Trigonometry - Trigonometric Functions

If $\;$ $\text{cosec} \theta - \sin \theta = a^3$, $\;$ $\sec \theta - \cos \theta = b^3$,
then prove that $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Given: $\;$ $\text{cosec } \theta - \sin \theta = a^3$

i.e. $\;$ $\dfrac{1}{\sin \theta} - \sin \theta = a^3$

i.e. $\;$ $a^3 = \dfrac{1 - \sin^2 \theta}{\sin \theta}$

i.e. $\;$ $a^3 = \dfrac{\cos^2 \theta}{\sin \theta}$ $\;\;\; \cdots \; (1)$

and $\;$ $\sec \theta - \cos \theta = b^3$

i.e. $\;$ $\dfrac{1}{\cos \theta} - \cos \theta = b^3$

i.e. $\;$ $b^3 = \dfrac{1 - \cos^2 \theta}{\cos \theta}$

i.e. $\;$ $b^3 = \dfrac{\sin^2 \theta}{\cos \theta}$ $\;\;\; \cdots \; (2)$

Dividing equations $(1)$ and $(2)$ gives

$\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \div \dfrac{\cos^2 \theta}{\sin \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^2 \theta}{\cos \theta} \times \dfrac{\sin \theta}{\cos^2 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \dfrac{\sin^3 \theta}{\cos^3 \theta}$

i.e. $\;$ $\dfrac{b^3}{a^3} = \tan^3 \theta$

i.e. $\;$ $\dfrac{b}{a} = \tan \theta$

$\implies$ $\sin \theta = \dfrac{b}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (3)$

and $\;$ $\cos \theta = \dfrac{a}{\sqrt{a^2 + b^2}}$ $\;\;\; \cdots \; (4)$

Substituting the values of $\sin \theta$ and $\cos \theta$ from equations $(3)$ and $(4)$ in equation $(1)$ gives

$a^3 = \dfrac{a^2}{a^2 + b^2} \div \dfrac{b}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^3 = \dfrac{a^2}{a^2 + b^2} \times \dfrac{\sqrt{a^2 + b^2}}{b}$

i.e. $\;$ $a \; b = \dfrac{1}{\sqrt{a^2 + b^2}}$

i.e. $\;$ $a^2 \; b^2 \left(a^2 + b^2\right) = 1$

Hence proved

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right) = \dfrac{\sin \beta}{\sin \alpha}$

LHS $= \dfrac{\sin \left(2 \alpha + \beta\right)}{\sin \alpha} - 2 \cos \left(\alpha + \beta\right)$

$= \dfrac{\sin \left[\left(\alpha + \alpha\right) + \beta\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha }$

$= \dfrac{\sin \left[\alpha + \left(\alpha + \beta\right)\right] - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \alpha \cos \left(\alpha + \beta\right) + \cos \alpha \sin \left(\alpha + \beta\right) - 2 \sin \alpha \cos \left(\alpha + \beta\right)}{\sin \alpha}$

$= \dfrac{\sin \left(\alpha + \beta\right) \cos \alpha - \cos \left(\alpha + \beta\right) \sin \alpha}{\sin \alpha}$

$= \dfrac{\sin \left[\left(\alpha + \beta\right) - \alpha\right]}{\sin \alpha}$

$= \dfrac{\sin \beta}{\sin \alpha} = $ RHS

Trigonometry - Identity Transformation of Trigonometric Expressions

Prove the identity:
$\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right) = 2 \sec 2 \alpha$

LHS $=\sec \left(\dfrac{\pi}{4} + \alpha\right) \sec \left(\dfrac{\pi}{4} - \alpha\right)$

$= \dfrac{1}{\cos \left(\dfrac{\pi}{4} + \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\left(\dfrac{\pi}{2} - \dfrac{\pi}{4}\right) + \alpha \right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\cos \left[\dfrac{\pi}{2} - \left(\dfrac{\pi}{4} - \alpha\right)\right] \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{1}{\sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{2 \sin \left(\dfrac{\pi}{4} - \alpha\right) \cos \left(\dfrac{\pi}{4} - \alpha\right)}$

$= \dfrac{2}{\sin \left[2 \left(\dfrac{\pi}{4} - \alpha\right)\right]}$

$= \dfrac{2}{\sin \left(\dfrac{\pi}{2} - 2 \alpha\right)}$

$= \dfrac{2}{\cos 2 \alpha}$

$= 2 \sec 2 \alpha = $ RHS

Algebra - Word Problems: Derivation of Equations

Two cyclists left the same point simultaneously and traveled in the same direction. The speed of the first was $15$ km/h and that of the second was $12$ km/h. Half an hour later, another cyclist left the same point and traveled in the same direction. Some time later, he overtook the second cyclist and another hour and a half later he overtook the first cyclist. Find the speed of the third cyclist.

Let cyclists $C_1$ and $C_2$ leave point $A$ simultaneously and travel in the same direction.

After $\dfrac{1}{2}$ hr, let cyclist $C_3$ start from point $A$.

Speed of $C_1 = u_1 = 15$ kmph

and speed of $C_2 = u_2 = 12$ kmph

Let $C_1$ and $C_2$ start from point $A$ at time $t = 0$

In time $= \dfrac{1}{2}$ hr, distance covered by $C_1 = 15 \times \dfrac{1}{2} = 7.5$ km

In time $= \dfrac{1}{2}$ hr, distance covered by $C_2 = 12 \times \dfrac{1}{2} = 6$ km

Let speed of third cyclist $C_3 = u_3$ kmph

Let $C_3$ overtake $C_2$ after a time $T$ hr

Distance covered by $C_2$ in time $T$ $= \left(12 \times T\right)$ km

$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,

distance covered by $C_2$ in time $T$ $= \left(12T + 6\right)$ km $\;\;\; \cdots \; (1)$

Distance covered by $C_3$ in time $T$ $= u_3 \times T$ km $\;\;\; \cdots \; (2)$

$\because \;$ $C_3$ overtakes $C_2$, we have from equations $(1)$ and $(2)$

$12 T + 6 = u_3 T$ $\implies$ $T = \dfrac{6}{u_3 - 12}$ $\;\;\; \cdots \; (3)$

Now, $C_3$ overtakes $C_1$ after a time $\left(T + 1.5\right)$ hr

Distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr is $= 15 \left(T + 1.5\right) = \left(15 T + 22.5\right)$ km

$\because \;$ $C_3$ starts $\dfrac{1}{2}$ hr later,

distance covered by $C_1$ in time $\left(T + 1.5\right)$ hr

$= \left(15T + 22.5 + 7.5\right) = 15 T + 30$ km $\;\;\; \cdots \; (4)$

Distance covered by $C_3$ in time $\left(T + 1.5\right)$ hr

$= u_3 \times \left(T + 1.5\right) = u_3 T + 1.5 u_3$ km $\;\;\; \cdots \; (5)$

$\because \;$ $C_3$ overtakes $C_1$, we have from equations $(4)$ and $(5)$

$15 T + 30 = u_3 T + 1.5 u_3$

i.e. $\;$ $T = \dfrac{1.5 u_3 - 30}{15 - u_3}$ $\;\;\; \cdots \; (6)$

$\therefore \;$ We have from equations $(3)$ and $(6)$,

$\dfrac{6}{u_3 - 12} = \dfrac{1.5 u_3 - 30}{15 - u_3}$

i.e. $\;$ $1.5 u_3^2 - 42 u_3 + 270 = 0$ $\;\;\; \cdots \; (7)$

Solving quadratic equation $(7)$ gives $\;$ $u_3 = 10$ $\;$ or $\;$ $u_3 = 18$

Speed of $C_3$ has to be greater than the speeds of $C_1$ and $C_2$ for $C_3$ to overtake them.

$\implies$ $u_3 = 10$ is not an acceptable solution.

$\therefore \;$ Speed of cyclist $C_3$ $= u_3 = 18$ kmph

Algebra - Word Problems: Derivation of Equations

Two people left simultaneously two points: one left point $A$ for point $B$ and the other left $B$ for $A$. Each of them walked at a constant speed and, having arrived at the point of destination, went back at once. First time they met $12$ km from $B$, and the second time, six hours after the first meeting, $6$ km from $A$ . Find the distance between A and B and the speeds of the two people.

Let person $P_1$ start from point $A$ and person $P_2$ start from point $B$ simultaneously

Let speed of $P_1$ $= u$ kmph

and speed of $P2$ $= v$ kmph

Let distance $AB = x$ km

For first meet at point $M_1$, $12$ km from point $B$:

Let $P_1$ and $P_2$ meet for the first time at point $M_1$

Then, distance $\left(BM_1\right) = 12$ km and distance $\left(M_1 A \right) = x - 12$ km

Time taken by $P_1$ to cover distance $M_1 A$ $= t_1 = \dfrac{M_1 A}{u} = \dfrac{x - 12}{u}$ hr $\;\;\; \cdots \; (1)$

Time taken by $P_2$ to cover distance $BM_1$ $= t_2 = \dfrac{B M_1}{v} = \dfrac{12}{v}$ hr $\;\;\; \cdots \; (2)$

Since $P_1$ and $P_2$ meet at point $M_1$, $t_1 = t_2$

$\therefore \;$ We have from equations $(1)$ and $(2)$

$\dfrac{x - 12}{u} = \dfrac{12}{v}$

i.e. $\;$ $\dfrac{u}{v} = \dfrac{x - 12}{12}$ $\;\;\; \cdots \; (3)$

For second meet at point $M_2$, $6$ km from point $A$:

Let $P_1$ and $P_2$ meet for the second time at point $M_2$

Then, distance $\left(AM_2\right) = 6$ km and distance $\left(M_2 B \right) = x - 6$ km

Time taken by $P_1$ to meet at $M_2$ $= t_3 = \dfrac{AB + M_2 B}{u} = \dfrac{x + x - 6}{u} = \dfrac{2x - 6}{u}$ hr $\;\;\; \cdots \; (4)$

Time taken by $P_2$ to meet at $M_2$ $= t_4 = \dfrac{AB + A M_2}{v} = \dfrac{x + 6}{v}$ hr $\;\;\; \cdots \; (5)$

Since $P_1$ and $P_2$ meet at point $M_2$, $t_3 = t_4$

$\therefore \;$ We have from equations $(4)$ and $(5)$

$\dfrac{2x - 6}{u} = \dfrac{x + 6}{v}$

i.e. $\;$ $\dfrac{u}{v} = \dfrac{2x - 6}{x + 6}$ $\;\;\; \cdots \; (6)$

$\therefore \;$ We have from equations $(3)$ and $(6)$

$\dfrac{x - 12}{12} = \dfrac{2x - 6}{x + 6}$

i.e. $\;$ $x^2 - 12x + 6x - 72 = 24x - 72$

i.e. $\;$ $x^2 - 30x = 0$

i.e. $\;$ $x \left(x - 30 \right) = 0$

i.e. $\;$ $x = 0$ $\;$ or $\;$ $x = 30$

Since distance between points $A$ and $B$ cannot be $0$ km

$\therefore \;$ Distance between points $A$ and $B$ is $= x = 30$ km $\;\;\; \cdots \; (7)$

Also, second meet time $=$ first meet time $+ 6$ hr

$\therefore \;$ We have from equations $(1)$ and $(4)$, for $P_1$

$\dfrac{2x - 6}{u} = \dfrac{x - 12}{u} + 6$

i.e. $\;$ $\dfrac{\left(2 \times 30 \right) - 6}{u} = \dfrac{30 - 12}{u} + 6$ $\;\;\;$ [in view of equation $(7)$]

i.e. $\;$ $\dfrac{54}{u} = \dfrac{18}{u} + 6$

i.e. $\;$ $\dfrac{36}{u} = 6$ $\implies$ $u = 6$ $\;\;\; \cdots \; (8)$

and we have from equations $(2)$ and $(5)$, for $P_2$

$\dfrac{x + 6}{v} = \dfrac{12}{v} + 6$

i.e. $\;$ $\dfrac{30 + 6}{v} = \dfrac{12}{v} + 6$ $\;\;\;$ [in view of equation $(7)$]

i.e. $\;$ $\dfrac{36}{v} - \dfrac{12}{v} = 6$

i.e. $\;$ $\dfrac{24}{v} = 6$ $\implies$ $v = 4$ $\;\;\; \cdots \; (9)$

$\therefore \;$ Speed of $P_1$ $= u = 6$ kmph

and speed of $P_2$ $= v = 4$ kmph

Algebra - Word Problems: Derivation of Equations

Point C is at a distance of $12$ km from point B down the river. A fisher left point A, which is somewhat farther up the river than point B, for point C in a boat. He arrived at C four hours later and covered the return trip in six hours. Fixing a motor to the boat and thus trebling its speed relative to the water, the fisher covered the distance from A to B in $45$ minutes. Determine the speed of the river flow, considering it to be constant.

Direction A to B to C is downstream

Direction C to B to A is upstream

Let speed of boat in still water $= u$ kmph

Let speed of river flow $= v$ kmph

Downstream speed of boat $= u + v$ kmph $\;\;\; \cdots \; (1)$

Upstream speed of boat $= u - v$ kmph $\;\;\; \cdots \; (2)$

Let distance AB $= x$ km

Given: Distance BC $= 12$ km

Then, distance AC $= x + 12$ km

Time taken by the boat to arrive from point A to point C (downstream) $= 4$ hours

Therefore, downstream speed of boat $= \dfrac{x + 12}{4}$ kmph $\;\;\; \cdots \; (3)$

Therefore, we have from equations $(1)$ and $(3)$

$u + v = \dfrac{x + 12}{4}$

i.e. $\;$ $4u + 4v = x + 12$ $\implies$ $x = 4u + 4v - 12$ $\;\;\; \cdots \; (4)$

Time taken by the boat to arrive from point C to point A (upstream) $= 6$ hours

Therefore, upstream speed of boat $= \dfrac{x + 12}{6}$ kmph $\;\;\; \cdots \; (5)$

Therefore, we have from equations $(2)$ and $(5)$

$u - v = \dfrac{x + 12}{6}$

i.e. $\;$ $6u - 6v = x + 12$ $\implies$ $x = 6u - 6v - 12$ $\;\;\; \cdots \; (6)$

Therefore, we have from equations $(4)$ and $(6)$

$4u + 4v - 12 = 6u - 6v - 12$

i.e. $\;$ $2u = 10 v$ $\implies$ $u = 5v$ $\;\;\; \cdots \; (7)$

After fixing motor, new speed of boat $= 3u = 3 \times 5v = 15v$ kmph [in view of equation $(7)$]

Therefore, speed of boat downstream from A to B $= 15v + v = 16v$ kmph

Time taken by the boat to cover distance AB $= \dfrac{x}{16v}$ hr $\;\;\; \cdots \; (8)$

Given: Time now taken to cover distance AB $= 45$ minutes $= \dfrac{45}{60} = \dfrac{3}{4}$ hr $\;\;\; \cdots \; (9)$

We have from equations $(8)$ and $(9)$

$\dfrac{x}{16v} = \dfrac{3}{4}$ $\implies$ $x = \dfrac{3}{4} \times 16v = 12v$ km $\;\;\; \cdots \; (10)$

In view of equations $(7)$ and $(10)$, equation $(4)$ becomes

$12v = 20v + 4v - 12$

i.e. $\;$ $12v = 12$ $\implies$ $v = 1$ kmph

i.e. Speed of river flow $= v = 1$ kmph

Algebra - Word Problems: Derivation of Equations

If a steamer and a motor-launch go down stream, then the steamer covers the distance from A to B $1.5$ times as fast as the motor-launch, the latter lagging behind the steamer $8$ km more each hour. Now if they go up stream, then the steamer covers the distance from B to A twice as fast as the motor-launch. Find the speeds of the steamer and the motor-launch in still water.

Direction A to B is downstream

Direction B to A is upstream

Let speed of steamer in still water $= u$ kmph

Let speed of motor-launch in still water $= v$ kmph

Let speed of stream $= s$ kmph

Speed of steamer downstream $= \left(u + s\right)$ kmph

Speed of motor-launch downstream $= \left(v + s\right)$ kmph

Given: Steamer downstream speed $=$ motor-launch downstream speed $\times 1.5$

i.e. $\;$ $u + s = \left(v + s\right) \times 1.5$

i.e. $\;$ $u + s = 1.5 v + 1.5s$

i.e. $\;$ $u - 1.5 v = 0.5 s$

i.e. $\;$ $s = 2u - 3v$ $\;\;\; \cdots \; (1)$

In view of equation $(1)$,

speed of steamer downstream $= u + 2u - 3v = 3u - 3v$ kmph $\;\;\; \cdots \; (2)$

speed ofmotor-launch downstream $= v + 2u - 3v = 2u - 2v$ kmph $\;\;\; \cdots \; (3)$

speed of steamer upstream $= \left(u - s\right) = u - 2u + 3v = 3v - u$ kmph $\;\;\; \cdots \; (4)$

speed of motor-launch upstream $= \left(v - s\right) = v - 2u + 3v = 4v - 2u$ kmph $\;\;\; \cdots \; (5)$

Given: Steamer upstream speed $=$ motor-launch upstream speed $\times 2$

i.e. $\;$ $3v - u = \left(4v - 2u\right) \times 2$

i.e. $\;$ $3v - u = 8v - 4u$

i.e. $\;$ $3u = 5v$

i.e. $\;$ $v = 0.6 u$ $\;\;\; \cdots \; (6)$

In view of equation $(6)$, we have

from equation $(2)$, speed of steamer downstream $= 3u - \left(3 \times 0.6u\right) = 1.2u$ kmph $\;\;\; \cdots \; (7)$

from equation $(3)$, speed of motor-launch downstream $= 2u - \left(2 \times 0.6u\right) = 0.8u$ kmph $\;\;\; \cdots \; (8)$

Let distance between points A and B be $= d$ km

While going downstream,

time taken by steamer to cover distance AB $= \dfrac{d}{1.2u}$ hr

time taken by motor-launch to cover distance AB $= \dfrac{d}{0.8u}$ hr

$\therefore \;$ Excess time taken by motor-launch to cover AB $= T = \dfrac{d}{0.8u} - \dfrac{d}{1.2u} = \dfrac{d}{2.4u}$ hr $\;\;\; \cdots \; (9)$

For every hour, the motor-launch lags the steamer by $8$ km

$\therefore \;$ For time $T$, the motor-launch lags the steamer by $= \dfrac{d}{2.4u} \times 8 = \dfrac{d}{0.3u}$ km $\;\;\; \cdots \; (10)$

Now, in time $T = \dfrac{d}{2.4u}$ hr,

the steamer covers downstream a distance $= d_1 = 1.2u \times \dfrac{d}{2.4u} = \dfrac{d}{2}$ km

and the motor-launch covers downstream a distance $= d_2 = 0.8u \times \dfrac{d}{2.4u} = \dfrac{d}{3}$ km

$\therefore \;$ In time $T$, the motor-launch lags the steamer by $= d_1 - d_2 = \dfrac{d}{2} - \dfrac{d}{3} = \dfrac{d}{6}$ km $\;\;\; \cdots \; (11)$

We have from equations $(10)$ and $(11)$

$\dfrac{d}{0.3u} = \dfrac{d}{6}$ $\implies$ $u = 20$ kmph

$\therefore \;$We have from equation $(6)$, $v = 0.6 \times 20 = 12$ kmph

Therefore, speed of steamer in still water $= 20$ kmph

and speed of motor-launch in still water $= 12$ kmph