Appendix ASolutions to the Exercises

Solution

(a) The statement that $\sqrt2$ is irrational, that is, that it cannot be expressed as a fraction (a ratio of two integers), is an assertion that is clearly determined to be either correct or incorrect. Hence it is a proposition, and it is in fact true (the irrationality of $\sqrt2$ is a classical fact that can be proved by contradiction).

(b) “Difficult” is perceived differently from person to person, so it is not clearly determined whether the assertion is correct or incorrect. Hence it is not a proposition.

(c) “For every natural number $n$, $n^2+n$ is even” is an assertion whose truth or falsity is clearly determined, so it is a proposition. As for its truth value, we have the factorization $n^2+n=n(n+1)$. Since $n$ and $n+1$ are two consecutive integers, one of them is always even and the other is odd, and the product of an even number and an odd number is even. Therefore $n(n+1)=n^2+n$ is even for every natural number $n$, and this proposition is true.

Solution

If “$x$ is a multiple of 4,” then for some integer $m$ we can write $x=4m=2(2m)$, so $x$ is also a multiple of 2. Hence the proposition “$x$ is a multiple of 4 $\Rightarrow$ $x$ is a multiple of 2” is true, and “$x$ is a multiple of 4” is a sufficient condition for “$x$ is a multiple of 2.”

The converse proposition “$x$ is a multiple of 2 $\Rightarrow$ $x$ is a multiple of 4” is false. As a counterexample, take $x=6$. Since $6=2\times3$, it is a multiple of 2, but $6\div4=1$ with remainder $2$, so it is not a multiple of 4. Hence “$x$ is a multiple of 2” is not a necessary condition for “$x$ is a multiple of 4.” To summarize: “being a multiple of 4” is a sufficient condition but not a necessary condition for “being a multiple of 2,” and conversely “being a multiple of 2” is a necessary condition but not a sufficient condition for “being a multiple of 4.”

Solution

We factor $x^2-3x-4$. Looking for two integers whose product is $-4$ and whose sum is $-3$, we find $-4$ and $1$ ($-4\times1=-4$, $-4+1=-3$), so it factors as

$$ x^2-3x-4=(x-4)(x+1) $$

A product is $0$ only when one of the two factors is $0$, so

$$ x^2-3x-4=0 \iff (x-4)(x+1)=0 \iff x-4=0 \text{ or } x+1=0 \iff x=4 \text{ or } x=-1 $$

Every step of this chain of transformations holds in both directions ($\iff$), so the statements “$x^2-3x-4=0$” and “$x=4$ or $x=-1$” are equivalent.

Solution

The original proposition “$x\gt2 \Rightarrow x^2\gt4$”: If $x\gt2$, both sides are positive, so squaring preserves the inequality, and we obtain $x^2\gt2^2=4$. Hence it is true.

Converse: $x^2\gt4 \Rightarrow x\gt2$. This is false. A counterexample is $x=-3$: it satisfies $(-3)^2=9\gt4$, but it does not satisfy $-3\gt2$.

Inverse: $x\le2 \Rightarrow x^2\le4$. This is also false. A counterexample is again $x=-3$: it satisfies $-3\le2$, but it does not satisfy $(-3)^2=9\le4$.

Contrapositive: $x^2\le4 \Rightarrow x\le2$. This is true. From $x^2\le4$ we get $-2\le x\le2$ ($x^2\le4 \iff |x|\le2$), and this in particular includes $x\le2$. Since the original proposition is true, the contrapositive is also true (as discussed in Section 1.4 of the main text).

Solution

The contrapositive of the original proposition “if $n^2$ is a multiple of 3, then $n$ is a multiple of 3” is “if $n$ is not a multiple of 3, then $n^2$ is not a multiple of 3 either.” We prove this contrapositive.

When $n$ is not a multiple of 3, the remainder on dividing $n$ by 3 is either $1$ or $2$.

The case where the remainder is 1: For some integer $k$ we can write $n=3k+1$. Then

$$ n^2=(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1 $$

(here we expanded $(3k+1)^2$ and, using the fact that $9k^2+6k$ is a multiple of 3, factored out 3 to bring the expression into the form $3\times(\text{integer})+1$), so the remainder on dividing $n^2$ by 3 is $1$, not $0$. That is, $n^2$ is not a multiple of 3.

The case where the remainder is 2: For some integer $k$ we can write $n=3k+2$. Then

$$ n^2=(3k+2)^2=9k^2+12k+4=3(3k^2+4k+1)+1 $$

(we split $4$ into $3+1$ and then factored out 3: $9k^2+12k+4=9k^2+12k+3+1=3(3k^2+4k+1)+1$), so in this case too the remainder on dividing $n^2$ by 3 is $1$, not $0$. That is, $n^2$ is not a multiple of 3.

From the above, we have shown that when $n$ is not a multiple of 3 (that is, when the remainder is 1 or 2), $n^2$ is not a multiple of 3 either. This means that the contrapositive is true, so the original proposition “if $n^2$ is a multiple of 3, then $n$ is a multiple of 3” is also true.

∎

Solution

By definition, the truth set of condition $p$ is simply

$$ P=\{1,2,3,4,5\} $$

For condition $q$, “$x$ is a prime number that is at most 5,” we check which of the natural numbers at most 5, namely $1,2,3,4,5$, are prime numbers (integers greater than or equal to 2 that have no divisors other than 1 and themselves): $2,3,5$ are prime numbers, $1$ is not a prime number (it is excluded by the definition of a prime number), and $4=2\times2$ is not a prime number. Hence

$$ Q=\{2,3,5\} $$

Every element of $Q$ ($2,3,5$) is also contained in $P=\{1,2,3,4,5\}$, so $Q\subset P$ holds. On the other hand, $1$, an element of $P$, is not contained in $Q$ ($1$ is not a prime number), so $P\subset Q$ does not hold.

By Theorem 1.1 (Section 1.5 of the main text), $Q\subset P$ is the same as “$q\Rightarrow p$ is true,” that is, “$q$ is a sufficient condition for $p$.” Viewed from the side of $p$, this means that “$p$ is a necessary condition for $q$” (but not a sufficient condition; a counterexample is $x=1$, which satisfies $p$ but does not satisfy $q$).

Solution

Internal division point $P$. By Theorem 2.1, $BP:PC=AB:AC=5:8$. Also $BP+PC=BC=7$, and the ratio $5:8$ means “take $5$ parts and $8$ parts when $7$ is divided into $5+8=13$ equal parts,” so

$$ BP=7\times\frac{5}{13}=\frac{35}{13},\qquad PC=7\times\frac{8}{13}=\frac{56}{13} $$

(Check: $BP+PC=\frac{35}{13}+\frac{56}{13}=\frac{91}{13}=7=BC$, and $BP:PC=35:56=5:8=AB:AC$.)

External division point $Q$. Since $AB\neq AC$ ($5\neq8$), Theorem 2.2 applies, and $BQ:CQ=AB:AC=5:8$. Since $AB\lt AC$, the external division point $Q$ lies on the extension of side $BC$ beyond $B$ (the opposite end from the one in Example 2.2: $Q$ appears on the side of the shorter of the two sides through $A$—here, the side of $AB$). Using coordinates, let $B=0,C=7$, and let the position of $Q$ be $q$ ($q\lt0$). Then $BQ=-q$ and $CQ=7-q$, and we have

$$ \frac{BQ}{CQ}=\frac{-q}{7-q}=\frac{5}{8} $$

From this, $-8q=5(7-q)=35-5q$, that is, $-8q+5q=35$, $-3q=35$, $q=-\dfrac{35}{3}$. Hence

$$ BQ=-q=\frac{35}{3},\qquad CQ=7-q=7+\frac{35}{3}=\frac{56}{3} $$

(Check: $BQ:CQ=\frac{35}{3}:\frac{56}{3}=35:56=5:8=AB:AC$, and since $q=-\frac{35}{3}\lt0$, $Q$ indeed lies on the extension beyond $B$.)

Solution

In an equilateral triangle $ABC$ ($AB=BC=CA$), consider the median drawn from vertex $A$ to side $BC$ (the segment $AL$ going to the midpoint $L$ of $BC$). The triangles $\triangle ABL$ and $\triangle ACL$ are congruent, because their three pairs of sides are equal: $AB=AC$ (a property of an equilateral triangle), $BL=CL$ ($L$ is the midpoint), and the side $AL$ is common. Hence $\angle ALB=\angle ALC$, and since these two angles add up to $180^\circ$ at $L$ (a straight line), each of them is $90^\circ$. By the congruence we also have $\angle BAL=\angle CAL$.

That is, the median $AL$ from vertex $A$ is also, at the same time, (i) the altitude to side $BC$, because it is perpendicular to side $BC$; (ii) the perpendicular bisector of side $BC$, because $L$ is the midpoint of $BC$; and (iii) the angle bisector of $\angle A$, because $\angle BAL=\angle CAL$. In other words, in an equilateral triangle the four kinds of special lines—median, altitude, perpendicular bisector, and angle bisector—all coincide, as seen from vertex $A$, with one and the same line $AL$. The same thing happens for vertices $B,C$ as well.

Therefore, the centroid, the intersection of the three medians; the orthocenter, the intersection of the three altitudes; the circumcenter, the intersection of the perpendicular bisectors of the three sides; and the incenter, the intersection of the bisectors of the three angles, all refer to “the intersection of the same three lines (the lines corresponding to $AL,BM,CN$ from the respective vertices),” so all four points coincide at one and the same point.

Solution

We have $BM=\dfrac{BC}{2}=4$. By the median theorem (Apollonius's theorem; Theorem 2.4),

$$ AB^2+AC^2=2\bigl(AM^2+BM^2\bigr) $$ $$ 7^2+5^2=2\bigl(AM^2+4^2\bigr) $$ $$ 49+25=2\,AM^2+32 $$ $$ 74-32=2\,AM^2 $$ $$ AM^2=\frac{42}{2}=21 $$

Since $AM>0$, $AM=\sqrt{21}$ ($\sqrt{21}$ is not an integer, but this is the correct answer. $\sqrt{21}\approx4.58$).

Solution

The geometric difference. Ceva's theorem concerns the situation in which the $3$ segments $AP,BQ,CR$—each joining a vertex of the triangle to the opposite side (at an internal division point)—meet at a single point inside the triangle. As in Figure 2.11, $R,P,Q$ are normally internal division points of the three sides, respectively. Menelaus's theorem, in contrast, concerns the situation in which a single straight line that does not pass through any vertex of the triangle cuts the three sides (or rather, the lines containing them). As in Figure 2.12, this line crosses the boundary of the triangle, so the number of points among $R,P,Q$ that lie on an extension (external division points) is always odd ($1$ or $3$ of them), and never even ($0$ or $2$)—the line cannot cut all three sides at internal division points (if it could, the line would pass only through the interior of the triangle and could never get out, a contradiction).

Why the same mnemonic can be used for both. Both theorems assert an equation of the same form, $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$, and the content of this equation is built by the same rule: “going once around the triangle in the order $A\to R\to B$, $B\to P\to C$, $C\to Q\to A$, read the numerators and denominators alternately.” In Ceva's theorem the proof uses a chain of area ratios starting from the point $X$ (the point of intersection), and in Menelaus's theorem a chain of ratios of distances to the line $\ell$, so the contents of the two proofs are entirely different; nevertheless, both have the same structure: “when the $3$ ratios are multiplied in the order in which we go once around along the sides, they cancel exactly and give $1$.” That is precisely why at least the “going-around order” used for memorization can be shared by both.

Solution

The medians satisfy Ceva's relation. In the case of medians, $R,P,Q$ are the midpoints of the sides $AB,BC,CA$, respectively, so $AR=RB$, $BP=PC$, $CQ=QA$, and we have

$$ \frac{AR}{RB}=1,\qquad \frac{BP}{PC}=1,\qquad \frac{CQ}{QA}=1 $$

Therefore

$$ \frac{AR}{RB}\times\frac{BP}{PC}\times\frac{CQ}{QA}=1\times1\times1=1 $$

and the relation of Ceva's theorem is satisfied.

An alternative proof of Theorem 2.3. As stated in the Math Note in Section 2.4.1, the converse of Ceva's theorem also holds: if the relation $\dfrac{AR}{RB}\times\dfrac{BP}{PC}\times\dfrac{CQ}{QA}=1$ holds, then $AP,BQ,CR$ meet at a single point. As we have just shown, the medians $AL,BM,CN$ ($R,P,Q$ of this exercise correspond to $N,L,M$, respectively) satisfy this relation, so by the converse of Ceva's theorem, the three lines $AL,BM,CN$ meet at a single point. This is exactly the statement of Theorem 2.3 (the three medians meet at a single point, namely the centroid); we have reached the same conclusion by a route different from the proof in Section 2.2.1 that used the midpoint theorem.

Solution

Circumradius $R$. As shown in Example 2.6, in a right triangle the circumcenter is the midpoint of the hypotenuse, and the circumradius equals half the hypotenuse. The hypotenuse is $AB=13$, so

$$ R=\frac{AB}{2}=\frac{13}{2} $$

This is the circumradius.

Inradius $r$. Since $\angle C=90^\circ$, the area is $S=\dfrac12\times CA\times CB=\dfrac12\times5\times12=30$. The semiperimeter is $s=\dfrac{5+12+13}{2}=15$. Using the same formula $r=\dfrac{S}{s}$ as in Example 2.7, we get

$$ r=\frac{30}{15}=2 $$

This is the inradius.

Solution

By Theorem 3.1 ($\sin^2\theta+\cos^2\theta=1$),

$$ \cos^2\theta=1-\sin^2\theta=1-\left(\frac13\right)^2=1-\frac19=\frac89 $$

Since $\theta$ is acute, $\cos\theta\gt0$. Hence

$$ \cos\theta=\sqrt{\frac89}=\frac{2\sqrt2}{3} $$

(We used $\sqrt8=2\sqrt2$ and $\sqrt9=3$.) Next, by Theorem 3.2 ($\tan\theta=\sin\theta/\cos\theta$),

$$ \tan\theta=\frac{\sin\theta}{\cos\theta}=\frac{1/3}{2\sqrt2/3}=\frac{1}{2\sqrt2}=\frac{\sqrt2}{4} $$

(We rationalized the denominator: $\dfrac{1}{2\sqrt2}=\dfrac{1\times\sqrt2}{2\sqrt2\times\sqrt2}=\dfrac{\sqrt2}{4}$.) Finally, we check that Theorem 3.3 holds. The left-hand side is

$$ 1+\tan^2\theta=1+\left(\frac{\sqrt2}{4}\right)^2=1+\frac{2}{16}=1+\frac18=\frac98 $$

and the right-hand side is

$$ \frac{1}{\cos^2\theta}=\frac{1}{8/9}=\frac98 $$

Both sides equal $\dfrac98$, so they agree; hence $1+\tan^2\theta=1/\cos^2\theta$ indeed holds.

Solution

Substituting $a=10$ and $A=30°$ into the law of sines $\dfrac{a}{\sin A}=2R$ gives

$$ 2R=\frac{10}{\sin30°}=\frac{10}{1/2}=20 \quad\Longrightarrow\quad R=10 $$

Next, substituting $R=10$ and $B=45°$ into $b=2R\sin B$ (the law of sines $\dfrac{b}{\sin B}=2R$ solved for $b$) gives

$$ b=2\times10\times\sin45°=20\times\frac{\sqrt2}{2}=10\sqrt2 $$

This is the required length ($10\sqrt2\approx14.14$).

Solution

Of the three sides $a=7,b=8,c=5$, the longest is $b=8$. The largest angle of a triangle is the one opposite the longest side, so $\angle B$ is the largest angle, and it suffices to determine whether it is acute or obtuse. The law of cosines solved for $B$ is

$$ \cos B=\frac{c^2+a^2-b^2}{2ca} $$

and substituting the numerical values gives

$$ \cos B=\frac{5^2+7^2-8^2}{2\times5\times7}=\frac{25+49-64}{70}=\frac{10}{70}=\frac17 $$

Since $\cos B=\dfrac17\gt0$, $\angle B$ is acute. Because the largest angle is acute, the other two angles ($A,C$) are smaller than it, and therefore all the angles are acute. Hence this triangle is an acute triangle.

Solution

First, substitute $a=6,b=4,C=60°$ into the law of cosines $c^2=a^2+b^2-2ab\cos C$.

$$ c^2=6^2+4^2-2\times6\times4\times\cos60°=36+16-48\times\frac12=52-24=28 $$

Since $c\gt0$, $c=\sqrt{28}=2\sqrt7$ (because $28=4\times7$, we have $\sqrt{28}=2\sqrt7$).

Next, solving the law of sines $\dfrac{a}{\sin A}=\dfrac{c}{\sin C}$ for $\sin A$ gives $\sin A=\dfrac{a\sin C}{c}$. Using $\sin60°=\dfrac{\sqrt3}{2}$ and substituting the numerical values, we obtain

$$ \sin A=\frac{6\times\dfrac{\sqrt3}{2}}{2\sqrt7}=\frac{3\sqrt3}{2\sqrt7}=\frac{3\sqrt3\times\sqrt7}{2\sqrt7\times\sqrt7}=\frac{3\sqrt{21}}{14} $$

(We rationalized the denominator.) Hence $\sin A=\dfrac{3\sqrt{21}}{14}$ ($\approx0.982$).

Solution

Substituting $l=4\pi$ and $r=6$ into the arc-length formula $l=r\theta$ gives

$$ 4\pi=6\theta \quad\Longrightarrow\quad \theta=\frac{4\pi}{6}=\frac{2\pi}{3}\ (\mathrm{rad}) $$

To convert to degree measure, we use the formula in Chapter 3 ($180°=\pi\,\mathrm{rad}$, that is, $1\,\mathrm{rad}=180°/\pi$).

$$ \frac{2\pi}{3}\,\mathrm{rad}=\frac{2\pi}{3}\times\frac{180°}{\pi}=\frac{2}{3}\times180°=120° $$

(The $\pi$ cancels.) Hence the central angle is $\dfrac{2\pi}{3}\,\mathrm{rad}=120°$. By the formula $S=\dfrac12 r^2\theta$, the area of the sector is computed as follows.

$$ S=\frac12\times6^2\times\frac{2\pi}{3}=\frac12\times36\times\frac{2\pi}{3}=18\times\frac{2\pi}{3}=12\pi $$

This is the area of the sector.

Solution

Substitute $A=90°$ into the law of cosines $a^2=b^2+c^2-2bc\cos A$. By Definition 3.2 (the definition of general angles via the unit circle), $\cos90°=0$ (the point on the unit circle corresponding to $90°$ is $(0,1)$, and its x-coordinate is $\cos90°$). Hence, leaving $b,c$ as symbols in the substitution, we obtain

$$ a^2=b^2+c^2-2bc\cos90°=b^2+c^2-2bc\times0=b^2+c^2-0=b^2+c^2 $$

Since $b,c$ remain arbitrary symbols (standing for the lengths of the two sides enclosing $\angle A=90°$), this shows that the Pythagorean theorem $a^2=b^2+c^2$ holds not only for particular numerical values but for every right triangle with $\angle A=90°$.

Solution

In Formula 4.1 (the formula in Chapter 4), with $a=-4,b=6,m=3,n=2$, the internal division point $P$ is

$$ x = \frac{na+mb}{m+n} = \frac{2\times(-4)+3\times6}{3+2} = \frac{-8+18}{5} = \frac{10}{5} = 2 $$

Hence $P(2)$. By Formula 4.2 (the formula in Chapter 4), the external division point $Q$ (it can be defined because $m=3\neq n=2$) is

$$ x = \frac{-na+mb}{m-n} = \frac{-2\times(-4)+3\times6}{3-2} = \frac{8+18}{1} = 26 $$

Hence $Q(26)$.

Solution

In Formula 4.3 (the formula in Chapter 4), let $x_1=2,y_1=5,x_2=8,y_2=-1,m=2,n=3$. The internal division point $P$ is

$$ P\left(\frac{3\times2+2\times8}{2+3},\ \frac{3\times5+2\times(-1)}{2+3}\right) = \left(\frac{6+16}{5},\ \frac{15-2}{5}\right) = \left(\frac{22}{5},\ \frac{13}{5}\right) $$

The external division point $Q$ (it can be defined because $m=2\neq n=3$; the denominator is $m-n=-1$) is

$$ Q\left(\frac{-3\times2+2\times8}{2-3},\ \frac{-3\times5+2\times(-1)}{2-3}\right) = \left(\frac{-6+16}{-1},\ \frac{-15-2}{-1}\right) = \left(\frac{10}{-1},\ \frac{-17}{-1}\right) = (-10,\ 17) $$

Hence the internal division point is $\left(\dfrac{22}{5},\dfrac{13}{5}\right)$ and the external division point is $(-10,17)$.

Solution

(i) By Formula 4.4 (the formula in Chapter 4),

$$ G\left(\frac{2+5+(-4)}{3},\ \frac{-1+4+3}{3}\right) = \left(\frac{3}{3},\ \frac{6}{3}\right) = (1,\ 2) $$

(ii) We check this by way of the midpoint $N$ of side $CA$. We have $N=\left(\dfrac{-4+2}{2},\dfrac{3+(-1)}{2}\right)=(-1,\ 1)$. The centroid $G$ divides the median $BN$ internally in the ratio $BG:GN=2:1$, so by Formula 4.3 (taking the $1$st point to be $B$ and the $2$nd point to be $N$, with $m=2,n=1$),

$$ G\left(\frac{1\times5+2\times(-1)}{2+1},\ \frac{1\times4+2\times1}{2+1}\right) = \left(\frac{5-2}{3},\ \frac{4+2}{3}\right) = \left(\frac{3}{3},\ \frac{6}{3}\right) = (1,\ 2) $$

Both (i) and (ii) give $G(1,2)$, so the two results agree.

Solution

In Formula 4.6 (the formula in Chapter 4), with $x_1=0,y_1=0,a=5,b=-12,c=26$, we have

$$ d = \frac{|5\times0-12\times0+26|}{\sqrt{5^2+(-12)^2}} = \frac{|26|}{\sqrt{25+144}} = \frac{26}{\sqrt{169}} = \frac{26}{13} = 2 $$

Hence the distance is $2$ ($5$-$12$-$13$ is, along with $3$-$4$-$5$, a well-known triple of integers that are the side lengths of a right triangle).

Solution

In Formula 4.6 (the formula in Chapter 4), let $a=3,b=4,c=-12$. The distance from the point $(1,1)$ is

$$ d_1 = \frac{|3\times1+4\times1-12|}{\sqrt{3^2+4^2}} = \frac{|3+4-12|}{\sqrt{9+16}} = \frac{|-5|}{5} = \frac{5}{5} = 1 $$

The distance from the origin $(0,0)$ is

$$ d_2 = \frac{|3\times0+4\times0-12|}{\sqrt{3^2+4^2}} = \frac{|-12|}{5} = \frac{12}{5} = 2.4 $$

Since $d_1=1$ and $d_2=2.4$, we have $d_1\lt d_2$, so the point $(1,1)$ is closer to the line $3x+4y-12=0$. (Remark: $3\times1+4\times1-12=-5\lt0$ and $3\times0+4\times0-12=-12\lt0$, so both points lie on the same side of the line, namely the side containing the origin.)

Solution

In Formula 4.2 (the formula in Chapter 4), with $a=1,b=5,m=1,n=1$, we have

$$ x = \frac{-na+mb}{m-n} = \frac{-1\times1+1\times5}{1-1} = \frac{4}{0} $$

Because the denominator is $0$, the value of $x$ is not determined (this computation is meaningless). That is, Formula 4.2 cannot be used when $m=n$.

Geometrically, this can be explained as follows. To divide externally in the ratio $m:n=1:1$ means to look for “a point outside the segment $AB$ that is equidistant from the points $A,B$.” But on the number line, the only point equidistant from the points $A,B$ is the exact middle of $A$ and $B$, that is, the midpoint, and it is a point inside the segment $AB$. There is no point outside the segment that is equidistant from the points $A,B$ (a point outside is always closer to one of $A,B$). Therefore, when $m=n$ the external division point does not exist, and the fact that the denominator of Formula 4.2 is $0$ is simply this fact of non-existence reflected directly in the formula.

Solution

Since $\alpha$ is acute and $\sin\alpha=\dfrac35$, we have $\cos\alpha>0$, and from $\cos^2\alpha=1-\sin^2\alpha=1-\dfrac{9}{25}=\dfrac{16}{25}$ we get $\cos\alpha=\dfrac45$. Similarly, since $\beta$ is acute and $\cos\beta=\dfrac{5}{13}$, we have $\sin\beta>0$, and from $\sin^2\beta=1-\cos^2\beta=1-\dfrac{25}{169}=\dfrac{144}{169}$ we get $\sin\beta=\dfrac{12}{13}$.

Substituting into Formula 5.3 (the addition formula for sine), we obtain

$$ \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta=\frac35\cdot\frac{5}{13}+\frac45\cdot\frac{12}{13}=\frac{15}{65}+\frac{48}{65}=\frac{63}{65} . $$

Substituting into Formula 5.2 (the addition formula for cosine), we obtain

$$ \cos(\alpha-\beta)=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\frac45\cdot\frac{5}{13}+\frac35\cdot\frac{12}{13}=\frac{20}{65}+\frac{36}{65}=\frac{56}{65} . $$

Hence $\sin(\alpha+\beta)=\dfrac{63}{65}$ and $\cos(\alpha-\beta)=\dfrac{56}{65}$.

Solution

Writing $22.5^\circ=\dfrac{45^\circ}{2}$, substitute $\alpha=45^\circ$ into Formula 5.12 (the half-angle formula for the square of the tangent). Since $\cos45^\circ=\dfrac{\sqrt2}{2}$,

$$ \tan^2 22.5^\circ=\frac{1-\cos45^\circ}{1+\cos45^\circ}=\frac{1-\dfrac{\sqrt2}{2}}{1+\dfrac{\sqrt2}{2}}=\frac{2-\sqrt2}{2+\sqrt2} . $$

To rationalize the denominator, multiply the numerator and the denominator by $2-\sqrt2$.

$$ \tan^2 22.5^\circ=\frac{(2-\sqrt2)^2}{(2+\sqrt2)(2-\sqrt2)}=\frac{4-4\sqrt2+2}{4-2}=\frac{6-4\sqrt2}{2}=3-2\sqrt2 . $$

Here we notice that $3-2\sqrt2=2-2\sqrt2+1=(\sqrt2-1)^2$ (indeed, $(\sqrt2-1)^2=2-2\sqrt2+1$). Since $0^\circ\lt22.5^\circ\lt90^\circ$ (first quadrant), we have $\tan22.5^\circ>0$, so taking the positive square root gives

$$ \tan22.5^\circ=\sqrt{(\sqrt2-1)^2}=\sqrt2-1\approx0.4142 . $$

Solution

Writing $3\theta=2\theta+\theta$, we use Formula 5.1 (the addition formula for cosine).

$$ \cos3\theta=\cos(2\theta+\theta)=\cos2\theta\cos\theta-\sin2\theta\sin\theta . $$

Since our goal is an expression in $\cos\theta$ alone, we substitute $\cos2\theta=2\cos^2\theta-1$ (the form of Formula 5.8 written with $\cos\theta$ alone) and $\sin2\theta=2\sin\theta\cos\theta$ (Formula 5.7).

$$ \cos3\theta=(2\cos^2\theta-1)\cos\theta-(2\sin\theta\cos\theta)\sin\theta=2\cos^3\theta-\cos\theta-2\sin^2\theta\cos\theta . $$

A $\sin^2\theta$ still remains, so we eliminate it by substituting the basic identity $\sin^2\theta=1-\cos^2\theta$.

$$ \begin{aligned} \cos3\theta&=2\cos^3\theta-\cos\theta-2(1-\cos^2\theta)\cos\theta\\ &=2\cos^3\theta-\cos\theta-2\cos\theta+2\cos^3\theta\\ &=4\cos^3\theta-3\cos\theta=-3\cos\theta+4\cos^3\theta . \end{aligned} $$

From the first line to the second, we expanded using the distributive law ($-2(1-\cos^2\theta)\cos\theta=-2\cos\theta+2\cos^3\theta$); from the second line to the third, we combined the $\cos^3\theta$ terms ($2\cos^3\theta+2\cos^3\theta=4\cos^3\theta$) and the $\cos\theta$ terms ($-\cos\theta-2\cos\theta=-3\cos\theta$) separately. We have therefore shown that

$$ \cos3\theta=-3\cos\theta+4\cos^3\theta $$

This agrees with Formula 5.14.

Solution

Let $A=75^\circ,B=15^\circ$ and substitute into Formula 5.21 (the sum-to-product formula for the sum of cosines). Since $\dfrac{A+B}{2}=\dfrac{90^\circ}{2}=45^\circ$ and $\dfrac{A-B}{2}=\dfrac{60^\circ}{2}=30^\circ$,

$$ \cos75^\circ+\cos15^\circ=2\cos45^\circ\cos30^\circ=2\cdot\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}=\frac{\sqrt6}{2}\approx1.2247 . $$

Solution

$2\sin\theta-2\cos\theta$ is the case $a=2,b=-2$. Then $r=\sqrt{a^2+b^2}=\sqrt{4+4}=\sqrt8=2\sqrt2$.

$$ \cos\alpha=\frac{a}{r}=\frac{2}{2\sqrt2}=\frac{\sqrt2}{2},\qquad \sin\alpha=\frac{b}{r}=\frac{-2}{2\sqrt2}=-\frac{\sqrt2}{2} . $$

Since $\cos\alpha>0,\sin\alpha\lt0$, the angle $\alpha$ lies in the fourth quadrant, and the angle satisfying these conditions is $\alpha=-\dfrac{\pi}{4}$. Hence, by Formula 5.23,

$$ 2\sin\theta-2\cos\theta=2\sqrt2\,\sin\left(\theta-\frac{\pi}{4}\right) . $$

Since $-1\le\sin\left(\theta-\dfrac{\pi}{4}\right)\le1$, the maximum value is $2\sqrt2$ and the minimum value is $-2\sqrt2$.

The maximum value $2\sqrt2$ is attained when $\theta-\dfrac{\pi}{4}=\dfrac{\pi}{2}$, that is, $\theta=\dfrac{3\pi}{4}$. The minimum value $-2\sqrt2$ is attained when $\theta-\dfrac{\pi}{4}=-\dfrac{\pi}{2}$, that is, $\theta=-\dfrac{\pi}{4}$ (both lie within the range $-\pi\le\theta\lt\pi$).

Solution

Substitute $\alpha=15^\circ,\beta=75^\circ$ into Formula 5.18 (the product-to-sum formula for $\sin\times\sin$).

$$ \sin15^\circ\sin75^\circ=-\frac12\{\cos(15^\circ+75^\circ)-\cos(15^\circ-75^\circ)\}=-\frac12\{\cos90^\circ-\cos(-60^\circ)\} . $$

Since $\cos90^\circ=0$ and $\cos(-60^\circ)=\cos60^\circ=\dfrac12$ (cosine is an even function),

$$ \sin15^\circ\sin75^\circ=-\frac12\left(0-\frac12\right)=-\frac12\cdot\left(-\frac12\right)=\frac14 . $$

Solution

Setting $a=3,\ b=5,\ c=10$ in Formula 6.1 (main text), we get $\log_35=\dfrac{\log_{10}5}{\log_{10}3}$. Substituting the given approximate values,

$$ \log_35 = \frac{\log_{10}5}{\log_{10}3} = \frac{0.6990}{0.4771} = 1.465\cdots $$

Hence $\log_35\approx1.465$.

Solution

Since $3^2=9$, Definition 6.1 (main text) gives $\log_39=2$. Setting $a=3,\ b=9$ in Formula 6.2 (main text), $\log_ab=\dfrac1{\log_ba}$, we get

$$ \log_93 = \frac{1}{\log_39} = \frac12 $$

Indeed, $9^{1/2}=\sqrt9=3$, so this is correct.

Solution

(a) We find the $x$ satisfying $8^x=16$. Since $8=2^3,\ 16=2^4$, we have $8^x=2^{3x}=16=2^4$, so $3x=4$ and $x=\dfrac43$. Hence $\log_816=\dfrac43$.

(b) Since $8=2^3,\ 16=2^4$, setting $a=2,\ m=3,\ b=2,\ n=4$ in Formula 6.3 (main text), $\log_{a^m}b^n=\dfrac nm\log_ab$, we get

$$ \log_816 = \log_{2^3}2^4 = \frac43\log_22 = \frac43\times1 = \frac43 $$

(We used $\log_22=1$.) Parts (a) and (b) agree. As a check, we may also verify that $8^{4/3}=(2^3)^{4/3}=2^4=16$.

Solution

Setting $a=2,\ b=7$ in Formula 6.4 (main text), $a^{\log_ab}=b$, we immediately get

$$ 2^{\log_27} = 7 $$

By definition, $\log_27$ is precisely the number that tells us “to what power $2$ must be raised to get $7$,” so if we raise $2$ to that very power once more, we naturally get back to $7$. This is why the answer is reasonable.

Solution

First we find the values of the logarithms we need. Since $9=3^2,\ 27=3^3$, Formula 6.3 (main text) gives

$$ \log_927 = \log_{3^2}3^3 = \frac32\log_33 = \frac32 $$

Also, $\log_93$ has the value $\log_93=\dfrac12$, as found in Exercise 6.2 (this can also be checked directly by applying Formula 6.3 to $9=3^2,\ 3=3^1$: $\log_93=\log_{3^2}3^1=\dfrac12\log_33=\dfrac12$). Using these values,

$$ a^{\log_cb} = 3^{\log_927} = 3^{3/2} = 3\sqrt3 \approx 5.196, \qquad b^{\log_ca} = 27^{\log_93} = 27^{1/2} = \sqrt{27} = 3\sqrt3 \approx 5.196 $$

Both sides are equal to $3\sqrt3$, so they agree, and this confirms that Formula 6.5 holds.

Solution

We use Formula 6.2 (main text), $\log_ab=\dfrac1{\log_ba}$, with the roles of $a$ and $b$ interchanged. (Formula 6.2 holds under the assumptions $a>0,a\neq1,b>0,b\neq1$, which are symmetric in $a,b$, so it holds in the same form after interchanging $a\leftrightarrow b$.) This gives

$$ \log_ba = \frac{1}{\log_ab} $$

Substituting this into $\log_ab\cdot\log_ba$, we obtain

$$ \log_ab\cdot\log_ba = \log_ab\cdot\frac{1}{\log_ab} = 1 $$

(That $\log_ab\neq0$ is guaranteed by $b\neq1$: if $\log_ab=0$, then $a^0=1=b$, which contradicts $b\neq1$.) Hence we have shown that $\log_ab\cdot\log_ba=1$.

∎

Solution

The midpoint $M$ of side $BC$ is the point that divides segment $BC$ internally in the ratio $1:1$, so by Formula 7.1 (position vector of an internal division point, with $m=n=1$),

$$ \vec{OM}=\frac{1\cdot\vec{OB}+1\cdot\vec{OC}}{1+1}=\frac{\vec{OB}+\vec{OC}}{2} $$

Next, since the centroid $G$ is the point that divides the median $AM$ internally in the ratio $2:1$, we substitute $A,M$ into Formula 7.1 (with $m=2,n=1$).

$$ \vec{OG}=\frac{1\cdot\vec{OA}+2\cdot\vec{OM}}{2+1}=\frac{\vec{OA}+2\vec{OM}}{3} $$

Substituting the value $\vec{OM}=\dfrac{\vec{OB}+\vec{OC}}{2}$ found above, we have $2\vec{OM}=\vec{OB}+\vec{OC}$, and so

$$ \vec{OG}=\frac{\vec{OA}+(\vec{OB}+\vec{OC})}{3}=\frac{\vec{OA}+\vec{OB}+\vec{OC}}{3} $$

and we have obtained the formula to be shown. This result also shows that the same formula is obtained whichever median is used to find the centroid (the formula is symmetric in $A,B,C$), which also supports the fact that the three medians really do meet at a single point (the centroid).

∎

Solution

When $\vec a,\vec b$ are linearly independent, Theorem 7.5 (the fundamental theorem for plane vectors) says that the way of expressing a vector as a linear combination of $\vec a,\vec b$ is unique. The given equality

$$ (2x-1)\vec a+(y+2)\vec b=3\vec a+(-1)\vec b $$

can be regarded as expressing the same vector (the right-hand side) as a linear combination of $\vec a,\vec b$ in two ways. By the uniqueness of the decomposition, the coefficients of $\vec a$ must be equal to each other, and the coefficients of $\vec b$ must be equal to each other.

$$ 2x-1=3, \qquad y+2=-1 $$

Solving these: from $2x-1=3$ we get $2x=4$, so $x=2$; from $y+2=-1$ we get $y=-3$.

Therefore $x=2$, $y=-3$.

Solution

(a) For $A(-1,3)$, $B(4,-2)$, by the idea behind Formula 7.2,

$$ \vec{AB}=(4-(-1),\ -2-3)=(5,-5) $$

The magnitude is, by the Pythagorean theorem,

$$ |\vec{AB}|=\sqrt{5^2+(-5)^2}=\sqrt{25+25}=\sqrt{50}=5\sqrt2 $$

($\sqrt{50}=\sqrt{25\times2}=\sqrt{25}\sqrt2=5\sqrt2$.)

(b) By Formula 7.1, the internal division point ($m=3,n=2$) is $\dfrac{2\vec a+3\vec b}{5}$. Computing in components,

$$ \frac{2(-1,3)+3(4,-2)}{5}=\frac{(-2,6)+(12,-6)}{5}=\frac{(10,0)}{5}=(2,0) $$

Hence the internal division point is $(2,0)$.

(c) By Formula 7.1, the external division point ($m=3,n=2$) is $\dfrac{-2\vec a+3\vec b}{3-2}=-2\vec a+3\vec b$.

$$ -2(-1,3)+3(4,-2)=(2,-6)+(12,-6)=(14,-12) $$

Hence the external division point is $(14,-12)$.

Solution

By Theorem 7.6 (perpendicularity and the inner product), $\vec a\perp\vec b\iff\vec a\cdot\vec b=0$. Computing $\vec a\cdot\vec b$ in components by Formula 7.3 (I),

$$ \vec a\cdot\vec b=2\times3+k\times(-4)=6-4k $$

The condition for this to be $0$ is

$$ 6-4k=0 \quad\Longrightarrow\quad k=\frac64=\frac32 $$

In this case $\vec a=\left(2,\dfrac32\right)$, so by Theorem 7.7 (IV) ($|\vec a|^2=\vec a\cdot\vec a$),

$$ |\vec a|=\sqrt{2^2+\left(\frac32\right)^2}=\sqrt{4+\frac94}=\sqrt{\frac{16}4+\frac94}=\sqrt{\frac{25}4}=\frac52 $$

(We rewrote $4=\frac{16}{4}$ with a common denominator before adding.) Therefore $k=\dfrac32$ and $|\vec a|=\dfrac52$.

Solution

A direction vector is $\vec{AB}=(4-1,\ 6-2)=(3,4)$. One vector perpendicular to $(p,q)$ is $(q,-p)$ (indeed, $(p,q)\cdot(q,-p)=pq-qp=0$), so we can take $\vec n=(4,-3)$ as a normal vector perpendicular to $\vec{AB}=(3,4)$.

Substituting the point $A(1,2)$ and $\vec n=(4,-3)$ into Formula 7.5, we get

$$ 4(x-1)-3(y-2)=0 $$

Expanding the parentheses and simplifying (here $4(x-1)=4x-4$, $-3(y-2)=-3y+6$, and the constant term is $-4+6=2$), we get

$$ 4x-4-3y+6=0 \quad\Longrightarrow\quad 4x-3y+2=0 $$

As a check, substituting the point $B(4,6)$ into the left-hand side gives $4\times4-3\times6+2=16-18+2=0$, so $B$ indeed lies on this line as well. Therefore, the required equation of the line is $4x-3y+2=0$.

Solution

Line. Substituting the point $A(2,-3,1)$ and the direction vector $\vec d=(1,2,-2)$ into Formula 7.7, we get

$$ \frac{x-2}{1}=\frac{y-(-3)}{2}=\frac{z-1}{-2}, \qquad\text{that is}\qquad \frac{x-2}{1}=\frac{y+3}{2}=\frac{z-1}{-2} $$

Plane. The plane passing through the same point $A(2,-3,1)$ with normal vector $\vec d=(1,2,-2)$ is, by Formula 7.6,

$$ 1\cdot(x-2)+2\cdot(y-(-3))+(-2)\cdot(z-1)=0 $$

Expanding the parentheses and simplifying (we have $x-2$, $2(y+3)=2y+6$, $-2(z-1)=-2z+2$, and the constant term is $-2+6+2=6$), we get

$$ x-2+2y+6-2z+2=0 \quad\Longrightarrow\quad x+2y-2z+6=0 $$

Sphere. Substituting the center $(0,1,-2)$ and the radius $r=3$ into Formula 7.8, we get

$$ (x-0)^2+(y-1)^2+(z-(-2))^2=3^2, \qquad\text{that is}\qquad x^2+(y-1)^2+(z+2)^2=9 $$

Solution

The focus $(0,3)$ lies on the $y$-axis, so the equation has the form $x^2=4py$ of Section 8.1.2 of the main text. From the focus $(0,p)=(0,3)$ we get $p=3$. Hence the equation of the parabola is

$$ x^2=4\times3\times y=12y $$

The directrix is $y=-p=-3$.

Solution

Comparing with the standard form $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, we get $a^2=16,b^2=7$. Since $16\gt7$, the major axis lies along the $x$-axis, and $a=\sqrt{16}=4$, $b=\sqrt7$. By the formula $c=\sqrt{a^2-b^2}$,

$$ c=\sqrt{a^2-b^2}=\sqrt{16-7}=\sqrt9=3 $$

so the foci are $(\pm3,0)$. The length of the major axis is $2a=8$, and the length of the minor axis is $2b=2\sqrt7$.

Solution

Comparing with the standard form $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$, we get $a^2=9$, that is, $a=3$. By Definition 8.3 (and the starting point of the derivation, $PF-PF'=\pm2a$), for any point $P$ on the hyperbola the absolute value of the difference of the distances is always equal to $2a$, so

$$ |PF-PF'|=2a=2\times3=6 $$

As in the case of the ellipse in Example 8.4, the answer follows from the definition alone, without finding the coordinates of $P$ explicitly.

Solution

By the definition, $\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a$. Moving one of the radicals to the other side, we get

$$ \sqrt{(x+c)^2+y^2}=2a-\sqrt{(x-c)^2+y^2} $$

Next, we square both sides (both sides are nonnegative, so this is an equivalent transformation). Expanding the right-hand side in the form $(A-B)^2=A^2-2AB+B^2$, we get

$$ (x+c)^2+y^2=4a^2-4a\sqrt{(x-c)^2+y^2}+(x-c)^2+y^2 $$

Canceling $y^2$ on both sides and simplifying with $(x+c)^2-(x-c)^2=4cx$ (which follows by expanding), we obtain

$$ 4cx=4a^2-4a\sqrt{(x-c)^2+y^2} $$

Dividing both sides by $4$ and leaving only the radical term on the left-hand side, we get

$$ a\sqrt{(x-c)^2+y^2}=a^2-cx $$

Squaring both sides once more gives

$$ a^2\{(x-c)^2+y^2\}=a^4-2a^2cx+c^2x^2 $$

Expanding the left-hand side gives $a^2x^2-2a^2cx+a^2c^2+a^2y^2$. Canceling the term $-2a^2cx$ common to both sides and simplifying, we obtain

$$ (a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2) $$

Since $a\gt c\gt0$, we have $a^2-c^2\gt0$, so we can let $b=\sqrt{a^2-c^2}$ (so $b^2=a^2-c^2$). Substituting this gives $b^2x^2+a^2y^2=a^2b^2$, and dividing both sides by $a^2b^2$ ($\neq0$) yields

$$ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$

This is exactly the same result as in Section 8.2.2 of the main text.

∎

Solution

The difference in sign comes from the first stage of the derivation, the step where we compute the difference between the two squares $(x\pm c)^2$. For the ellipse, we start from the sum of the distances to the foci, $\sqrt{(x+c)^2+y^2}+\sqrt{(x-c)^2+y^2}=2a$, and along the way we use the plus relation $(x+c)^2-(x-c)^2=+4cx$. For the hyperbola, on the other hand, we start from the difference of the distances, $\sqrt{(x-c)^2+y^2}-\sqrt{(x+c)^2+y^2}=\pm2a$, so in the course of the manipulation the minus relation $(x-c)^2-(x+c)^2=-4cx$ appears.

This reversal of sign is carried over to the sign of the coefficient of $x^2$ in the final equation. For the ellipse we get $(a^2-c^2)x^2+a^2y^2=a^2(a^2-c^2)$ (since $a\gt c$, we have $a^2-c^2\gt0$, and the coefficients of $x^2,y^2$ have the same sign), and for the hyperbola we get $(c^2-a^2)x^2-a^2y^2=a^2(c^2-a^2)$ (since $c\gt a$, we have $c^2-a^2\gt0$, but the coefficient of $y^2$ remains $-a^2$). This is why, after substituting $b^2$, the standard forms appear with opposite signs, as $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. The root cause lies in the difference between the definitions themselves: whether it is the “sum of distances” or the “difference of distances.”

Solution

Substituting $y=x$ into $y^2=4px$, we get

$$ x^2=4px \quad\Longrightarrow\quad x^2-4px=0 \quad\Longrightarrow\quad x(x-4p)=0 $$

(we factored $x$ out of the left-hand side). Hence $x=0$ or $x=4p$. Since $y=x$, the intersection points are

$$ (0,0),\quad(4p,4p) $$

That is, there are two intersection points (since $p\neq0$, these two points are distinct).