \(2\ \sin (100^{\circ})\sin (20^{\circ})\)

    Révèle la réponse

    \(\cos (80^{\circ})-\cos (120^{\circ})\)

  36. \(2\ \sin (-100^{\circ})\sin (-20^{\circ})\)

  37. \(\sin (213^{\circ})\cos (8^{\circ})\)

    Révèle la réponse

    \(\frac{1}{2}(\sin (221^{\circ})+\sin (205^{\circ}))\)

  38. \(2\ \cos (56^{\circ})\cos (47^{\circ})\)

  39. \(\sin (76^{\circ})+\sin (14^{\circ})\)

    Révèle la réponse

    \(\sqrt{2}\ \cos (31^{\circ})\)

  40. \(\cos (58^{\circ})-\cos (12^{\circ})\)

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Comment faire: Sum-to-Product and Product-to-Sum Formulas

  1. Express products as sums.
  2. Express sums as products.
  3. Write the formula for the product of cosines.
  4. Substitute the given angles into the formula.
  5. Simplify.
  6. From the sum and difference identities, we can derive the product-to-sum formulas and the sum-to-product formulas for sine and cosine.
  7. 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
  8. We can also derive the sum-to-product identities from the product-to-sum identities using substitution.

Questions posées par les gens

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.

Certaines parties de cette page sont adaptées de OpenStax Algebra and Trigonometry 2e (CC BY-NC-SA 4.0), OpenStax Precalculus 2e (CC BY-NC-SA 4.0). Condensés et réexpliqués ici ; les erreurs sont les nôtres. Comme sa source est autorisée à des fins non commerciales, cette leçon est gratuite, ne vend rien et est partagée avec la même licence.

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