Solutions 52: Sum and Difference Identities

Solutions to Odd-Numbered Answers

1. The cofunction identities apply to complementary angles. Viewing the two acute angles of a right triangle, if one of those angles measures [latex]x[/latex], the second angle measures [latex]\frac{\pi }{2}-x[/latex]. Then [latex]\sin x=\cos \left(\frac{\pi }{2}-x\right)[/latex]. The same holds for the other cofunction identities. The key is that the angles are complementary.

3. [latex]\sin \left(-x\right)=-\sin x[/latex], so [latex]\sin x[/latex] is odd. [latex]\cos \left(-x\right)=\cos \left(0-x\right)=\cos x[/latex], so [latex]\cos x[/latex] is even.

5. [latex]\frac{\sqrt{2}+\sqrt{6}}{4}[/latex]

7. [latex]\frac{\sqrt{6}-\sqrt{2}}{4}[/latex]

9. [latex]-2-\sqrt{3}[/latex]

11. [latex]-\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x[/latex]

13. [latex]-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x[/latex]

15. [latex]\csc \theta[/latex]

17. [latex]\cot x[/latex]

19. [latex]\tan \left(\frac{x}{10}\right)[/latex]

21. [latex]\sin \left(a-b\right)=\left(\frac{4}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{3}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)=\frac{4 - 6\sqrt{2}}{15}[/latex]
[latex]\cos \left(a+b\right)=\left(\frac{3}{5}\right)\left(\frac{1}{3}\right)-\left(\frac{4}{5}\right)\left(\frac{2\sqrt{2}}{3}\right)=\frac{3 - 8\sqrt{2}}{15}[/latex]

23. [latex]\frac{\sqrt{2}-\sqrt{6}}{4}[/latex]

25. [latex]\sin x[/latex]

Graph of y=sin(x) from -2pi to 2pi.

27. [latex]\cot \left(\frac{\pi }{6}-x\right)[/latex]

Graph of y=cot(pi/6 - x) from -2pi to pi - in comparison to the usual y=cot(x) graph, this one is reflected across the x-axis and shifted by pi/6.

29. [latex]\cot \left(\frac{\pi }{4}+x\right)[/latex]

Graph of y=cot(pi/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi/4.

31. [latex]\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}}[/latex]

Graph of y = sin(x) / rad2 + cos(x) / rad2 - it looks like the sin curve shifted by pi/4.

 

33. They are the same.

35. They are the different, try [latex]g\left(x\right)=\sin \left(9x\right)-\cos \left(3x\right)\sin \left(6x\right)[/latex].

37. They are the same.

39. They are the different, try [latex]g\left(\theta \right)=\frac{2\tan \theta }{1-{\tan }^{2}\theta }[/latex].

41. They are different, try [latex]g\left(x\right)=\frac{\tan x-\tan \left(2x\right)}{1+\tan x\tan \left(2x\right)}[/latex].

43. [latex]-\frac{\sqrt{3}-1}{2\sqrt{2}},\text{ or }-0.2588[/latex]

45. [latex]\frac{1+\sqrt{3}}{2\sqrt{2}}[/latex], or 0.9659

47. [latex]\begin{array}{c}\tan \left(x+\frac{\pi }{4}\right)=\\ \frac{\tan x+\tan \left(\frac{\pi }{4}\right)}{1-\tan x\tan \left(\frac{\pi }{4}\right)}=\\ \frac{\tan x+1}{1-\tan x\left(1\right)}=\frac{\tan x+1}{1-\tan x}\end{array}[/latex]

 

49. [latex]\begin{array}{c}\frac{\cos \left(a+b\right)}{\cos a\cos b}=\\ \frac{\cos a\cos b}{\cos a\cos b}-\frac{\sin a\sin b}{\cos a\cos b}=1-\tan a\tan b\end{array}[/latex]

51. [latex]\begin{array}{c}\frac{\cos \left(x+h\right)-\cos x}{h}=\\ \frac{\cos x\mathrm{cosh}-\sin x\mathrm{sinh}-\cos x}{h}=\\ \frac{\cos x\left(\mathrm{cosh}-1\right)-\sin x\mathrm{sinh}}{h}=\cos x\frac{\cos h - 1}{h}-\sin x\frac{\sin h}{h}\end{array}[/latex]

53. True

55. True. Note that [latex]\sin \left(\alpha +\beta \right)=\sin \left(\pi -\gamma \right)[/latex] and expand the right hand side.