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מדריכי לימוד > Precalculus II

Sum-to-Product and Product-to-Sum Formulas

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.

Expressing Products as Sums for Cosine

We can derive the product-to-sum formula from the sum and difference identities for cosine. If we add the two equations, we get:
cos⁡αcos⁡β+sin⁡αsin⁡β=cos⁡(α−β)+cos⁡αcos⁡β−sin⁡αsin⁡β=cos⁡(α+β)‾\begin{array}{l}{\begin{array}{c}\cos \alpha \cos \beta +\sin \alpha \sin \beta =\cos \left(\alpha -\beta \right)\\\underline{ +\cos \alpha \cos \beta -\sin \alpha \sin \beta =\cos \left(\alpha +\beta \right)}\end{array}}\end{array}\\
2cos⁡αcos⁡β=cos⁡(α−β)+cos⁡(α+β)\begin{array}{l} 2\cos \alpha \cos \beta =\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\hfill \end{array}\\
Then, we divide by 22 to isolate the product of cosines:
cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]\\

How To: Given a product of cosines, express as a sum.

  1. Write the formula for the product of cosines.
  2. Substitute the given angles into the formula.
  3. Simplify.

Example 1: Writing the Product as a Sum Using the Product-to-Sum Formula for Cosine

Write the following product of cosines as a sum: 2cos⁡(7x2)cos⁡3x22\cos \left(\frac{7x}{2}\right)\cos \frac{3x}{2}\\.

Solution

We begin by writing the formula for the product of cosines:
cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]\\
We can then substitute the given angles into the formula and simplify.
2cos⁡(7x2)cos⁡(3x2)=(2)(12)[cos⁡(7x2−3x2)+cos⁡(7x2+3x2)] =[cos⁡(4x2)+cos⁡(10x2)] =cos⁡2x+cos⁡5x\begin{array}{l}2\cos \left(\frac{7x}{2}\right)\cos \left(\frac{3x}{2}\right)=\left(2\right)\left(\frac{1}{2}\right)\left[\cos \left(\frac{7x}{2}-\frac{3x}{2}\right)+\cos \left(\frac{7x}{2}+\frac{3x}{2}\right)\right]\hfill \\ \text{ }=\left[\cos \left(\frac{4x}{2}\right)+\cos \left(\frac{10x}{2}\right)\right]\hfill \\ \text{ }=\cos 2x+\cos 5x\hfill \end{array}\\

Try It 1

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

Expressing the Product of Sine and Cosine as a Sum

Next, we will derive the product-to-sum formula for sine and cosine from the sum and difference formulas for sine. If we add the sum and difference identities, we get:
 sin⁡(α+β)=sin⁡αcos⁡β+cos⁡αsin⁡β+ sin⁡(α−β)=sin⁡αcos⁡β−cos⁡αsin⁡β‾sin⁡(α+β)+sin⁡(α−β)=2sin⁡αcos⁡β\begin{array}{l}{\begin{array}{l}\begin{array}{l} \hfill \\ \text{ }\sin \left(\alpha +\beta \right)=\sin \alpha \cos \beta +\cos \alpha \sin \beta \hfill \end{array}\hfill \\\underline{ +\text{ }\sin \left(\alpha -\beta \right)=\sin \alpha \cos \beta -\cos \alpha \sin \beta} \hfill \end{array}}\\ \sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)=2\sin \alpha \cos \beta \end{array}\\
Then, we divide by 2 to isolate the product of cosine and sine:
sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]\\

Example 2: Writing the Product as a Sum Containing only Sine or Cosine

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

Solution

Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.
sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]sin⁡(4θ)cos⁡(2θ)=12[sin⁡(4θ+2θ)+sin⁡(4θ−2θ)]=12[sin⁡(6θ)+sin⁡(2θ)]\begin{array}{l}\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]\hfill \\ \sin \left(4\theta \right)\cos \left(2\theta \right)=\frac{1}{2}\left[\sin \left(4\theta +2\theta \right)+\sin \left(4\theta -2\theta \right)\right]\hfill \\ =\frac{1}{2}\left[\sin \left(6\theta \right)+\sin \left(2\theta \right)\right]\hfill \end{array}\\

Try It 2

Use the product-to-sum formula to write the product as a sum: sin⁡(x+y)cos⁡(x−y)\sin \left(x+y\right)\cos \left(x-y\right)\\. Solution

Expressing Products of Sines in Terms of Cosine

Expressing the product of sines in terms of cosine is also derived from the sum and difference identities for cosine. In this case, we will first subtract the two cosine formulas:
 cos⁡(α−β)=cos⁡αcos⁡β+sin⁡αsin⁡β− cos⁡(α+β)=−(cos⁡αcos⁡β−sin⁡αsin⁡β)‾cos⁡(α−β)−cos⁡(α+β)=2sin⁡αsin⁡β\begin{array}{l}{\begin{array}{l}\begin{array}{l}\hfill \\ \text{ }\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta +\sin \alpha \sin \beta \hfill \end{array}\hfill \\\underline{ -\text{ }\cos \left(\alpha +\beta \right)=-\left(\cos \alpha \cos \beta -\sin \alpha \sin \beta \right)}\hfill \end{array}}\hfill \\ \cos \left(\alpha -\beta \right)-\cos \left(\alpha +\beta \right)=2\sin \alpha \sin \beta \hfill \end{array}\\
Then, we divide by 2 to isolate the product of sines:
sin⁡αsin⁡β=12[cos⁡(α−β)−cos⁡(α+β)]\sin \alpha \sin \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)-\cos \left(\alpha +\beta \right)\right]\\
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:
cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]\\
sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]\\
sin⁡αsin⁡β=12[cos⁡(α−β)−cos⁡(α+β)]\sin \alpha \sin \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)-\cos \left(\alpha +\beta \right)\right]\\
cos⁡αsin⁡β=12[sin⁡(α+β)−sin⁡(α−β)]\cos \alpha \sin \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)-\sin \left(\alpha -\beta \right)\right]\\

Example 3: Express the Product as a Sum or Difference

Write cos⁡(3θ)cos⁡(5θ)\cos \left(3\theta \right)\cos \left(5\theta \right)\\ as a sum or difference.

Solution

We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.
 cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]cos⁡(3θ)cos⁡(5θ)=12[cos⁡(3θ−5θ)+cos⁡(3θ+5θ)] =12[cos⁡(2θ)+cos⁡(8θ)]Use even-odd identity.\begin{array}{ll}\text{ }\cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]\hfill & \hfill \\ \cos \left(3\theta \right)\cos \left(5\theta \right)=\frac{1}{2}\left[\cos \left(3\theta -5\theta \right)+\cos \left(3\theta +5\theta \right)\right]\hfill & \hfill \\ \text{ }=\frac{1}{2}\left[\cos \left(2\theta \right)+\cos \left(8\theta \right)\right] \hfill & \text{Use even-odd identity}.\hfill \end{array}\\

Try It 3

Use the product-to-sum formula to evaluate cos⁡11π12cos⁡π12\cos \frac{11\pi }{12}\cos \frac{\pi }{12}\\. Solution

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 u+v2=α\frac{u+v}{2}=\alpha and u−v2=β\frac{u-v}{2}=\beta . Then,
α+β=u+v2+u−v2 =2u2 =uα−β=u+v2−u−v2 =2v2 =v\begin{array}{l}\alpha +\beta =\frac{u+v}{2}+\frac{u-v}{2}\hfill \\ \text{ }=\frac{2u}{2}\hfill \\ \text{ }=u\hfill \\ \hfill \\ \alpha -\beta =\frac{u+v}{2}-\frac{u-v}{2}\hfill \\ \text{ }=\frac{2v}{2}\hfill \\ \text{ }=v\hfill \end{array}
Thus, replacing α\alpha and β\beta in the product-to-sum formula with the substitute expressions, we have
 sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)] sin⁡(u+v2)cos⁡(u−v2)=12[sin⁡u+sin⁡v]Substitute for(α+β) and (α−β)2sin⁡(u+v2)cos⁡(u−v2)=sin⁡u+sin⁡v\begin{array}{lll}\text{ }\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]\hfill & \hfill & \hfill \\ \text{ }\sin \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right)=\frac{1}{2}\left[\sin u+\sin v\right] \hfill & \hfill & \text{Substitute for}\left(\alpha +\beta \right)\text{ and }\left(\alpha -\beta \right)\hfill \\ 2\sin \left(\frac{u+v}{2}\right)\cos \left(\frac{u-v}{2}\right)=\sin u+\sin v\hfill & \hfill & \hfill \end{array}
The other sum-to-product identities are derived similarly.

A General Note: Sum-to-Product Formulas

The sum-to-product formulas are as follows:
sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)\sin \alpha +\sin \beta =2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)
sin⁡α−sin⁡β=2sin⁡(α−β2)cos⁡(α+β2)\sin \alpha -\sin \beta =2\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)
cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)
cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2)\cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)

Example 4: Writing the Difference of Sines as a Product

Write the following difference of sines expression as a product: sin⁡(4θ)−sin⁡(2θ)\sin \left(4\theta \right)-\sin \left(2\theta \right).

Solution

We begin by writing the formula for the difference of sines.
sin⁡α−sin⁡β=2sin⁡(α−β2)cos⁡(α+β2)\sin \alpha -\sin \beta =2\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)
Substitute the values into the formula, and simplify.
sin⁡(4θ)−sin⁡(2θ)=2sin⁡(4θ−2θ2)cos⁡(4θ+2θ2) =2sin⁡(2θ2)cos⁡(6θ2) =2sin⁡θcos⁡(3θ)\begin{array}{l}\sin \left(4\theta \right)-\sin \left(2\theta \right)=2\sin \left(\frac{4\theta -2\theta }{2}\right)\cos \left(\frac{4\theta +2\theta }{2}\right)\hfill \\ \text{ }=2\sin \left(\frac{2\theta }{2}\right)\cos \left(\frac{6\theta }{2}\right)\hfill \\ \text{ }=2\sin \theta \cos \left(3\theta \right)\hfill \end{array}

Try It 4

Use the sum-to-product formula to write the sum as a product: sin⁡(3θ)+sin⁡(θ)\sin \left(3\theta \right)+\sin \left(\theta \right). Solution

Example 5: Evaluating Using the Sum-to-Product Formula

Evaluate cos⁡(15∘)−cos⁡(75∘)\cos \left({15}^{\circ }\right)-\cos \left({75}^{\circ }\right).

Solution

We begin by writing the formula for the difference of cosines.
cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)\cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)
Then we substitute the given angles and simplify.
cos⁡(15∘)−cos⁡(75∘)=−2sin⁡(15∘+75∘2)sin⁡(15∘−75∘2) =−2sin⁡(45∘)sin⁡(−30∘) =−2(22)(−12) =22\begin{array}{l}\cos \left({15}^{\circ }\right)-\cos \left({75}^{\circ }\right)=-2\sin \left(\frac{{15}^{\circ }+{75}^{\circ }}{2}\right)\sin \left(\frac{{15}^{\circ }-{75}^{\circ }}{2}\right)\hfill \\ \text{ }=-2\sin \left({45}^{\circ }\right)\sin \left(-{30}^{\circ }\right)\hfill \\ \text{ }=-2\left(\frac{\sqrt{2}}{2}\right)\left(-\frac{1}{2}\right)\hfill \\ \text{ }=\frac{\sqrt{2}}{2}\hfill \end{array}

Example 6: Proving an Identity

Prove the identity:
cos⁡(4t)−cos⁡(2t)sin⁡(4t)+sin⁡(2t)=−tan⁡t\frac{\cos \left(4t\right)-\cos \left(2t\right)}{\sin \left(4t\right)+\sin \left(2t\right)}=-\tan t

Solution

We will start with the left side, the more complicated side of the equation, and rewrite the expression until it matches the right side.
cos⁡(4t)−cos⁡(2t)sin⁡(4t)+sin⁡(2t)=−2sin⁡(4t+2t2)sin⁡(4t−2t2)2sin⁡(4t+2t2)cos⁡(4t−2t2) =−2sin⁡(3t)sin⁡t2sin⁡(3t)cos⁡t =−)2‾)sin⁡(3t)‾sin⁡t)2‾)sin⁡(3t)‾cos⁡t =−sin⁡tcos⁡t =−tan⁡t\begin{array}{l}\frac{\cos \left(4t\right)-\cos \left(2t\right)}{\sin \left(4t\right)+\sin \left(2t\right)}=\frac{-2\sin \left(\frac{4t+2t}{2}\right)\sin \left(\frac{4t - 2t}{2}\right)}{2\sin \left(\frac{4t+2t}{2}\right)\cos \left(\frac{4t - 2t}{2}\right)}\hfill \\ \text{ }=\frac{-2\sin \left(3t\right)\sin t}{2\sin \left(3t\right)\cos t}\hfill \\ \text{ }=\frac{-\overline{)2}\overline{)\sin \left(3t\right)}\sin t}{\overline{)2}\overline{)\sin \left(3t\right)}\cos t}\hfill \\ \text{ }=-\frac{\sin t}{\cos t}\hfill \\ \text{ }=-\tan t\hfill \end{array}

Analysis of the Solution

Recall that verifying trigonometric identities has its own set of rules. The procedures for solving an equation are not the same as the procedures for verifying an identity. When we prove an identity, we pick one side to work on and make substitutions until that side is transformed into the other side.

Example 7: Verifying the Identity Using Double-Angle Formulas and Reciprocal Identities

Verify the identity csc⁡2θ−2=cos⁡(2θ)sin⁡2θ{\csc }^{2}\theta -2=\frac{\cos \left(2\theta \right)}{{\sin }^{2}\theta }.

Solution

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.
cos⁡(2θ)sin⁡2θ=1−2sin⁡2θsin⁡2θ =1sin⁡2θ−2sin⁡2θsin⁡2θ =csc⁡2θ−2\begin{array}{l}\frac{\cos \left(2\theta \right)}{{\sin }^{2}\theta }=\frac{1 - 2{\sin }^{2}\theta }{{\sin }^{2}\theta }\hfill \\ \text{ }=\frac{1}{{\sin }^{2}\theta }-\frac{2{\sin }^{2}\theta }{{\sin }^{2}\theta }\hfill \\ \text{ }={\csc }^{2}\theta -2\hfill \end{array}

Try It 5

Verify the identity tan⁡θcot⁡θ−cos⁡2θ=sin⁡2θ\tan \theta \cot \theta -{\cos }^{2}\theta ={\sin }^{2}\theta . Solution

Key Equations

Product-to-sum Formulas cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]sin⁡αsin⁡β=12[cos⁡(α−β)−cos⁡(α+β)]cos⁡αsin⁡β=12[sin⁡(α+β)−sin⁡(α−β)]\begin{array}{l}\hfill \\ \cos \alpha \cos \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)+\cos \left(\alpha +\beta \right)\right]\hfill \\ \sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right]\hfill \\ \sin \alpha \sin \beta =\frac{1}{2}\left[\cos \left(\alpha -\beta \right)-\cos \left(\alpha +\beta \right)\right]\hfill \\ \cos \alpha \sin \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)-\sin \left(\alpha -\beta \right)\right]\hfill \end{array}
Sum-to-product Formulas sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)sin⁡α−sin⁡β=2sin⁡(α−β2)cos⁡(α+β2)cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)cos⁡α+cos⁡β=2cos⁡(α+β2)cos⁡(α−β2)\begin{array}{l}\hfill \\ \sin \alpha +\sin \beta =2\sin \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)\hfill \\ \sin \alpha -\sin \beta =2\sin \left(\frac{\alpha -\beta }{2}\right)\cos \left(\frac{\alpha +\beta }{2}\right)\hfill \\ \cos \alpha -\cos \beta =-2\sin \left(\frac{\alpha +\beta }{2}\right)\sin \left(\frac{\alpha -\beta }{2}\right)\hfill \\ \cos \alpha +\cos \beta =2\cos \left(\frac{\alpha +\beta }{2}\right)\cos \left(\frac{\alpha -\beta }{2}\right)\hfill \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.

Glossary

product-to-sum formula
a trigonometric identity that allows the writing of a product of trigonometric functions as a sum or difference of trigonometric functions
sum-to-product formula
a trigonometric identity that allows, by using substitution, the writing of a sum of trigonometric functions as a product of trigonometric functions

Section Exercises

1. Starting with the product to sum formula sin⁡αcos⁡β=12[sin⁡(α+β)+sin⁡(α−β)]\sin \alpha \cos \beta =\frac{1}{2}\left[\sin \left(\alpha +\beta \right)+\sin \left(\alpha -\beta \right)\right], explain how to determine the formula for cos⁡αsin⁡β\cos \alpha \sin \beta . 2. Explain two different methods of calculating cos⁡(195∘)cos⁡(105∘)\cos \left(195^\circ \right)\cos \left(105^\circ \right), one of which uses the product to sum. Which method is easier? 3. Explain a situation where we would convert an equation from a sum to a product and give an example. 4. Explain a situation where we would convert an equation from a product to a sum, and give an example. For the following exercises, rewrite the product as a sum or difference. 5. 16sin⁡(16x)sin⁡(11x)16\sin \left(16x\right)\sin \left(11x\right) 6. 20cos⁡(36t)cos⁡(6t)20\cos \left(36t\right)\cos \left(6t\right) 7. 2sin⁡(5x)cos⁡(3x)2\sin \left(5x\right)\cos \left(3x\right) 8. 10cos⁡(5x)sin⁡(10x)10\cos \left(5x\right)\sin \left(10x\right) 9. sin⁡(−x)sin⁡(5x)\sin \left(-x\right)\sin \left(5x\right) 10. sin⁡(3x)cos⁡(5x)\sin \left(3x\right)\cos \left(5x\right) For the following exercises, rewrite the sum or difference as a product. 11. cos⁡(6t)+cos⁡(4t)\cos \left(6t\right)+\cos \left(4t\right) 12. sin⁡(3x)+sin⁡(7x)\sin \left(3x\right)+\sin \left(7x\right) 13. cos⁡(7x)+cos⁡(−7x)\cos \left(7x\right)+\cos \left(-7x\right) 14. sin⁡(3x)−sin⁡(−3x)\sin \left(3x\right)-\sin \left(-3x\right) 15. cos⁡(3x)+cos⁡(9x)\cos \left(3x\right)+\cos \left(9x\right) 16. sin⁡h−sin⁡(3h)\sin h-\sin \left(3h\right) For the following exercises, evaluate the product for the following using a sum or difference of two functions. Evaluate exactly. 17. cos⁡(45∘)cos⁡(15∘)\cos \left(45^\circ \right)\cos \left(15^\circ \right) 18. cos⁡(45∘)sin⁡(15∘)\cos \left(45^\circ \right)\sin \left(15^\circ \right) 19. sin⁡(−345∘)sin⁡(−15∘)\sin \left(-345^\circ \right)\sin \left(-15^\circ \right) 20. sin⁡(195∘)cos⁡(15∘)\sin \left(195^\circ \right)\cos \left(15^\circ \right) 21. sin⁡(−45∘)sin⁡(−15∘)\sin \left(-45^\circ \right)\sin \left(-15^\circ \right) For the following exercises, evaluate the product using a sum or difference of two functions. Leave in terms of sine and cosine. 22. cos⁡(23∘)sin⁡(17∘)\cos \left(23^\circ \right)\sin \left(17^\circ \right) 23. 2sin⁡(100∘)sin⁡(20∘)2\sin \left(100^\circ \right)\sin \left(20^\circ \right) 24. 2sin⁡(−100∘)sin⁡(−20∘)2\sin \left(-100^\circ \right)\sin \left(-20^\circ \right) 25. sin⁡(213∘)cos⁡(8∘)\sin \left(213^\circ \right)\cos \left(8^\circ \right) 26. 2cos⁡(56∘)cos⁡(47∘)2\cos \left(56^\circ \right)\cos \left(47^\circ \right) For the following exercises, rewrite the sum as a product of two functions. Leave in terms of sine and cosine. 27. sin⁡(76∘)+sin⁡(14∘)\sin \left(76^\circ \right)+\sin \left(14^\circ \right) 28. cos⁡(58∘)−cos⁡(12∘)\cos \left(58^\circ \right)-\cos \left(12^\circ \right) 29. sin⁡(101∘)−sin⁡(32∘)\sin \left(101^\circ \right)-\sin \left(32^\circ \right) 30. cos⁡(100∘)+cos⁡(200∘)\cos \left(100^\circ \right)+\cos \left(200^\circ \right) 31. sin⁡(−1∘)+sin⁡(−2∘)\sin \left(-1^\circ \right)+\sin \left(-2^\circ \right) For the following exercises, prove the identity. 32. cos⁡(a+b)cos⁡(a−b)=1−tan⁡atan⁡b1+tan⁡atan⁡b\frac{\cos \left(a+b\right)}{\cos \left(a-b\right)}=\frac{1-\tan a\tan b}{1+\tan a\tan b} 33. 4sin⁡(3x)cos⁡(4x)=2sin⁡(7x)−2sin⁡x4\sin \left(3x\right)\cos \left(4x\right)=2\sin \left(7x\right)-2\sin x 34. 6cos⁡(8x)sin⁡(2x)sin⁡(−6x)=−3sin⁡(10x)csc⁡(6x)+3\frac{6\cos \left(8x\right)\sin \left(2x\right)}{\sin \left(-6x\right)}=-3\sin \left(10x\right)\csc \left(6x\right)+3 35. sin⁡x+sin⁡(3x)=4sin⁡xcos⁡2x\sin x+\sin \left(3x\right)=4\sin x{\cos }^{2}x 36. 2(cos⁡3x−cos⁡xsin⁡2x)=cos⁡(3x)+cos⁡x2\left({\cos }^{3}x-\cos x{\sin }^{2}x\right)=\cos \left(3x\right)+\cos x 37. 2tan⁡xcos⁡(3x)=sec⁡x(sin⁡(4x)−sin⁡(2x))2\tan x\cos \left(3x\right)=\sec x\left(\sin \left(4x\right)-\sin \left(2x\right)\right) 38. cos⁡(a+b)+cos⁡(a−b)=2cos⁡acos⁡b\cos \left(a+b\right)+\cos \left(a-b\right)=2\cos a\cos b For the following exercises, rewrite the sum as a product of two functions or the product as a sum of two functions. Give your answer in terms of sines and cosines. Then evaluate the final answer numerically, rounded to four decimal places. 39. cos⁡(58∘)+cos⁡(12∘)\cos \left({58}^{\circ }\right)+\cos \left({12}^{\circ }\right) 40. sin⁡(2∘)−sin⁡(3∘)\sin \left({2}^{\circ }\right)-\sin \left({3}^{\circ }\right) 41. cos⁡(44∘)−cos⁡(22∘)\cos \left({44}^{\circ }\right)-\cos \left({22}^{\circ }\right) 42. cos⁡(176∘)sin⁡(9∘)\cos \left({176}^{\circ }\right)\sin \left({9}^{\circ }\right) 43. sin⁡(−14∘)sin⁡(85∘)\sin \left(-{14}^{\circ }\right)\sin \left({85}^{\circ }\right) For the following exercises, algebraically determine whether each of the given expressions is a true identity. If it is not an identity, replace the right-hand side with an expression equivalent to the left side. Verify the results by graphing both expressions on a calculator. 44. 2sin⁡(2x)sin⁡(3x)=cos⁡x−cos⁡(5x)2\sin \left(2x\right)\sin \left(3x\right)=\cos x-\cos \left(5x\right) 45. cos⁡(10θ)+cos⁡(6θ)cos⁡(6θ)−cos⁡(10θ)=cot⁡(2θ)cot⁡(8θ)\frac{\cos \left(10\theta \right)+\cos \left(6\theta \right)}{\cos \left(6\theta \right)-\cos \left(10\theta \right)}=\cot \left(2\theta \right)\cot \left(8\theta \right) 46. sin⁡(3x)−sin⁡(5x)cos⁡(3x)+cos⁡(5x)=tan⁡x\frac{\sin \left(3x\right)-\sin \left(5x\right)}{\cos \left(3x\right)+\cos \left(5x\right)}=\tan x 47. 2cos⁡(2x)cos⁡x+sin⁡(2x)sin⁡x=2sin⁡x2\cos \left(2x\right)\cos x+\sin \left(2x\right)\sin x=2\sin x 48. sin⁡(2x)+sin⁡(4x)sin⁡(2x)−sin⁡(4x)=−tan⁡(3x)cot⁡x\frac{\sin \left(2x\right)+\sin \left(4x\right)}{\sin \left(2x\right)-\sin \left(4x\right)}=-\tan \left(3x\right)\cot x For the following exercises, simplify the expression to one term, then graph the original function and your simplified version to verify they are identical. 49. sin⁡(9t)−sin⁡(3t)cos⁡(9t)+cos⁡(3t)\frac{\sin \left(9t\right)-\sin \left(3t\right)}{\cos \left(9t\right)+\cos \left(3t\right)} 50. 2sin⁡(8x)cos⁡(6x)−sin⁡(2x)2\sin \left(8x\right)\cos \left(6x\right)-\sin \left(2x\right) 51. sin⁡(3x)−sin⁡xsin⁡x\frac{\sin \left(3x\right)-\sin x}{\sin x} 52. cos⁡(5x)+cos⁡(3x)sin⁡(5x)+sin⁡(3x)\frac{\cos \left(5x\right)+\cos \left(3x\right)}{\sin \left(5x\right)+\sin \left(3x\right)} 53. sin⁡xcos⁡(15x)−cos⁡xsin⁡(15x)\sin x\cos \left(15x\right)-\cos x\sin \left(15x\right) For the following exercises, prove the following sum-to-product formulas. 54. sin⁡x−sin⁡y=2sin⁡(x−y2)cos⁡(x+y2)\sin x-\sin y=2\sin \left(\frac{x-y}{2}\right)\cos \left(\frac{x+y}{2}\right) 55. cos⁡x+cos⁡y=2cos⁡(x+y2)cos⁡(x−y2)\cos x+\cos y=2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right) For the following exercises, prove the identity. 56. sin⁡(6x)+sin⁡(4x)sin⁡(6x)−sin⁡(4x)=tan⁡(5x)cot⁡x\frac{\sin \left(6x\right)+\sin \left(4x\right)}{\sin \left(6x\right)-\sin \left(4x\right)}=\tan \left(5x\right)\cot x 57. cos⁡(3x)+cos⁡xcos⁡(3x)−cos⁡x=−cot⁡(2x)cot⁡x\frac{\cos \left(3x\right)+\cos x}{\cos \left(3x\right)-\cos x}=-\cot \left(2x\right)\cot x 58. cos⁡(6y)+cos⁡(8y)sin⁡(6y)−sin⁡(4y)=cot⁡ycos⁡(7y)sec⁡(5y)\frac{\cos \left(6y\right)+\cos \left(8y\right)}{\sin \left(6y\right)-\sin \left(4y\right)}=\cot y\cos \left(7y\right)\sec \left(5y\right) 59. cos⁡(2y)−cos⁡(4y)sin⁡(2y)+sin⁡(4y)=tan⁡y\frac{\cos \left(2y\right)-\cos \left(4y\right)}{\sin \left(2y\right)+\sin \left(4y\right)}=\tan y 60. sin⁡(10x)−sin⁡(2x)cos⁡(10x)+cos⁡(2x)=tan⁡(4x)\frac{\sin \left(10x\right)-\sin \left(2x\right)}{\cos \left(10x\right)+\cos \left(2x\right)}=\tan \left(4x\right) 61. cos⁡x−cos⁡(3x)=4sin⁡2xcos⁡x\cos x-\cos \left(3x\right)=4{\sin }^{2}x\cos x 62. (cos⁡(2x)−cos⁡(4x))2+(sin⁡(4x)+sin⁡(2x))2=4sin⁡2(3x){\left(\cos \left(2x\right)-\cos \left(4x\right)\right)}^{2}+{\left(\sin \left(4x\right)+\sin \left(2x\right)\right)}^{2}=4{\sin }^{2}\left(3x\right) 63. tan⁡(π4−t)=1−tan⁡t1+tan⁡t\tan \left(\frac{\pi }{4}-t\right)=\frac{1-\tan t}{1+\tan t}

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