Trigonometry: properties of a triangle (sine, cosine, tangent, projection, half-angle laws); solution of triangles

(b) $\cos A$

By the cosine rule, $a^2=b^2+c^2-2bc\cos A$, so $\dfrac{b^2+c^2-a^2}{2bc}=\cos A$.

(a) $2\sin A\cos A$

This is the standard double-angle formula for sine. The other three options are all equal to $\cos 2A$ (since $\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A$), not $\sin 2A$.

a. Let the angles of $\triangle PQR$ be $P,Q,R$ and the opposite sides be $p,q,r$ respectively. Put [ x=\frac R2,\qquad y=\frac{P-Q}{2}. ] Then [ \cos P+\cos Q=2\cos\frac{P+Q}{2}\cos\frac{P-Q}{2} =2\cos\left(\frac{\pi-R}{2}\right)\cos y =2\sin x\cos y. ] Also, [ \cos R=\cos 2x=1-2\sin^2x. ] Hence [ \cos P+\cos Q+\cos R =1+2\sin x(\cos y-\sin x). ] Given that [ \cos P+\cos Q+\cos R=\frac32, ] we get [ 2\sin x(\cos y-\sin x)=\frac12. ] Since $\cos y\le 1$, [ 2\sin x(\cos y-\sin x)\le 2\sin x(1-\sin x). ] But for any real $t$, [ 2t(1-t)=\frac12-2\left(t-\frac12\right)^2\le \frac12. ] Thus the left-hand side is at most $\frac12$, and since it is given to be exactly $\frac12$, equality must hold throughout. Therefore [ \sin x=\frac12,\qquad \cos y=1. ] So [ x=\frac{\pi}{6}\quad\Rightarrow\quad R=60^\circ, ] and [ y=0\quad\Rightarrow\quad P=Q. ] Since $P+Q=120^\circ$, we get [ P=Q=R=60^\circ. ] Hence the triangle is equilateral, so its sides are equal: [ p=q=r. ] Therefore the sides are in A.P. with common difference $0$.

b. Given [ \vec a=2\hat i-\hat j+3\hat k,\qquad \vec b=3\hat i+2\hat j-\hat k. ] Now, [ \vec a\cdot \vec b=(2)(3)+(-1)(2)+(3)(-1)=6-2-3=1. ] Also, [ |\vec b|=\sqrt{3^2+2^2+(-1)^2}=\sqrt{14}. ] The projection of $\vec a$ on $\vec b$ is [ \frac{\vec a\cdot \vec b}{|\vec b|} =\frac{1}{\sqrt{14}}. ] [ \boxed{\text{Projection of } \vec a \text{ on } \vec b=\frac{1}{\sqrt{14}}} ]

If vector projection is required, it is [ \frac{\vec a\cdot \vec b}{|\vec b|^2}\vec b =\frac{1}{14}(3\hat i+2\hat j-\hat k). ]

(b) equilateral

If $\vec{AB}\cdot\vec{BC}=\vec{BC}\cdot\vec{CA}=\vec{CA}\cdot\vec{AB}$, then from the first two,

$\vec{BC}\cdot(\vec{AB}-\vec{CA})=0 \Rightarrow \vec{BC}\cdot(\vec{AB}+\vec{AC})=0,$

i.e. the median from A is perpendicular to BC, so $AB=AC$. Repeating with the other pairs gives $BC=CA$ and $CA=AB$. Hence all three sides are equal and the triangle is equilateral.

Take $a=2,\ b=\sqrt6,\ A=45^\circ$. By the sine rule, $\sin B=\frac{b\sin A}{a}=\frac{\sqrt6\cdot\frac{\sqrt2}{2}}{2}=\frac{\sqrt3}{2}\Rightarrow B=60^\circ\ \text{or}\ 120^\circ.$ Both are admissible ($A+B<180^\circ$), so there are two triangles (the ambiguous case).

Case I: $B=60^\circ$. $C=75^\circ$, and $c=\frac{a\sin C}{\sin A}=\frac{2\sin75^\circ}{\sin45^\circ}=\frac{2\cdot\frac{\sqrt6+\sqrt2}{4}}{\frac{\sqrt2}{2}}=\sqrt3+1.$

Case II: $B=120^\circ$. $C=15^\circ$, and $c=\frac{2\sin15^\circ}{\sin45^\circ}=\frac{2\cdot\frac{\sqrt6-\sqrt2}{4}}{\frac{\sqrt2}{2}}=\sqrt3-1.$

In triangle $ABC$, prove by vector method that $\vec{AB}+\vec{BC}+\vec{CA}=\vec{0}.$

Let $O$ be the origin and let the position vectors of the vertices be

$\vec{OA}=\vec a,\qquad \vec{OB}=\vec b,\qquad \vec{OC}=\vec c.$

Then

$\vec{AB}=\vec{OB}-\vec{OA}=\vec b-\vec a,$

$\vec{BC}=\vec{OC}-\vec{OB}=\vec c-\vec b,$

$\vec{CA}=\vec{OA}-\vec{OC}=\vec a-\vec c.$

Now,

Rearranging,

$=\vec b-\vec b+\vec c-\vec c+\vec a-\vec a=\vec 0.$

Hence,

$\vec{AB}+\vec{BC}+\vec{CA}=\vec{0}.$

Thus, the sum of vectors forming a closed loop in a triangle is zero. $\blacksquare$

The semiperimeter is $s=\frac{a+b+c}{2}.$ It is directly needed in the derivation of the half-angle formulae and Heron’s formula, not in the projection formulae. The projection formulae $a=b\cos C+c\cos B,\qquad b=c\cos A+a\cos C,\qquad c=a\cos B+b\cos A$ are obtained by resolving the sides of $\triangle ABC$ or by using vector dot products; no semiperimeter is required.

For the half-angle formulae, start from the cosine rule: $\cos A=\frac{b^2+c^2-a^2}{2bc}.$ Then

Since $a+b-c=2(s-c),\qquad a-b+c=2(s-b),$ we get $1-\cos A=\frac{2(s-b)(s-c)}{bc}.$ Hence $\sin^2\frac A2=\frac{1-\cos A}{2}=\frac{(s-b)(s-c)}{bc},$ so $\sin\frac A2=\sqrt{\frac{(s-b)(s-c)}{bc}}.$ Similarly,

Thus $\cos^2\frac A2=\frac{1+\cos A}{2}=\frac{s(s-a)}{bc},$ so $\cos\frac A2=\sqrt{\frac{s(s-a)}{bc}}.$ Therefore, $\tan\frac A2=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}.$ The same semiperimeter gives Heron’s formula:

Thus the concept of semiperimeter is crucial for the half-angle formulae and Heron’s formula; it is not required for the projection formulae.

(a) $\sqrt{7}$

Using the cosine rule, $c^2=a^2+b^2-2ab\cos C=2^2+3^2-2(2)(3)\cos60^\circ=4+9-6=7,$ so $c=\sqrt7$.

To prove: $\dfrac{\sin P}{\sin Q} = \dfrac{p}{q}$

In $\triangle PQR$, take $p = QR$ (side opposite $P$), $q = PR$ (side opposite $Q$), $r = PQ$ (side opposite $R$).

Step 1. Drop $PM \perp QR$. In the right triangles $PMQ$ and $PMR$,

$PM = PQ\sin Q = r\sin Q \quad\text{and}\quad PM = PR\sin R = q\sin R$

$\therefore\ r\sin Q = q\sin R \implies \frac{q}{\sin Q} = \frac{r}{\sin R} \qquad ...(i)$

Step 2. Drop $QN \perp PR$. In the right triangles $QNP$ and $QNR$,

$QN = QR\sin R = p\sin R \quad\text{and}\quad QN = PQ\sin P = r\sin P$

$\therefore\ p\sin R = r\sin P \implies \frac{p}{\sin P} = \frac{r}{\sin R} \qquad ...(ii)$

Step 3. From (i) and (ii),

$\frac{p}{\sin P} = \frac{q}{\sin Q} = \frac{r}{\sin R}$

Taking the first two ratios and cross-multiplying,

$p\sin Q = q\sin P \implies \boxed{\frac{\sin P}{\sin Q} = \frac{p}{q}}$

Hence proved. (This is the sine rule: in any triangle the sides are proportional to the sines of the opposite angles.)

Given:
$a=3,\quad b=4,\quad c=5$

Here, (a) is opposite (A), (b) opposite (B), and (c) opposite (C).

Using the cosine rule, $\cos C=\frac{a^2+b^2-c^2}{2ab}$ $=\frac{3^2+4^2-5^2}{2\cdot3\cdot4}$ $=\frac{9+16-25}{24}=0$ $\therefore C=90^\circ$

Again, $\cos A=\frac{b^2+c^2-a^2}{2bc}$ $=\frac{4^2+5^2-3^2}{2\cdot4\cdot5}$ $=\frac{16+25-9}{40}=\frac{32}{40}=\frac45$ $\therefore A=\cos^{-1}\left(\frac45\right)=36^\circ52'12''$

Now, $B=180^\circ-(A+C)$ $=180^\circ-(36^\circ52'12''+90^\circ)$ $=53^\circ7'48''$

Hence the triangle is solved as: $A=36^\circ52'12'',\quad B=53^\circ7'48'',\quad C=90^\circ$

(d) $[-1,1]$

$\sin^{-1}x$ is defined only when $-1\le x\le1$, because $\sin\theta$ takes values only in $[-1,1]$.

(d) $2n\pi\pm\dfrac\pi3,\ n\in I$

$2\cos\theta-1=0\Rightarrow\cos\theta=\dfrac12=\cos\dfrac\pi3$, and $\cos\theta=\cos\alpha$ gives $\theta=2n\pi\pm\alpha$.

Using $\cot^{-1}3=\tan^{-1}\frac13$,

[ \tan^{-1}\frac12+\cot^{-1}3=\tan^{-1}\frac12+\tan^{-1}\frac13 ]

[ =\tan^{-1}\left(\frac{\frac12+\frac13}{1-\frac12\cdot\frac13}\right) ]

[ =\tan^{-1}\left(\frac{\frac56}{1-\frac16}\right) =\tan^{-1}\left(\frac{\frac56}{\frac56}\right) =\tan^{-1}(1) ]

Hence,

[ \tan^{-1}x=\tan^{-1}(1) ]

[ \boxed{x=1} ]

The question is incomplete as given — the expression to be shown is missing. Please provide the full statement.

The sine law states that in any triangle $ABC$, the ratio of the length of a side to the sine of its opposite angle is constant.

That is,

$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R$

where $a,b,c$ are the sides opposite to angles $A,B,C$ respectively, and $R$ is the circumradius of the triangle.

The projection of vector $\vec{a}$ on vector $\vec{b}$ is

$\text{proj}_{\vec{b}}\,\vec{a} = \frac{\vec{a}\cdot\vec{b}}{|\vec{b}|}$

(If the projection is to be given as a vector, it is $\dfrac{\vec{a}\cdot\vec{b}}{|\vec{b}|^2}\,\vec{b}$.)

Using the cosine rule,

$c^2=a^2+b^2-2ab\cos C=2^2+3^2-2(2)(3)\cos60^\circ$

$c^2=4+9-12\times\tfrac12=13-6=7$

$c=\sqrt7\text{ units}$

Given: A triangle $ABC$ with sides $a$, $b$, $c$ opposite to angles $A$, $B$, $C$ respectively.

To prove: $\dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}$

Proof:

The area of $\triangle ABC$ can be written in three ways:

$\Delta = \frac{1}{2}bc\sin A \quad \cdots (1)$

$\Delta = \frac{1}{2}ca\sin B \quad \cdots (2)$

$\Delta = \frac{1}{2}ab\sin C \quad \cdots (3)$

From (1) and (2),

$\frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B$

Dividing both sides by $\dfrac{1}{2}abc$,

$\frac{\sin A}{a} = \frac{\sin B}{b} \quad \cdots (4)$

From (2) and (3),

$\frac{1}{2}ca\sin B = \frac{1}{2}ab\sin C$

Dividing both sides by $\dfrac{1}{2}abc$,

$\frac{\sin B}{b} = \frac{\sin C}{c} \quad \cdots (5)$

From (4) and (5),

$\boxed{\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}}$

Hence proved.

Given:
$a=5,\quad b=7,\quad C=60^\circ$

Using the cosine law, $c^2=a^2+b^2-2ab\cos C$ $c^2=5^2+7^2-2(5)(7)\cos60^\circ$ $c^2=25+49-70\left(\frac12\right)=74-35=39$ $c=\sqrt{39}\approx 6.24\text{ units}$

Now, using the sine law, $\frac{a}{\sin A}=\frac{c}{\sin C}$ $\sin A=\frac{a\sin C}{c}=\frac{5\sin60^\circ}{\sqrt{39}}$ $\sin A=\frac{5\cdot \frac{\sqrt3}{2}}{\sqrt{39}}=\frac{5}{2\sqrt{13}}$ $A=\sin^{-1}\left(\frac{5}{2\sqrt{13}}\right)\approx 43.9^\circ$

Since $A+B+C=180^\circ$ $B=180^\circ-60^\circ-43.9^\circ=76.1^\circ$

Hence, $\boxed{c=\sqrt{39}\approx 6.24\text{ units},\quad A\approx 43.9^\circ,\quad B\approx 76.1^\circ}$

(a) Let $A=\sin^{-1}\frac35$ and $B=\sin^{-1}\frac5{13}$, both acute. Then $\sin A=\frac35,\ \cos A=\frac45$ and $\sin B=\frac5{13},\ \cos B=\frac{12}{13}$.

$\cos(A+B)=\cos A\cos B-\sin A\sin B=\frac45\cdot\frac{12}{13}-\frac35\cdot\frac5{13}=\frac{48-15}{65}=\frac{33}{65}.$

(b) For $A(1,4,6)$, $B(-2,5,1)$, $C(1,-1,1)$: $\overrightarrow{AB}=(-3,1,-5),\qquad \overrightarrow{AC}=(0,-5,-5).$

$\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\sqrt{900+225+225}=\sqrt{1350}=15\sqrt6.$ $\text{Area}=\frac12\left|\overrightarrow{AB}\times\overrightarrow{AC}\right|=\frac{15\sqrt6}{2}\approx18.37\text{ square units}.$

(d) $-1 \leq x \leq 1$

The domain of $\sin^{-1}x$ is the range of $\sin x$, which is $[-1, 1]$, so $\sin^{-1}x$ is defined only when $-1 \leq x \leq 1$.

(a) $2n\pi \pm \dfrac{2\pi}{3}$

$\cos\theta=-\tfrac12=\cos\dfrac{2\pi}{3}$, and $\cos\theta=\cos\alpha$ has the general solution $\theta=2n\pi\pm\alpha$. With $\alpha=\dfrac{2\pi}{3}$, this gives $\theta=2n\pi\pm\dfrac{2\pi}{3}$, $n\in\mathbb Z$ (covering both $\dfrac{2\pi}{3}$ and $\dfrac{4\pi}{3}$ in each turn).