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מדריכי לימוד > Precalculus II

Solutions for Sum and Difference Identities

Solutions to Try Its

1. 2+64\frac{\sqrt{2}+\sqrt{6}}{4} 2. 2−64\frac{\sqrt{2}-\sqrt{6}}{4} 3. 1−31+3\frac{1-\sqrt{3}}{1+\sqrt{3}} 4. cos⁡(5π14)\cos \left(\frac{5\pi }{14}\right) 5. tan⁡(π−θ)=tan⁡(π)−tan⁡θ1+tan⁡(π)tan⁡θ =0−tan⁡θ1+0⋅tan⁡θ =−tan⁡θ\begin{array}{l}\tan \left(\pi -\theta \right)=\frac{\tan \left(\pi \right)-\tan \theta }{1+\tan \left(\pi \right)\tan \theta }\hfill \\ \text{ }=\frac{0-\tan \theta }{1+0\cdot \tan \theta }\hfill \\ \text{ }=-\tan \theta \hfill \end{array}

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 xx, the second angle measures π2−x\frac{\pi }{2}-x. Then sin⁡x=cos⁡(π2−x)\sin x=\cos \left(\frac{\pi }{2}-x\right). The same holds for the other cofunction identities. The key is that the angles are complementary. 3. sin⁡(−x)=−sin⁡x\sin \left(-x\right)=-\sin x, so sin⁡x\sin x is odd. cos⁡(−x)=cos⁡(0−x)=cos⁡x\cos \left(-x\right)=\cos \left(0-x\right)=\cos x, so cos⁡x\cos x is even. 5. 2+64\frac{\sqrt{2}+\sqrt{6}}{4} 7. 6−24\frac{\sqrt{6}-\sqrt{2}}{4} 9. −2−3-2-\sqrt{3} 11. −22sin⁡x−22cos⁡x-\frac{\sqrt{2}}{2}\sin x-\frac{\sqrt{2}}{2}\cos x 13. −12cos⁡x−32sin⁡x-\frac{1}{2}\cos x-\frac{\sqrt{3}}{2}\sin x 15. csc⁡θ\csc \theta 17. cot⁡x\cot x 19. tan⁡(x10)\tan \left(\frac{x}{10}\right) 21. sin⁡(a−b)=(45)(13)−(35)(223)=4−6215\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} cos⁡(a+b)=(35)(13)−(45)(223)=3−8215\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} 23. 2−64\frac{\sqrt{2}-\sqrt{6}}{4} 25. sin⁡x\sin x Graph of y=sin(x) from -2pi to 2pi. 27. cot⁡(π6−x)\cot \left(\frac{\pi }{6}-x\right) 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. cot⁡(π4+x)\cot \left(\frac{\pi }{4}+x\right) Graph of y=cot(pi/4 + x) - in comparison to the usual y=cot(x) graph, this one is shifted by pi/4. 31. sin⁡x2+cos⁡x2\frac{\sin x}{\sqrt{2}}+\frac{\cos x}{\sqrt{2}} 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 g(x)=sin⁡(9x)−cos⁡(3x)sin⁡(6x)g\left(x\right)=\sin \left(9x\right)-\cos \left(3x\right)\sin \left(6x\right). 37. They are the same. 39. They are the different, try g(θ)=2tan⁡θ1−tan⁡2θg\left(\theta \right)=\frac{2\tan \theta }{1-{\tan }^{2}\theta }. 41. They are different, try g(x)=tan⁡x−tan⁡(2x)1+tan⁡xtan⁡(2x)g\left(x\right)=\frac{\tan x-\tan \left(2x\right)}{1+\tan x\tan \left(2x\right)}. 43. −3−122, or −0.2588-\frac{\sqrt{3}-1}{2\sqrt{2}},\text{ or }-0.2588 45. 1+322\frac{1+\sqrt{3}}{2\sqrt{2}}, or 0.9659 47. tan⁡(x+π4)=tan⁡x+tan⁡(π4)1−tan⁡xtan⁡(π4)=tan⁡x+11−tan⁡x(1)=tan⁡x+11−tan⁡x\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}   49. cos⁡(a+b)cos⁡acos⁡b=cos⁡acos⁡bcos⁡acos⁡b−sin⁡asin⁡bcos⁡acos⁡b=1−tan⁡atan⁡b\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} 51. cos⁡(x+h)−cos⁡xh=cos⁡xcosh−sin⁡xsinh−cos⁡xh=cos⁡x(cosh−1)−sin⁡xsinhh=cos⁡xcos⁡h−1h−sin⁡xsin⁡hh\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} 53. True 55. True. Note that sin⁡(α+β)=sin⁡(π−γ)\sin \left(\alpha +\beta \right)=\sin \left(\pi -\gamma \right) and expand the right hand side.

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