Corrected exercises: trigonometric functions and their derivatives (201-103)

$f'(t)=6\cos(3t)\cdot 3=18\cos(3t)$. At $t=\frac{\pi}{3}$, $3t=\pi$ and $\cos\pi=-1$: $f'\left(\frac{\pi}{3}\right)=-18$. Pitfall: forgetting the $3$ from the inside and answering $-6$.

$g'(t)=-\sin\left(2t+\frac{\pi}{3}\right)\cdot 2=-2\sin\left(2t+\frac{\pi}{3}\right)$. At $t=\frac{\pi}{12}$, the angle is $\frac{\pi}{6}+\frac{\pi}{3}=\frac{\pi}{2}$ and $\sin\frac{\pi}{2}=1$: $g'\left(\frac{\pi}{12}\right)=-2$. The phase shift $\frac{\pi}{3}$ is a constant: it leaves no factor.

$h'(t)=\frac{4}{\cos^{2}(4t)}$. At $t=\frac{\pi}{16}$, $4t=\frac{\pi}{4}$ and $\cos^{2}\frac{\pi}{4}=\frac{1}{2}$: $h'\left(\frac{\pi}{16}\right)=\frac{4}{\frac{1}{2}}=8$. Check with the other form: $4\left(1+\tan^{2}\frac{\pi}{4}\right)=4\times 2=8$.

$f'(t)=2\cos t-3\cdot(-\sin t)=2\cos t+3\sin t$. At $t=\frac{\pi}{6}$: $f'\left(\frac{\pi}{6}\right)=2\cdot\frac{\sqrt{3}}{2}+3\cdot\frac{1}{2}=\sqrt{3}+1.5\approx 3.232$. Pitfall: the sign, $-3\cos t$ differentiates to $+3\sin t$.

Product: $g'(t)=2t\sin t+t^{2}\cos t$. At $t=\frac{\pi}{2}$, $\sin=1$ and $\cos=0$: $g'\left(\frac{\pi}{2}\right)=2\cdot\frac{\pi}{2}=\pi\approx 3.142$. Pitfall: writing $(t^{2}\sin t)'=2t\cos t$, by multiplying together the derivatives of the two factors.

Quotient: $h'(t)=\frac{\cos t\,(1+\cos t)-\sin t\cdot(-\sin t)}{(1+\cos t)^{2}}=\frac{\cos t+\cos^{2}t+\sin^{2}t}{(1+\cos t)^{2}}=\frac{1+\cos t}{(1+\cos t)^{2}}=\frac{1}{1+\cos t}$, thanks to $\sin^{2}t+\cos^{2}t=1$. At $t=\frac{\pi}{3}$: $h'\left(\frac{\pi}{3}\right)=\frac{1}{1.5}=\frac{2}{3}\approx 0.667$.

Power first, $f(t)=(\cos t)^{3}$: $f'(t)=3\cos^{2}t\cdot(-\sin t)=-3\cos^{2}t\sin t$. At $t=\frac{\pi}{4}$: $f'\left(\frac{\pi}{4}\right)=-3\cdot\frac{1}{2}\cdot\frac{\sqrt{2}}{2}=-\frac{3\sqrt{2}}{4}\approx -1.061$. Pitfall: confusing $\cos^{3}t$ and $\cos(t^{3})$.

$g'(t)=\cos\left(\sqrt{t}\right)\cdot\frac{1}{2\sqrt{t}}$. At $t=\frac{\pi^{2}}{9}$, $\sqrt{t}=\frac{\pi}{3}$ and $\cos\frac{\pi}{3}=\frac{1}{2}$: $g'\left(\frac{\pi^{2}}{9}\right)=\frac{1}{2}\cdot\frac{1}{\frac{2\pi}{3}}=\frac{3}{4\pi}\approx 0.239$.

Product of two composites: $h'(t)=2\cos(2t)\cos(3t)+\sin(2t)\cdot(-3\sin(3t))=2\cos(2t)\cos(3t)-3\sin(2t)\sin(3t)$. At $t=\frac{\pi}{6}$, $2t=\frac{\pi}{3}$ and $3t=\frac{\pi}{2}$: the first term vanishes because $\cos\frac{\pi}{2}=0$, and $h'\left(\frac{\pi}{6}\right)=-3\cdot\frac{\sqrt{3}}{2}\cdot 1=-\frac{3\sqrt{3}}{2}\approx -2.598$.

$f'(t)=10\cos(2t)$, then $f''(t)=-20\sin(2t)=-4f(t)$: the second derivative of a sine with angular frequency $2$ is $-2^{2}$ times the function. At $t=\frac{\pi}{12}$, $\sin\frac{\pi}{6}=\frac{1}{2}$: $f''\left(\frac{\pi}{12}\right)=-10$. Pitfall: a single factor $2$ instead of two, which gives $-5$.

The bridge between the two units is $180^{\circ}=\pi$ rad. So $150^{\circ}=150\times\dfrac{\pi}{180}=\dfrac{5\pi}{6}\approx 2.618$ rad. In the other direction, $\dfrac{7\pi}{6}=\dfrac{7}{6}\times 180^{\circ}=210^{\circ}$. Sanity check: $150^{\circ}$ is past the right angle but short of the straight angle, so its measure in radians must fall between $\dfrac{\pi}{2}\approx 1.571$ and $\pi\approx 3.142$, which $2.618$ satisfies. The classic pitfall is to multiply by $\dfrac{180}{\pi}$ when you should divide: you can always catch it by checking that the result lands in the right interval. A botched conversion does not cost one mark, it spoils the whole model that follows.

$\dfrac{2\pi}{3}=120^{\circ}$ is in the second quadrant, where cosine is NEGATIVE; its reference angle is $\dfrac{\pi}{3}$, so $\cos\dfrac{2\pi}{3}=-\dfrac{1}{2}=-0.5$. $\dfrac{5\pi}{4}=225^{\circ}$ is in the third quadrant, where sine is negative, with $\dfrac{\pi}{4}$ as reference: $\sin\dfrac{5\pi}{4}=-\dfrac{\sqrt{2}}{2}\approx -0.7071$. Finally $\tan\dfrac{\pi}{3}=\dfrac{\sin(\pi/3)}{\cos(\pi/3)}=\dfrac{\sqrt{3}/2}{1/2}=\sqrt{3}\approx 1.7321$. The method comes down to two moves, and only one of them is a calculation: find the reference angle, then attach the sign of the quadrant.

The Pythagorean identity $\sin^{2}\theta+\cos^{2}\theta=1$ gives $\cos^{2}\theta=1-\dfrac{9}{25}=\dfrac{16}{25}$, so $\cos\theta=\pm\dfrac{4}{5}$. The quadrant decides: in the second quadrant cosine is negative, so $\cos\theta=-\dfrac{4}{5}=-0.8$. Then $\tan\theta=\dfrac{\sin\theta}{\cos\theta}=\dfrac{3/5}{-4/5}=-\dfrac{3}{4}=-0.75$. The pitfall is costly: forgetting the sign amounts to placing the terminal point in the first quadrant, hence answering $0.8$ and $0.75$, two values that look right and are both wrong. The identity gives the SQUARE, never the sign.

These three identities can be read off the figure without computing anything. $\pi-x$ is the mirror image of $x$ in the vertical axis: the y-coordinate stays, the x-coordinate changes sign, so $\sin(\pi-x)=\sin x$. $\pi+x$ is the reflection through the centre: both coordinates change sign, so $\cos(\pi+x)=-\cos x$. Finally $-x$ is the reflection in the horizontal axis, so $\sin(-x)=-\sin x$: sine is odd, cosine is even. These three equalities come back in exercise 5, where they are used to prove a derivative.

The software read 30 RADIANS, not 30 degrees. Check: $30-4\times 2\pi\approx 30-25.133=4.867$ rad, an angle in the fourth quadrant, where sine is negative; and indeed $\sin(4.867)\approx -0.988$. To get $\sin(30^{\circ})$, type $\sin\left(30\times\dfrac{\pi}{180}\right)=\sin\dfrac{\pi}{6}=0.5$. Remember the warning rule: every calculating package, every spreadsheet and every programming language works in radians by default. A report that announces a falling index where the phenomenon is rising almost always comes from this confusion of units.

$\sin(0.2)=0.198669$, so $\dfrac{\sin(0.2)}{0.2}=0.993347\approx 0.9933$. Then $\sin(0.05)=0.0499792$, so $\dfrac{\sin(0.05)}{0.05}=0.999583\approx 0.9996$. The trend is clear: dividing the input by 4 divided the GAP FROM 1 by about 16. That is the signature of convergence in $x^{2}$, and the approximation $\sin x\approx x-\dfrac{x^{3}}{6}$ confirms it, since it gives $\dfrac{\sin x}{x}\approx 1-\dfrac{x^{2}}{6}$. Careful: the calculator must be in radians, otherwise these two numbers are about $0.0175$ and nothing converges to 1.

$\displaystyle\lim_{x\to 0}\dfrac{\sin x}{x}=1$. On the figure, the line $y=x$ and the curve $y=\sin x$ visibly part ways from $x=1$ on, but stay glued together between $-0.3$ and $0.3$: saying that the quotient tends to 1 means that sine and its argument become proportional with a ratio of 1, in other words that $\sin x\approx x$ near zero. Geometrically, it is the arc merging with its chord as the angle gets small, and this is true ONLY if the angle is measured in radians, that is, by an arc length.

Multiply top and bottom by the conjugate $1+\cos x$: $\dfrac{1-\cos x}{x}=\dfrac{(1-\cos x)(1+\cos x)}{x(1+\cos x)}=\dfrac{1-\cos^{2}x}{x(1+\cos x)}=\dfrac{\sin^{2}x}{x(1+\cos x)}=\dfrac{\sin x}{x}\times\dfrac{\sin x}{1+\cos x}$. The first factor tends to 1, the second to $\dfrac{0}{2}=0$, so the product tends to 0. Check at $x=0.01$: $1-\cos(0.01)=0.0000499996$, divided by $0.01$ this gives $0.0050$. The conjugate method is the one from the chapter on algebraic limits; only the Pythagorean identity changes.

For the first, force the same argument on top and bottom: $\dfrac{\sin 5x}{3x}=\dfrac{5}{3}\times\dfrac{\sin 5x}{5x}$, and since $5x\to 0$, the second factor tends to 1: the limit is $\dfrac{5}{3}\approx 1.6667$. For the second, build two reference quotients: $\dfrac{\sin 5x}{\sin 3x}=\dfrac{\sin 5x}{5x}\times\dfrac{3x}{\sin 3x}\times\dfrac{5}{3}$, which gives $1\times 1\times\dfrac{5}{3}\approx 1.6667$. Remember the rule: when the two sines carry different coefficients, the limit is the RATIO of those coefficients, never 1. Answering 1 by reflex is the most common mistake on this page.

$0.1^{\circ}=0.1\times\dfrac{\pi}{180}=0.0017453$ rad, whose sine is $0.0017453$. The quotient is therefore $\dfrac{0.0017453}{0.1}=0.017453$, and the limit is $\dfrac{\pi}{180}\approx 0.01745$, not 1. Direct consequence: in degrees, the derivative of sine is no longer cosine but $\dfrac{\pi}{180}\cos x$, and this stray factor would contaminate every formula of the chapter, every chain rule and every optimization problem. The radian is not a teacher's convention: it is the only unit that keeps the table of derivatives clean.

Read directly: $A=8$, so the amplitude is 8 units. $\omega=\dfrac{\pi}{6}$, so the period is $T=\dfrac{2\pi}{\pi/6}=12$ time units, which makes it the standard annual model when $t$ counts months. The midline is the line $y=k=30$. The maximum is $k+A=38$, the minimum $k-A=22$. Check: the average of the two is indeed $\dfrac{38+22}{2}=30$, and their half-difference is $\dfrac{38-22}{2}=8$. These two formulas, $k=\dfrac{\max+\min}{2}$ and $A=\dfrac{\max-\min}{2}$, will be used in reverse in exercise 4, where the data supply the maximum and the minimum.

Factor $\omega$ out of the argument: $\dfrac{\pi t}{6}+\dfrac{\pi}{3}=\dfrac{\pi}{6}\left(t+2\right)$, so $g(t)=12\sin\left(\dfrac{\pi}{6}(t+2)\right)+40$ and $t_0=-2$. The shift is NEGATIVE, which means the cycle runs two months ahead of the model without a phase shift. The maximum is reached when the argument equals $\dfrac{\pi}{2}$: $\dfrac{\pi}{6}(t+2)=\dfrac{\pi}{2}$ gives $t+2=3$, so $t=1$, and $g(1)=52$. The pitfall is to read $t_0=+2$ because the statement shows a plus: the standard form is written $t-t_0$, so a plus in the argument means a negative $t_0$, exactly as for the parabola.

The tangent has period $\pi$, not $2\pi$: it repeats its values twice as fast as sine, because the simultaneous sign change of sine and cosine leaves their quotient unchanged. Above all, it has no amplitude: it starts at $-\infty$ and climbs to $+\infty$ on each interval, with a vertical asymptote at $\dfrac{\pi}{2}+k\pi$. A real seasonal phenomenon is bounded, sales do not become infinite in December, so the tangent is never used as a cyclical model. It appears in this course for another reason, as a result of differentiation and in the equations of the type $\tan u=c$ of exercise 10.

Secant is the reciprocal of cosine, $\sec x=\dfrac{1}{\cos x}$; cosecant is the reciprocal of sine, $\csc x=\dfrac{1}{\sin x}$; cotangent is the reciprocal of tangent, $\cot x=\dfrac{\cos x}{\sin x}$. Watch the vocabulary: the prefix co does not carry over, CO-secant answers to SINE. These three functions appear only in some course outlines and will not be used further here, where everything is written with the three basic functions.

$\dfrac{A}{k}$ measures the weight of the season relative to the usual level, and it is this proportion, not the raw amplitude, that a manager compares from one series to another. For $f$: $\dfrac{8}{30}=0.2667$, or $26.7$ percent. For $g$: $\dfrac{12}{40}=0.30$, or 30 percent. Series $g$ is therefore the more seasonal, even though both have the same period. Comparing the amplitudes alone, 8 against 12, would give the same conclusion here by chance; it would flip as soon as the average levels diverge, for instance between an amplitude of 12 on a level of 200 and an amplitude of 8 on a level of 30.

The chapter's two formulas are used here in reverse from exercise 3: $A=\dfrac{\max-\min}{2}=\dfrac{140-80}{2}=30$ and $k=\dfrac{\max+\min}{2}=\dfrac{140+80}{2}=110$, both in thousands of dollars. The pitfall is to take $A=140$, that is, to confuse the amplitude with the maximum value: the amplitude is measured from the AXIS, not from zero. This confusion returns in exercise 8, statement d), and it then throws the derivative off by a factor of 4 in this example.

The phenomenon is annual and the data are monthly, so the period is $T=12$ months. From this, $\omega=\dfrac{2\pi}{T}=\dfrac{2\pi}{12}=\dfrac{\pi}{6}\approx 0.5236$ radians per month. The units deserve a second look: $\omega$ is not a pure number, it is a number of radians PER MONTH, and that is what will make $A\omega$ a number of thousands of dollars per month in exercise 6. A student who writes $\omega=12$ has swapped the period and the angular frequency; the curve would then oscillate about twice a month.

The maximum of a sine is reached when its argument equals $\dfrac{\pi}{2}$ modulo $2\pi$. At $t=12$: $\dfrac{\pi}{6}\times 12+\varphi=2\pi+\varphi$, which we want equal to $\dfrac{\pi}{2}+2\pi$, so $\varphi=\dfrac{\pi}{2}\approx 1.5708$. The model is written $f(t)=30\sin\left(\dfrac{\pi t}{6}+\dfrac{\pi}{2}\right)+110$, which is also $f(t)=110+30\cos\left(\dfrac{\pi t}{6}\right)$, since a sine shifted by a quarter turn IS a cosine. Both forms are correct; keeping the first avoids switching functions in the middle of the chapter.