| M141: Calculus with A&T | Spring 2014 | |
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Homework Help
Unit One |
Section 1.3
| 2. | Use $s = \alpha r$. Here 10$\pi$ = $\alpha \times 8$, so $\alpha = \frac{5}{4}\pi$. |
| 4. | The radius is $r$ = 50 cm. The arc length is $s$=30cm. Use $s = \alpha r$ to determine $\alpha$. |
| 6. | I suggest filling in the blanks on the unit circle hand out prior to completing this exercise. Then use the fact that the angle $-3\pi/2$ is the same as the angle $\pi/2$, an angle of $-\pi/3$ is the same as the angle $5\pi/3$, and so on. |
| 8. | If $\tan x = 2$, then $\frac{\sin x}{\cos x} = 2 \Rightarrow \sin x = 2\cos x$. Use this fact and the Fundamental Theorem of Trigonometry ($\sin^2 x + \cos^2 x= 1$) to determine the value of the $\cos x$ is $1/\sqrt{5}$. |
| 35. |
Both exercises 35 and 36 rely on the fact that the sine function is an odd function (that is, $\sin(-B) = -\sin B$) and the cosine function is an even function (that is, $\cos (-B) = \cos B)$.
Now, $\cos (A-B) = \cos(A + -B) = \cos A \cos(-B) - \sin A \sin (-B) = \cos A \cos B - (\sin A)(-\sin B) = \cos A \cos B + \sin A \sin B $ |
| 35. |
See the help for Exercise 35 above.
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| 40. |
The answer is $-\sin x$. You can use either the sum or difference formula for sine.
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Section 1.6
| 66. | The inverse cosine function ($\cos^{-1} x$) has a range restricted to the interval $[0, \pi]$. Therefore, the answer to (b) is $\displaystyle{\frac{3\pi}{4}}$. |
Section 2.4
| 30. | Split the numerator on the minus and plus signs. That is $\displaystyle{\frac{x^2 - x + \sin x}{2x} = \frac{x^2}{2x} - \frac{x}{2x} + \frac{1}{2}\frac{\sin x}{x}}$. Use the difference and sum laws for limits because each of these three terms have limits as $x$ approaches zero. |
| 34. | Make a substitution $x = \sin h$. Knowing that $\displaystyle{\lim_{h \rightarrow 0} x = \lim_{h \rightarrow 0} \sin h = \sin 0 = 0}$, you can write $\displaystyle{\lim_{h \rightarrow 0}\frac{\sin (\sin h)}{\sin h} = \lim_{x \rightarrow 0}\frac{\sin x}{x} = 1}$ |
Section 3.5
| 30. | $\displaystyle{p' = \frac{\sec^2 q(1 + \tan q) - \sec^2 q \tan q}{(1 + \tan q)^2}}$ |