Trigonometry

Let

Since $A+B+C=180^\circ$, we have

Now,

Using

and

we get

Now,

So,

But since $x+y+z=90^\circ$,

Therefore,

Replacing $x,y,z$ by $\frac A2,\frac B2,\frac C2$, we get

Hence proved.

Since $A+B+C=\pi$, we have $\dfrac{A+C}2=\dfrac\pi2-\dfrac B2$.

$\text{LHS}=(\sin A+\sin C)-\sin B$

$\sin A+\sin C=2\sin\frac{A+C}2\cos\frac{A-C}2=2\cos\frac B2\cos\frac{A-C}2$

$\sin B=2\sin\frac B2\cos\frac B2$

So $\text{LHS}=2\cos\frac B2\left[\cos\frac{A-C}2-\sin\frac B2\right]$

Also $\sin\dfrac B2=\cos\dfrac{A+C}2$, hence $\cos\frac{A-C}2-\cos\frac{A+C}2=2\sin\frac A2\sin\frac C2$

$\text{LHS}=2\cos\frac B2\cdot2\sin\frac A2\sin\frac C2=4\sin\frac A2\cos\frac B2\sin\frac C2=\text{RHS}$

Hence proved.

$\cos A-\cos B=-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}=2\sin\frac{A+B}{2}\sin\frac{B-A}{2}$

First simplify the numerator:

Using $\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$,

Now simplify the denominator:

Using $\sin^2A-\sin^2B=\sin(A+B)\sin(A-B)$,

Therefore,

Cancelling $\sin(A-B)$,

Hence proved.

Given: $3\tan^2\theta - 9 = 0$

$3\tan^2\theta = 9$

$\tan^2\theta = 3$

$\tan\theta = \sqrt{3} \quad (\text{since } 0^\circ \le \theta \le 90^\circ)$

$\theta = 60^\circ \quad (\because \tan 60^\circ = \sqrt{3})$

Therefore, $\theta = 60^\circ$.

$2\sin A\cos B=\sin(A+B)+\sin(A-B)$

This follows by adding $\sin(A+B)=\sin A\cos B+\cos A\sin B$ and $\sin(A-B)=\sin A\cos B-\cos A\sin B$.

$\sin3A=3\sin A-4\sin^3A$

Proof: $\sin3A=\sin(2A+A)=\sin2A\cos A+\cos2A\sin A=2\sin A\cos^2A+(1-2\sin^2A)\sin A$. Putting $\cos^2A=1-\sin^2A$ gives $2\sin A-2\sin^3A+\sin A-2\sin^3A=3\sin A-4\sin^3A$.

$\sin A = \dfrac{2\tan\frac{A}{2}}{1+\tan^2\frac{A}{2}}$

This follows from the half-angle formula: writing $\sin A = 2\sin\frac{A}{2}\cos\frac{A}{2}$ and dividing numerator and denominator by $\cos^2\frac{A}{2}$ gives $\dfrac{2\tan\frac{A}{2}}{1+\tan^2\frac{A}{2}}$.

$\cos(A-B)-\cos(A+B)=2\sin A\sin B$

Using $\cos C-\cos D=-2\sin\dfrac{C+D}{2}\sin\dfrac{C-D}{2}$ with $C=A-B$, $D=A+B$: $-2\sin A\sin(-B)=2\sin A\sin B$

Using $\sin A\cos B=\dfrac{\sin(A+B)+\sin(A-B)}{2}$ with $A=75^\circ$, $B=15^\circ$: $\sin75^\circ\cos15^\circ=\frac{\sin90^\circ+\sin60^\circ}{2}=\frac{1+\frac{\sqrt3}{2}}{2}=\frac{2+\sqrt3}{4}$ Hence proved.

$\sin 2x-\sin x=0$

$\Rightarrow 2\sin x\cos x-\sin x=0$

$\Rightarrow \sin x(2\cos x-1)=0$

So either $\sin x=0$ or $2\cos x-1=0$.

• $\sin x=0 \Rightarrow x=0^\circ$ (in $0^\circ\le x\le90^\circ$)
• $2\cos x-1=0 \Rightarrow \cos x=\dfrac12 \Rightarrow x=60^\circ$

Hence, $\boxed{x=0^\circ,\ 60^\circ}$

Using the triple-angle formulae, $\cos 3\theta = 4\cos^3\theta - 3\cos\theta,\qquad \sin 3\theta = 3\sin\theta - 4\sin^3\theta.$

Now,

Hence proved.

Since $P+Q+R=180^\circ$, $P+Q-R=180^\circ-2R\Rightarrow\sin(P+Q-R)=\sin(180^\circ-2R)=\sin2R$ Similarly $\sin(Q+R-P)=\sin2P$ and $\sin(R+P-Q)=\sin2Q$.

So LHS $=\sin2P+\sin2Q+\sin2R$.

$\sin2P+\sin2Q=2\sin(P+Q)\cos(P-Q)=2\sin R\cos(P-Q)$ because $\sin(P+Q)=\sin(180^\circ-R)=\sin R$. Also $\sin2R=2\sin R\cos R$. Hence $\text{LHS}=2\sin R\left[\cos(P-Q)+\cos R\right]$ Since $\cos R=-\cos(P+Q)$, $\text{LHS}=2\sin R\left[\cos(P-Q)-\cos(P+Q)\right]=2\sin R\cdot2\sin P\sin Q=4\sin P\sin Q\sin R$ Hence proved.

Let the tower be $AB=H$ m (top $A$, foot $B$) and the pole $CD=20$ m, at horizontal distance $d$ m from the tower.

Angle of depression of the bottom of the pole is $60^\circ$: $\tan60^\circ=\frac{H}{d}\Rightarrow d=\frac{H}{\sqrt3}$

Angle of depression of the top of the pole is $30^\circ$: $\tan30^\circ=\frac{H-20}{d}\Rightarrow d=\sqrt3\,(H-20)$

Equating the two values of $d$: $\frac{H}{\sqrt3}=\sqrt3(H-20)\Rightarrow H=3H-60\Rightarrow H=30$

The height of the tower is 30 m.

$\cos A=\frac{1-\tan^2\frac A2}{1+\tan^2\frac A2}$

Writing $\cos A=\cos^2\frac A2-\sin^2\frac A2$ and dividing numerator and denominator by $\cos^2\frac A2$ gives this result.

$\text{LHS}=\tan A+\cot A=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}=\frac{\sin^2A+\cos^2A}{\sin A\cos A}=\frac{1}{\sin A\cos A}$

$\text{RHS}=2\operatorname{cosec}2A=\frac{2}{\sin2A}=\frac{2}{2\sin A\cos A}=\frac{1}{\sin A\cos A}$

$\therefore\ \tan A+\cot A=2\operatorname{cosec}2A$. Hence proved.

Given: $4\sin\theta=\operatorname{cosec}\theta$

Since $\operatorname{cosec}\theta=\dfrac{1}{\sin\theta}$, $4\sin\theta=\frac{1}{\sin\theta}$

Multiplying both sides by $\sin\theta$, $4\sin^2\theta=1$ $\sin^2\theta=\frac14$ $\sin\theta=\pm\frac12$

But $0^\circ\le\theta\le90^\circ$, so $\sin\theta\ge0$. Hence, $\sin\theta=\frac12=\sin30^\circ$

Therefore, $\boxed{\theta=30^\circ}$

Let the height of the tree be $h$ m.

The point with angle of elevation $45^\circ$ is nearer to the tree.
Let its distance from the foot of the tree be $x$ m.
Then the distance of the other point from the foot of the tree is $(x+20)$ m.

From the nearer point, $h=x\tan45^\circ=x$

From the farther point, $h=(x+20)\tan30^\circ=\frac{x+20}{\sqrt3}$

Equating the two values of $h$, $x=\frac{x+20}{\sqrt3}$ $\sqrt3x=x+20$ $x(\sqrt3-1)=20$ $x=\frac{20}{\sqrt3-1}$ $x=\frac{20(\sqrt3+1)}{3-1}=10(\sqrt3+1)$

Therefore, $h=10(\sqrt3+1)\text{ m}\approx 27.32\text{ m}$

So, the height of the tree is $10(\sqrt3+1)$ m, i.e. about 27.32 m.

$\sin A-\sin B=2\cos\frac{A+B}{2}\sin\frac{A-B}{2}$

Let $A$ be the observer on the roof of the house, $D$ the foot of the house ($AD=20$ m), $B$ the bottom and $C$ the top of the tower, and $E$ the point on the tower at the level of $A$.

Angle of depression of the bottom of the tower is $30^\circ$. In right-angled $\triangle ADB$: $\tan30^\circ=\frac{AD}{DB}\Rightarrow\frac{1}{\sqrt3}=\frac{20}{DB}\Rightarrow DB=20\sqrt3\text{ m}$ So $AE=DB=20\sqrt3$ m.

Angle of elevation of the top is $45^\circ$. In right-angled $\triangle AEC$: $\tan45^\circ=\frac{EC}{AE}\Rightarrow EC=20\sqrt3\text{ m}$

Height of the tower: $BC=BE+EC=20+20\sqrt3=20(1+\sqrt3)\approx54.64\text{ m}$

$\cos 2A = 2\cos^2 A - 1$

This follows from the double-angle identity $\cos 2A = \cos^2 A - \sin^2 A$, replacing $\sin^2 A$ with $1 - \cos^2 A$.

$\sin C-\sin D=2\cos\frac{C+D}{2}\sin\frac{C-D}{2}$

Let $\theta = 2x$, so that $x=\frac{\theta}{2}$.

Then,

Using $\sin 2x=2\sin x\cos x$ and $\cos 2x=2\cos^2x-1$,

Hence proved.

$\sqrt3\tan A-1=0\Rightarrow\tan A=\frac{1}{\sqrt3}=\tan30^\circ$

In $0^\circ\le A\le180^\circ$, $\tan A$ is positive only in the first quadrant, so $A=30^\circ$

Let $\tan\dfrac A2=t$. Then $\cot\left(\frac A2+\frac\pi4\right)=\frac{1-t}{1+t},\qquad\tan\left(\frac A2-\frac\pi4\right)=\frac{t-1}{1+t}$ $\text{LHS}=\frac{1-t}{1+t}-\frac{t-1}{1+t}=\frac{2(1-t)}{1+t}$

Now $\cos A=\dfrac{1-t^2}{1+t^2}$ and $\sin A=\dfrac{2t}{1+t^2}$, so $\text{RHS}=\frac{2\cos A}{1+\sin A}=\frac{2(1-t^2)}{1+t^2+2t}=\frac{2(1-t)(1+t)}{(1+t)^2}=\frac{2(1-t)}{1+t}$

LHS $=$ RHS. Hence proved.

Let the tower be of height $h$ m, at horizontal distance $x$ m from the temple. The pinnacle of the temple is 10 m above the ground.

From the bottom of the tower, the angle of elevation of the pinnacle is $30^\circ$: $\tan30^\circ=\frac{10}{x}\Rightarrow x=10\sqrt3\text{ m}$

From the top of the tower, the angle of depression of the pinnacle is $60^\circ$, and the vertical drop is $h-10$: $\tan60^\circ=\frac{h-10}{x}\Rightarrow\sqrt3=\frac{h-10}{10\sqrt3}\Rightarrow h-10=30\Rightarrow h=40$

The height of the tower is 40 m.

$\sin\theta = \dfrac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$

This follows from the half-angle formula: writing $\sin\theta = 2\sin\frac{\theta}{2}\cos\frac{\theta}{2}$ and dividing numerator and denominator by $\cos^2\frac{\theta}{2}$ gives $\sin\theta = \dfrac{2\tan\frac{\theta}{2}}{1+\tan^2\frac{\theta}{2}}$.

Let $\theta = 2x$, so that $\dfrac{\theta}{2}=x$.

Then, [ \sin\theta=\sin 2x=2\sin x\cos x ]

Now, [ \text{LHS}=\frac{\sin\theta+\cos\frac{\theta}{2}}{2\sin\frac{\theta}{2}+1} =\frac{2\sin x\cos x+\cos x}{2\sin x+1} ]

Taking $\cos x$ common in the numerator, [ =\frac{\cos x(2\sin x+1)}{2\sin x+1} ]

Cancelling $(2\sin x+1)$, [ =\cos x=\cos\frac{\theta}{2}=\text{RHS} ]

Hence proved.

Using $\sin(A+B)\sin(A-B)=\sin^2A-\sin^2B$ with $A=45^\circ$, $B=\theta$: $\sin(45^\circ+\theta)\sin(45^\circ-\theta)=\sin^245^\circ-\sin^2\theta=\frac12-\sin^2\theta=\frac{1-2\sin^2\theta}{2}=\frac{\cos2\theta}{2}$

Therefore $\frac{1}{\sin(45^\circ+\theta)\sin(45^\circ-\theta)}=\frac{2}{\cos2\theta}=2\sec2\theta$ Hence proved.

Given $A+B+C=\pi^c$, so $\dfrac{A+B}{2}=\dfrac\pi2-\dfrac C2$.

$\sin A+\sin B+\sin C=2\sin\frac{A+B}{2}\cos\frac{A-B}{2}+2\sin\frac C2\cos\frac C2$

Since $\sin\dfrac{A+B}{2}=\cos\dfrac C2$, $=2\cos\frac C2\left[\cos\frac{A-B}{2}+\sin\frac C2\right]$

Also $\sin\dfrac C2=\cos\dfrac{A+B}{2}$, so $\cos\frac{A-B}{2}+\cos\frac{A+B}{2}=2\cos\frac A2\cos\frac B2$

Hence $\sin A+\sin B+\sin C=2\cos\frac C2\cdot2\cos\frac A2\cos\frac B2=4\cos\frac A2\cos\frac B2\cos\frac C2$ Hence proved.

Let the observer's eye be at $A$, $AB=1.8$ m above the ground at $B$, and let the pole stand at horizontal distance $d$ from the observer, with foot $Q$ and top $P$. Let $h$ be the height of the pole.

Angle of depression of the foot is $30^\circ$: $\tan30^\circ=\frac{1.8}{d}\Rightarrow d=1.8\sqrt3\text{ m}$

Angle of elevation of the top is $60^\circ$; the top is $h-1.8$ m above the level of the eye: $\tan60^\circ=\frac{h-1.8}{d}\Rightarrow h-1.8=\sqrt3\times1.8\sqrt3=5.4$ $h=5.4+1.8=7.2\text{ m}$

The height of the pole is 7.2 m.

$\cos 2A=\cos^2A-\sin^2A$

Let $\theta=\dfrac{A}{3}$, so $A=3\theta$ and $\cos\theta=\dfrac12\left(p+\dfrac1p\right)$.

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

Now $4\cos^3\theta=4\cdot\frac18\left(p+\frac1p\right)^3=\frac12\left(p^3+3p+\frac3p+\frac1{p^3}\right)$ $3\cos\theta=\frac32\left(p+\frac1p\right)$

Subtracting, the terms $\frac32\left(p+\frac1p\right)$ cancel:

$\cos A=\frac12\left(p^3+\frac1{p^3}\right)$

Using $\operatorname{cosec}^2x=1+\cot^2x$:

$\cot^2x+1+\cot^2x=3$ $2\cot^2x=2\Rightarrow\cot^2x=1\Rightarrow\cot x=\pm1$

For $0^\circ\le x\le90^\circ$, $\cot x\ge0$, so $\cot x=1=\cot45^\circ$.

$\boxed{x=45^\circ}$

$\text{LHS}=\frac{\sin2\theta}{1+\cos2\theta}\cdot\frac{\cot\frac\theta2}{1+\cos\theta}$

Use $\sin2\theta=2\sin\theta\cos\theta$, $1+\cos2\theta=2\cos^2\theta$ and $1+\cos\theta=2\cos^2\frac\theta2$.

First factor: $\frac{2\sin\theta\cos\theta}{2\cos^2\theta}=\tan\theta$

Second factor: $\frac{\cos\frac\theta2/\sin\frac\theta2}{2\cos^2\frac\theta2}=\frac1{2\sin\frac\theta2\cos\frac\theta2}=\frac1{\sin\theta}$

Therefore $\text{LHS}=\tan\theta\cdot\frac1{\sin\theta}=\frac{\sin\theta}{\cos\theta\sin\theta}=\frac1{\cos\theta}=\sec\theta=\text{RHS}$

Hence proved.

Let the building be $AB$ of height $h$ m, with $B$ its foot. Let $C$ (nearer) and $D$ (farther) be the two points on the same side, $CD=10$ m, and let $BC=x$ m.

From $C$ (elevation $45^\circ$): $\tan45^\circ=\dfrac hx\Rightarrow h=x$.

From $D$ (elevation $30^\circ$): $\tan30^\circ=\dfrac h{x+10}\Rightarrow$ $\frac1{\sqrt3}=\frac h{x+10}\Rightarrow x+10=\sqrt3\,h$

Substituting $x=h$: $h+10=\sqrt3\,h\Rightarrow h(\sqrt3-1)=10$ $h=\frac{10}{\sqrt3-1}\cdot\frac{\sqrt3+1}{\sqrt3+1}=5(\sqrt3+1)\approx13.66$

The height of the building is $\boxed{5(\sqrt3+1)\approx13.66\text{ m}}$.