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Verify the identity $\text{\hspace{0.17em}}{\mathrm{csc}}^{2}\theta -2=\frac{\mathrm{cos}(2\theta )}{{\mathrm{sin}}^{2}\theta}.$
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.
Verify the identity $\text{\hspace{0.17em}}\mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\theta -{\mathrm{cos}}^{2}\theta ={\mathrm{sin}}^{2}\theta .$
$$\begin{array}{ccc}\hfill \mathrm{tan}\text{\hspace{0.17em}}\theta \text{\hspace{0.17em}}\mathrm{cot}\text{\hspace{0.17em}}\theta -{\mathrm{cos}}^{2}\theta & =& \left(\frac{\mathrm{sin}\text{\hspace{0.17em}}\theta}{\mathrm{cos}\text{\hspace{0.17em}}\theta}\right)\left(\frac{\mathrm{cos}\text{\hspace{0.17em}}\theta}{\mathrm{sin}\text{\hspace{0.17em}}\theta}\right)-{\mathrm{cos}}^{2}\theta \hfill \\ & =& 1-{\mathrm{cos}}^{2}\theta \hfill \\ & =& {\mathrm{sin}}^{2}\theta \hfill \end{array}$$
Access these online resources for additional instruction and practice with the product-to-sum and sum-to-product identities.
Product-to-sum Formulas | $\begin{array}{ccc}\hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}[\mathrm{cos}(\alpha -\beta )+\mathrm{cos}(\alpha +\beta )]\hfill \\ \hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}[\mathrm{sin}(\alpha +\beta )+\mathrm{sin}(\alpha -\beta )]\hfill \\ \hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}[\mathrm{cos}(\alpha -\beta )-\mathrm{cos}(\alpha +\beta )]\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta & =& \frac{1}{2}[\mathrm{sin}(\alpha +\beta )-\mathrm{sin}(\alpha -\beta )]\hfill \end{array}$ |
Sum-to-product Formulas | $\begin{array}{ccc}\hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha +\mathrm{sin}\text{\hspace{0.17em}}\beta & =& 2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha +\beta}{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta}{2}\right)\hfill \\ \hfill \mathrm{sin}\text{\hspace{0.17em}}\alpha -\mathrm{sin}\text{\hspace{0.17em}}\beta & =& 2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha -\beta}{2}\right)\mathrm{cos}\left(\frac{\alpha +\beta}{2}\right)\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha -\mathrm{cos}\text{\hspace{0.17em}}\beta & =& -2\text{\hspace{0.17em}}\mathrm{sin}\left(\frac{\alpha +\beta}{2}\right)\mathrm{sin}\left(\frac{\alpha -\beta}{2}\right)\hfill \\ \hfill \mathrm{cos}\text{\hspace{0.17em}}\alpha +\mathrm{cos}\text{\hspace{0.17em}}\beta & =& 2\text{\hspace{0.17em}}\mathrm{cos}\left(\frac{\alpha +\beta}{2}\right)\mathrm{cos}\left(\frac{\alpha -\beta}{2}\right)\hfill \end{array}$ |
Starting with the product to sum formula $\text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\beta =\frac{1}{2}[\mathrm{sin}(\alpha +\beta )+\mathrm{sin}(\alpha -\beta )],$ explain how to determine the formula for $\text{\hspace{0.17em}}\mathrm{cos}\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}\mathrm{sin}\text{\hspace{0.17em}}\beta .$
Substitute $\text{\hspace{0.17em}}\alpha \text{\hspace{0.17em}}$ into cosine and $\text{\hspace{0.17em}}\beta \text{\hspace{0.17em}}$ into sine and evaluate.
Provide two different methods of calculating $\text{\hspace{0.17em}}\mathrm{cos}(\mathrm{195\xb0})\mathrm{cos}(\mathrm{105\xb0}),$ one of which uses the product to sum. Which method is easier?
Describe a situation where we would convert an equation from a sum to a product and give an example.
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: $\text{\hspace{0.17em}}\frac{\mathrm{sin}(3x)+\mathrm{sin}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}=1.\text{\hspace{0.17em}}$ When converting the numerator to a product the equation becomes: $\text{\hspace{0.17em}}\frac{2\text{\hspace{0.17em}}\mathrm{sin}(2x)\mathrm{cos}\text{\hspace{0.17em}}x}{\mathrm{cos}\text{\hspace{0.17em}}x}=1$
Describe 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.
$16\text{\hspace{0.17em}}\mathrm{sin}(16x)\mathrm{sin}(11x)$
$8\left(\mathrm{cos}\left(5x\right)-\mathrm{cos}\left(27x\right)\right)$
$20\text{\hspace{0.17em}}\mathrm{cos}\left(36t\right)\mathrm{cos}\left(6t\right)$
$2\text{\hspace{0.17em}}\mathrm{sin}\left(5x\right)\mathrm{cos}\left(3x\right)$
$\mathrm{sin}\left(2x\right)+\mathrm{sin}\left(8x\right)$
$10\text{\hspace{0.17em}}\mathrm{cos}\left(5x\right)\mathrm{sin}\left(10x\right)$
$\mathrm{sin}\left(-x\right)\mathrm{sin}\left(5x\right)$
$\frac{1}{2}\left(\mathrm{cos}\left(6x\right)-\mathrm{cos}\left(4x\right)\right)$
$\mathrm{sin}\left(3x\right)\mathrm{cos}\left(5x\right)$
For the following exercises, rewrite the sum or difference as a product.
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