Solutions for Sum-to-Product and Product-to-Sum Formulas
Solutions to Try Its
1. [latex]\frac{1}{2}\left(\cos 6\theta +\cos 2\theta \right)\\[/latex] 2. [latex]\frac{1}{2}\left(\sin 2x+\sin 2y\right)\\[/latex] 3. [latex]\frac{-2-\sqrt{3}}{4}\\[/latex] 4. [latex]2\sin \left(2\theta \right)\cos \left(\theta \right)[/latex] 5. [latex]\begin{array}{l}\tan \theta \cot \theta -{\cos }^{2}\theta =\left(\frac{\sin \theta }{\cos \theta }\right)\left(\frac{\cos \theta }{\sin \theta }\right)-{\cos }^{2}\theta \hfill \\ =1-{\cos }^{2}\theta \hfill \\ ={\sin }^{2}\theta \hfill \end{array}\\[/latex]Solutions to Odd-Numbered Exercises
1. Substitute [latex]\alpha [/latex] into cosine and [latex]\beta [/latex] into sine and evaluate. 3. 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: [latex]\frac{\sin \left(3x\right)+\sin x}{\cos x}=1[/latex]. When converting the numerator to a product the equation becomes: [latex]\frac{2\sin \left(2x\right)\cos x}{\cos x}=1\\[/latex] 5. [latex]8\left(\cos \left(5x\right)-\cos \left(27x\right)\right)[/latex] 7. [latex]\sin \left(2x\right)+\sin \left(8x\right)[/latex] 9. [latex]\frac{1}{2}\left(\cos \left(6x\right)-\cos \left(4x\right)\right)[/latex] 11. [latex]2\cos \left(5t\right)\cos t[/latex] 13. [latex]2\cos \left(7x\right)[/latex] 15. [latex]2\cos \left(6x\right)\cos \left(3x\right)[/latex] 17. [latex]\frac{1}{4}\left(1+\sqrt{3}\right)[/latex] 19. [latex]\frac{1}{4}\left(\sqrt{3}-2\right)[/latex] 21. [latex]\frac{1}{4}\left(\sqrt{3}-1\right)[/latex] 23. [latex]\cos \left(80^\circ \right)-\cos \left(120^\circ \right)[/latex] 25. [latex]\frac{1}{2}\left(\sin \left(221^\circ \right)+\sin \left(205^\circ \right)\right)[/latex] 27. [latex]\sqrt{2}\cos \left(31^\circ \right)[/latex] 29. [latex]2\cos \left(66.5^\circ \right)\sin \left(34.5^\circ \right)[/latex] 31. [latex]2\sin \left(-1.5^\circ \right)\cos \left(0.5^\circ \right)[/latex] 33. [latex]{2}\sin \left({7x}\right){-2}\sin{ x}={ 2}\sin \left({4x}+{ 3x }\right)-{ 2 }\sin\left({4x } - { 3x }\right)=\\ {2}\left(\sin\left({ 4x }\right)\cos\left({ 3x }\right)+\sin\left({ 3x }\right)\cos\left({ 4x }\right)\right)-{ 2 }\left(\sin\left({ 4x }\right)\cos\left({ 3x }\right)-\sin \left({ 3x }\right)\cos\left({ 4x }\right)\right)=\\{2}\sin\left({ 4x }\right)\cos\left({ 3x }\right)+{2}\sin\left({ 3x }\right)\cos\left({ 4x }\right)-{ 2 }\sin\left({ 4x }\right)\cos\left({ 3x }\right)+{ 2 }\sin\left({ 3x }\right)\cos\left({ 4x }\right)=\\{ 4 }\sin\left({ 3x }\right)\cos\left({ 4x }\right)\\[/latex] 35. [latex]\sin x+\sin \left(3x\right)=2\sin \left(\frac{4x}{2}\right)\cos \left(\frac{-2x}{2}\right)=[/latex] [latex-display]2\sin \left(2x\right)\cos x=2\left(2\sin x\cos x\right)\cos x=[/latex-display] [latex-display]4\sin x{\cos }^{2}x[/latex-display] 37. [latex]2\tan x\cos \left(3x\right)=\frac{2\sin x\cos \left(3x\right)}{\cos x}=\frac{2\left(.5\left(\sin \left(4x\right)-\sin \left(2x\right)\right)\right)}{\cos x}[/latex] [latex-display]=\frac{1}{\cos x}\left(\sin \left(4x\right)-\sin \left(2x\right)\right)=\sec x\left(\sin \left(4x\right)-\sin \left(2x\right)\right)[/latex-display] 39. [latex]2\cos \left({35}^{\circ }\right)\cos \left({23}^{\circ }\right),\text{ 1}\text{.5081}[/latex] 41. [latex]-2\sin \left({33}^{\circ }\right)\sin \left({11}^{\circ }\right),\text{ }-0.2078[/latex] 43. [latex]\frac{1}{2}\left(\cos \left({99}^{\circ }\right)-\cos \left({71}^{\circ }\right)\right),\text{ }-0.2410[/latex] 45. It is an identity. 47. It is not an identity, but [latex]2{\cos }^{3}x[/latex] is. 49. [latex]\tan \left(3t\right)[/latex] 51. [latex]2\cos \left(2x\right)[/latex] 53. [latex]-\sin \left(14x\right)[/latex] 55. Start with [latex]\cos x+\cos y[/latex]. Make a substitution and let [latex]x=\alpha +\beta [/latex] and let [latex]y=\alpha -\beta [/latex], so [latex]\cos x+\cos y[/latex] becomes[latex]\cos \left(\alpha +\beta \right)+\cos \left(\alpha -\beta \right)=\cos \alpha \cos \beta -\sin \alpha \sin \beta +\cos \alpha \cos \beta +\sin \alpha \sin \beta =2\cos \alpha \cos \beta [/latex]
Since [latex]x=\alpha +\beta [/latex] and [latex]y=\alpha -\beta [/latex], we can solve for [latex]\alpha [/latex] and [latex]\beta [/latex] in terms of x and y and substitute in for [latex]2\cos \alpha \cos \beta [/latex] and get [latex]2\cos \left(\frac{x+y}{2}\right)\cos \left(\frac{x-y}{2}\right)[/latex]. 57. [latex]\frac{\cos \left(3x\right)+\cos x}{\cos \left(3x\right)-\cos x}=\frac{2\cos \left(2x\right)\cos x}{-2\sin \left(2x\right)\sin x}=-\cot \left(2x\right)\cot x[/latex] 59. [latex]\begin{array}{l}\frac{\cos \left(2y\right)-\cos \left(4y\right)}{\sin \left(2y\right)+\sin \left(4y\right)}=\frac{-2\sin \left(3y\right)\sin \left(-y\right)}{2\sin \left(3y\right)\cos y}=\\ \frac{2\sin \left(3y\right)\sin \left(y\right)}{2\sin \left(3y\right)\cos y}=\tan y\end{array}[/latex] 61. [latex]\begin{array}{l}\cos x-\cos \left(3x\right)=-2\sin \left(2x\right)\sin \left(-x\right)=\\ 2\left(2\sin x\cos x\right)\sin x=4{\sin }^{2}x\cos x\end{array}[/latex] 63. [latex]\tan \left(\frac{\pi }{4}-t\right)=\frac{\tan \left(\frac{\pi }{4}\right)-\tan t}{1+\tan \left(\frac{\pi }{4}\right)\tan \left(t\right)}=\frac{1-\tan t}{1+\tan t}[/latex]Licenses & Attributions
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- Precalculus. Provided by: OpenStax Authored by: OpenStax College. Located at: https://cnx.org/contents/[email protected]:1/Preface. License: CC BY: Attribution.
