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]2\sin \left(2x\right)\cos x=2\left(2\sin x\cos x\right)\cos x=[/latex]
[latex]4\sin x{\cos }^{2}x[/latex]

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]=\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]

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]