Question 1: Find the maximum and minimum values of each of the following trigonometrical expressions:

i) $12 \sin x - 5 \cos x$    ii) $12 \cos x + 5 \sin x + 4$

iii) $5 \cos x + 3 \sin \Big($ $\frac{\pi}{4}$ $- x \Big) + 4$    iv) $\sin x - \cos x + 1$

i) Let $f(x) = 12 \sin x - 5 \cos x$

We know that $-\sqrt{(12)^2 + (-5)^2} \leq f(x) \leq \sqrt{(12)^2 + (-5)^2}$

$\Rightarrow -\sqrt{169} \leq f(x) \leq \sqrt{169}$

$\Rightarrow - 13 \leq f(x) \leq 13$

Hence minimum and maximum value of $12 \sin x - 5 \cos x$ are $-13$ and $13$ respectively.

ii) Given $12 \cos x + 5 \sin x + 4$

Let $f(x) = 12 \cos x + 5 \sin x$

We know that $-\sqrt{(12)^2 + (5)^2} \leq f(x) \leq \sqrt{(12)^2 + (5)^2}$

$\Rightarrow -\sqrt{169} \leq f(x) \leq \sqrt{169}$

$\Rightarrow - 13 \leq f(x) \leq 13$

$\Rightarrow - 13+4 \leq f(x)+4 \leq 13+4$

$\Rightarrow - 9 \leq 12 \cos x + 5 \sin x + 4 \leq 17$

Hence minimum and maximum value of $12 \cos x + 5 \sin x + 4$ are $-9$ and $17$ respectively.

iii) Let $f(x) = 5 \cos x + 3 \sin \Big($ $\frac{\pi}{4}$ $- x \Big) + 4$

$\Rightarrow f(x) = 5 \cos x + 3 \Big[ \sin$ $\frac{\pi}{6}$ $\cos x - \cos$ $\frac{\pi}{6}$ $\sin x \Big] + 4$

$\Rightarrow f(x) = 5 \cos x + 3 \Big[$ $\frac{1}{2}$ $\cos x -$ $\frac{\sqrt{3}}{2}$ $\sin x \Big] + 4$

$\Rightarrow f(x) = (5 +$ $\frac{3}{2}$ $) \cos x -$ $\frac{3\sqrt{3}}{2}$ $\sin x + 4$

$\Rightarrow f(x) =$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $\sin x + 4$

We know

$-$ $\sqrt{ \Big( \frac{13}{2} \Big)^2 + \Big( \frac{-3\sqrt{3}}{2}\Big)^2 }$ $\leq$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $\sin x \leq$ $\sqrt{ \Big( \frac{13}{2} \Big)^2 + \Big( \frac{-3\sqrt{3}}{2} \Big)^2 }$

$-$ $\sqrt{ \frac{169}{4} + \frac{27}{4} }$ $\leq$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $\sin x \leq$ $\sqrt{ \frac{169}{4} + \frac{27}{4} }$

$-$ $\frac{14}{2}$ $\leq$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $\sin x \leq$ $\frac{14}{2}$

$- 7 \leq$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $\sin x \leq 7$

$- 7 + 4 \leq$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $+ 4 \sin x \leq 7 + 4$

$- 3 \leq$ $\frac{13}{2}$ $\cos x -$ $\frac{3\sqrt{3}}{2}$ $+ 4 \sin x \leq 11$

Hence minimum and maximum value of $5 \cos x + 3 \sin \Big($ $\frac{\pi}{4}$ $- x \Big) + 4$ are $-3$ and $11$ respectively.

iv) Given $\sin x - \cos x + 1$

Let $f(x) = \sin x - \cos x$

We know that $-\sqrt{(-1)^2 + (1)^2} \leq f(x) \leq \sqrt{(-1)^2 + (1)^2}$

$\Rightarrow -\sqrt{2} \leq f(x) \leq \sqrt{2}$

$\Rightarrow - \sqrt{2} \leq f(x) \leq \sqrt{2}$

$\Rightarrow - \sqrt{2} + 1 \leq f(x)+ 1 \leq \sqrt{2} + 1$

$\Rightarrow 1 - \sqrt{2} \leq \sin x - \cos x + 1 \leq \sqrt{2} + 1$

Hence minimum and maximum value of $\sin x - \cos x + 1$ are $1 - \sqrt{2}$ and $\sqrt{2} + 1$ respectively.

$\\$

Question 2: Reduce each of the following expressions to the sine and cosine of a single expression: i) $\sqrt{3} \sin x - \cos x$  ii) $\cos x - \sin x$  iii) $24 \cos x + 7 \sin x$

i) $f(x) = \sqrt{3} \sin x - \cos x$

Multiplying and dividing by $\sqrt{(\sqrt{3})^2 + (-1)^2}$ we get

$f(x) = \sqrt{(\sqrt{3})^2 + (-1)^2} \Big[$ $\frac{\sqrt{3} \sin x}{\sqrt{(\sqrt{3})^2 + (-1)^2}}$ $-$ $\frac{\cos x}{\sqrt{(\sqrt{3})^2 + (-1)^2}}$ $\Big]$

$\Rightarrow f(x) = \sqrt{(3+1)} \Big[$ $\frac{\sqrt{3} \sin x}{\sqrt{(3+1)}}$ $-$ $\frac{\cos x}{\sqrt{(3+1)}}$ $\Big]$

$\Rightarrow f(x) = 2 \Big[$ $\frac{\sqrt{3} \sin x}{2}$ $-$ $\frac{\cos x}{2}$ $\Big]$

$\Rightarrow f(x) = 2 \Big[$ $\frac{\sqrt{3}}{2}$ $\sin x -$ $\frac{1}{2}$ $\cos x \Big]$

$\Rightarrow f(x) = 2 \Big[ \sin x \cos$ $\frac{\pi}{6}$ $- \cos x \cos$ $\frac{\pi}{6}$ $\Big]$

$\Rightarrow f(x) = 2 \sin (x -$ $\frac{\pi}{6}$ $)$

Also

$f(x) = 2 \Big[$ $\frac{\sqrt{3} \sin x}{2}$ $-$ $\frac{\cos x}{2}$ $\Big]$

$\Rightarrow f(x) = - 2 \Big[$ $\frac{1}{2}$ $\cos x -$ $\frac{\sqrt{3}}{2}$ $\sin x \Big]$

$\Rightarrow f(x) = - 2 \Big[ \cos$ $\frac{\pi}{3}$ $\cos x - \cos$ $\frac{\pi}{3}$ $\sin x \Big]$

$\Rightarrow f(x) = -2 \cos (x +$ $\frac{\pi}{3}$ $)$

ii) $f(x) = \cos x - \sin x$

Multiplying and dividing by $\sqrt{(1)^2 + (1)^2}$ we get

$f(x) = \sqrt{(1)^2 + (1)^2} \Big[$ $\frac{ \cos x}{\sqrt{(1)^2 + (1)^2}}$ $-$ $\frac{\sin x}{\sqrt{(1)^2 + (1)^2}}$ $\Big]$

$\Rightarrow f(x) = \sqrt{(1+1)} \Big[$ $\frac{ \cos x}{\sqrt{(1+1)}}$ $-$ $\frac{\sin x}{\sqrt{(1+1)}}$ $\Big]$

$\Rightarrow f(x) = \sqrt{2} \Big[$ $\frac{ \cos x}{\sqrt{2}}$ $-$ $\frac{\sin x}{\sqrt{2}}$ $\Big]$

$\Rightarrow f(x) = \sqrt{2} \Big[$ $\frac{1}{\sqrt{2}}$ $\cos x -$ $\frac{1}{\sqrt{2}}$ $\sin x$ $\Big]$

$\Rightarrow f(x) = \sqrt{2} \Big[ \sin$ $\frac{\pi}{4}$ $\cos x - \cos$ $\frac{\pi}{4}$ $\sin x$ $\Big]$

$\Rightarrow f(x) = \sqrt{2} \sin ($ $\frac{\pi}{4}$ $- x )$

Similarly,

$\Rightarrow f(x) = \sqrt{2} \Big[$ $\frac{1}{\sqrt{2}}$ $\cos x -$ $\frac{1}{\sqrt{2}}$ $\sin x$ $\Big]$

$\Rightarrow f(x) = \sqrt{2} \Big[ \cos$ $\frac{\pi}{4}$ $\cos x - \sin$ $\frac{\pi}{4}$ $\sin x$ $\Big]$

$\Rightarrow f(x) = \sqrt{2} \cos ($ $\frac{\pi}{4}$ $+ x )$

iii) $f(x) = 24 \cos x +7 \sin x$

Multiplying and dividing by $\sqrt{(24)^2 + (7)^2}$ we get

$f(x) = \sqrt{(24)^2 + (7)^2} \Big[$ $\frac{ \cos x}{\sqrt{(24)^2 + (7)^2}}$ $-$ $\frac{\sin x}{\sqrt{(24)^2 + (7)^2}}$ $\Big]$

$\Rightarrow f(x) = 25 \Big[$ $\frac{ 24}{25}$ $\cos x +$ $\frac{7 }{25}$ $\sin x \Big]$

Let $\sin \alpha =$ $\frac{24}{25}$ and $\cos \alpha$ $\frac{7}{25}$

$\therefore f(x) = 25 [\sin \alpha \cos x + \cos \alpha \sin x ]$

$f(x) = 25 \sin ( x + \alpha)$   where $\tan \alpha =$ $\frac{\sin \alpha }{\cos \alpha}$ $=$ $\frac{24}{7}$

Similarly,

$\Rightarrow f(x) = 25 \Big[$ $\frac{ 24}{25}$ $\cos x +$ $\frac{7 }{25}$ $\sin x \Big]$

Let $\cos \alpha =$ $\frac{24}{25}$ and $\sin \alpha$ $\frac{7}{25}$

$\therefore f(x) = 25 [\cos \alpha \cos x + \sin \alpha \sin x ]$

$f(x) = 25 \sin ( x - \alpha)$   where $\tan \alpha =$ $\frac{\sin \alpha }{\cos \alpha}$ $=$ $\frac{7}{24}$

$\\$

Question 3: Show that $\sin 100^o - \sin 10^o$ is positive

Given $\sin 100^o - \sin 10^o$

$= \sqrt{2} \Big[$ $\frac{1}{\sqrt{2}}$ $\sin 100^o -$ $\frac{1}{\sqrt{2}}$ $\sin 10^o \Big]$

$= \sqrt{2} \Big[ \sin 45^o \sin 100^o - \cos 45^o \sin 10^o \Big]$

$= \sqrt{2} \sin (100^o - 45^o)$

$= \sqrt{2} \sin 55$ which is a positive number as $\sin 55^o$ is in first quadrant.

$\\$

Question 4: Prove that $(2\sqrt{3} + 3) \sin x + 2 \sqrt{3} \cos x$ lies between $-(2\sqrt{3}+ \sqrt{15})$ and $(2\sqrt{3}+ \sqrt{15})$

Given $(2\sqrt{3} + 3) \sin x + 2 \sqrt{3} \cos x$

Assume $a = (2\sqrt{3} + 3)$ and $b = 2 \sqrt{3}$

$\sqrt{ a^2 +b^2} = \sqrt{ (2\sqrt{3} + 3)^2 + (2 \sqrt{3})^2 }= \sqrt{12+ 9 + 12\sqrt{3} + 12 }= \sqrt{33+ 12\sqrt{3}}$

Dividing and multiplying by $\sqrt{33+ 12\sqrt{3}}$ we get

$\sqrt{33+ 12\sqrt{3}}$ $\Big[$ $\frac{(2\sqrt{3} + 3) \sin x}{\sqrt{33+ 12\sqrt{3}} }$ $+$ $\frac{2 \sqrt{3} \cos x}{\sqrt{33+ 12\sqrt{3}} }$ $\Big]$

Assume $\tan \phi =$ $\frac{a}{b}$ we get $\sin \phi =$ $\frac{a}{\sqrt{a^2 + b^2}}$, $\cos \phi =$ $\frac{b}{\sqrt{a^2 + b^2}}$

$\Rightarrow \sqrt{33+ 12\sqrt{3}} \ \ ( \sin \phi \sin x + \cos \phi \cos x) = \sqrt{33+ 12\sqrt{3}} \ \ \cos ( x - \phi)$

We know that the max. and minimum value of any cosine is $+ 1$ and $-1$

We know, $\sqrt{33+ 12\sqrt{3}} = \sqrt{15+12+6+12\sqrt{3}}$

We know $12 \sqrt{3} + 6 < 12 \sqrt{5}$ because value of $\sqrt{5} -\sqrt{3}$ is more that $0.5$

So we replace  $12 \sqrt{3} + 6$ with $12 \sqrt{5}$. The above inequality still holds true.

So range of above expression can be $\sqrt{15+12+12\sqrt{5}} = 2\sqrt{3}+ \sqrt{15}$

$-(2\sqrt{3}+ \sqrt{15}) < \sqrt{33+ 12\sqrt{3}} \ \ \cos ( x - \phi) < (2\sqrt{3}+ \sqrt{15})$