Trigonometric Equations

Applying sine rule in $\triangle P Q R$,

$ \begin{aligned} & \frac{\alpha}{\sin 30^{\circ}}=\frac{1}{\sin 80^{\circ}} \\\\ \Rightarrow & \alpha=\frac{1}{2 \sin 80^{\circ}} \end{aligned} $

Applying sine rule in $\triangle P R S$,

$ \begin{aligned} & \frac{\beta}{\sin 40^{\circ}}=\frac{1}{\sin \theta} \\\\ \Rightarrow \beta \sin \theta=& \sin 40^{\circ} \end{aligned} $

From (i) and (ii),

$ 4 \alpha \beta \sin \theta=4 \cdot \frac{1}{2 \sin 80^{\circ}} \cdot \sin 40^{\circ}=\frac{1}{\cos 40^{\circ}} $

Using $\cos 0^{\circ}=1, \cos 30^{\circ}=\frac{\sqrt{3}}{2}, \cos 45^{\circ}=\frac{1}{\sqrt{2}}$ and $\cos 60^{\circ}=\frac{1}{2}$

Only Options (A) and (B) are correct.

It is given, that for non-negative integers 'n',

$f(n) = {{\sum\limits_{k = 0}^n {\sin \left( {{{k + 1} \over {n + 2}}\pi } \right)} \sin \left( {{{k + 2} \over {n + 2}}\pi } \right)} \over {\sum\limits_{k = 0}^n {{{\sin }^2}\left( {{{k + 1} \over {n + 2}}\pi } \right)} }} $

$= {{\sum\limits_{k = 0}^n {\left( {\cos {\pi \over {n + 2}} - \cos \left( {{{2k + 3} \over {n + 2}}\pi } \right)} \right)} } \over {\sum\limits_{k = 0}^n {\left( {1 - \cos \left( {{{2k + 2} \over {n + 2}}\pi } \right)} \right)} }}$

[$ \because $ $2\sin A\sin B = \cos (A - B) - \cos (A + B)\,and\,2si{n^2}A = 1 - \cos 2A$]

$={{\left( {\cos \left( {{\pi \over {n + 2}}} \right)} \right)\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{3\pi } \over {n + 2}} + \cos {{5\pi } \over {n + 2}} + \cos {{7\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 3} \over {n + 2}}\pi } \right)} \right\}} } \over {\sum\limits_{k = 0}^n {1 - \left\{ {\cos {{2\pi } \over {n + 2}} + \cos {{4\pi } \over {n + 2}} + \cos {{6\pi } \over {n + 2}} + ... + \cos \left( {{{2n + 2} \over {n + 2}}\pi } \right)} \right\}} }}$

$={{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)} \over {(n + 1) - {{\sin \left( {{{n\pi } \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{{n + 3} \over {n + 2}}\pi } \right)}}$

[$ \because $ $\cos (\alpha ) + \cos (\alpha + \beta ) + cos(\alpha + 2\beta ) + ... + cos(\alpha + (n - 1)\beta ) = {{\sin \left( {{{n\beta } \over 2}} \right)} \over {\sin \left( {{\beta \over 2}} \right)}}\cos \left. {\left( {{{2\alpha + (n - 1)\beta } \over 2}} \right)} \right]$

$ = {{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) - {{\sin \left( {\pi - {\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {\pi + {\pi \over {n + 2}}} \right)} \over {(n + 1) - {{\sin \left( {\pi - {\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos (\pi )}}$

$ = {{(n + 1)\cos \left( {{\pi \over {n + 2}}} \right) + {{\sin \left( {{\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 1) + {{\sin \left( {{\pi \over {n + 2}}} \right)} \over {\sin \left( {{\pi \over {n + 2}}} \right)}}}}$

$ = {{(n + 2)\cos \left( {{\pi \over {n + 2}}} \right)} \over {(n + 2)}} = \cos \left( {{\pi \over {n + 2}}} \right)$

$ \Rightarrow f(n) = \cos \left( {{\pi \over {n + 2}}} \right)$

Now, $f(6) = \cos \left( {{\pi \over 8}} \right)$

$ \because $ $\alpha = \tan ({\cos ^{ - 1}}f((6))) = \tan \left( {{\pi \over 8}} \right)$

$\left\{ \matrix{ {\cos ^{ - 1}}\cos x = x \hfill \cr if\,x \in \left( {0,\,{\pi \over 2}} \right) \hfill \cr} \right\}$

$ = \sqrt 2 - 1$

$ \Rightarrow (\alpha + 1) = \sqrt 2 \Rightarrow {(\alpha + 1)^2} = 2 \Rightarrow {\alpha ^2} + 2\alpha + 1 = 2$

$ \Rightarrow {\alpha ^2} + 2\alpha - 1 = 0$

Now, $f(4) = \cos \left( {{\pi \over {4 + 2}}} \right) = \cos \left( {{\pi \over 6}} \right) = {{\sqrt 3 } \over 2}$,

Now, $\sin (7{\cos ^{ - 1}}f(5)) = \sin \left( {7{{\cos }^{ - 1}}\left( {\cos \left( {{\pi \over {5 + 2}}} \right)} \right)} \right) = \sin \left( {7\left( {{\pi \over 7}} \right)} \right) = \sin \pi = 0$

and Now, $\mathop {\lim }\limits_{n \to \infty } f(x) = \mathop {\lim }\limits_{n \to \infty } \cos {\pi \over {n + 2}} = \cos 0 = 1$

Hence, options (a), (b) and (c) are correct.

We have,

In $\Delta $PQR

$\angle PQR = 30^\circ $

$PQ = 10\sqrt 3 $

$QR = 10$



By cosine rule

$\cos 30^\circ = {{P{Q^2} + Q{R^2} - P{R^2}} \over {2PQ.QR}}$

$ \Rightarrow {{\sqrt 3 } \over 2} = {{300 + 100 - P{R^2}} \over {200\sqrt 3 }}$

$ \Rightarrow 300 = 300 + 100 - P{R^2}$

$ \Rightarrow PR = 10$

Since, $PR = QR = 10$

$ \therefore $ $\angle QPR = 30^\circ $ and $\angle QRP = 120^\circ $

Area of $\Delta PQR = {1 \over 2}PQ.QR.\sin 30^\circ $

$ = {1 \over 2} \times 10\sqrt 3 \times 10 \times {1 \over 2} = 25\sqrt 3 $

Radius of incircle of

$\Delta PQR = {{Area\,of\,\Delta PQR} \over {Semi - perimetre\,of\,\Delta PQR}}$

i.e. $r = {\Delta \over s} = {{25\sqrt 3 } \over {{{10\sqrt 3 + 10 + 10} \over 2}}} = {{25\sqrt 3 } \over {5(\sqrt 3 + 2)}}$

$ \Rightarrow r = 5\sqrt 3 (2 - \sqrt 3 ) = 10\sqrt 3 - 15$

and radius of circumcircle

$(R) = {{abc} \over {4\Delta }} = {{10\sqrt 3 \times 10 \times 10} \over {4 \times 25\sqrt 3 }} = 10$

$ \therefore $ Area of circumcircle of

$\Delta PQR = \pi {R^2} = 100\pi $

Hence, option (b), (c) and (d) are correct answer.

$2(\cos \beta - \cos \alpha ) + \cos \alpha \cos \beta = 1$

or $4(\cos \beta - \cos \alpha ) + 2\cos \alpha \cos \beta = 2$

$ \Rightarrow 1 - \cos \alpha + \cos \beta - \cos \alpha \cos \beta $

$ = 3 + 3\cos \alpha - 3\cos \beta - 3\cos \alpha \cos \beta $

$ \Rightarrow (1 - \cos \alpha )(1 + \cos \beta )$

$ = 3(1 + \cos \alpha )(1 - \cos \beta )$

$ \Rightarrow {{(1 - \cos \alpha )} \over {(1 + \cos \alpha )}} = {{3(1 - \cos \beta )} \over {1 + \cos \beta }}$

$ \Rightarrow {\tan ^2}{\alpha \over 2} = 3{\tan ^2}{\beta \over 2}$

$ \therefore $ $\tan {\alpha \over 2} \pm \sqrt 3 \tan {\beta \over 2} = 0$

Given, ,$f(x)=x \cdot \sin \pi x, x>0$

$\Rightarrow \quad f^{\prime}(x)=1 \cdot \sin \pi x+x \cdot \pi \cos \pi x$

Apply $f^{\prime}(x)=0$

$\begin{aligned} & \Rightarrow \sin \pi x+\pi x \cos \pi x=0 \\ & \Rightarrow \sin \pi x=-\pi x \cos \pi x \\ & \Rightarrow \frac{\sin \pi x}{\cos \pi x}=-\pi x \\ & \Rightarrow \tan \pi x=-\pi x \end{aligned}$

Draw the graph of $y=\tan \pi x$ and $y=-\pi x$ for $x>0$

It is clear that $y=\tan \pi x$ and $y=-\pi x$ intersect at a unique point if $\frac{1}{2} < x < 1$ as $\frac{3}{2} < x < 2$ as $\frac{5}{2} < x < 3$ or.....

So, $y=\tan \pi x$ and $y=-\pi x$ intersect at a unique point is $x \in\left(n+\frac{1}{2}, n+1\right)$ as $(n, n+1)$.

Hints:

The number of point of intersection of graphs of $y=f(x)$ and $y=g(x)$ is equal to the number of solution of $f(x)=g(x)$

Given, $\tan (2 \pi-\theta)>0$

$\begin{aligned} & \Rightarrow 0<(2 \pi-\theta)<\frac{\pi}{2} \text { or } \pi<(2 \pi-\theta)<\frac{3 \pi}{2} \\ & \Rightarrow \frac{3 \pi}{2}<\theta<2 \pi \text { or } \frac{\pi}{2}<\theta<\pi \quad \text{... (i)} \end{aligned}$

Also, $-1<\sin \theta<\frac{-\sqrt{3}}{2}$

$\Rightarrow \frac{3 \pi}{2}<\theta<\frac{5 \pi}{3} \quad \text{... (ii)}$

From (i) and (ii),

$\Rightarrow \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\quad \text{... (iii)}$

Now,

$\begin{aligned} & 2 \cos \theta(1-\sin \phi)=\sin ^2 \theta\left(\tan \frac{\theta}{2}+\cot \frac{\theta}{2}\right) \cos \phi-1 \\ & \Rightarrow 2 \cos \theta(1-\sin \phi) \\ & =\sin ^2 \theta\left(\frac{\sin \frac{\theta}{2}}{\cos \frac{\theta}{2}}+\frac{\cos \frac{\theta}{2}}{\sin \frac{\theta}{2}}\right) \cos \phi-1 \\ & \Rightarrow 2 \cos \theta(1-\sin \phi) \\ & =\sin ^2 \theta\left(\frac{\sin ^2 \frac{\theta}{2}+\cos ^2 \frac{\theta}{2}}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad 2 \cos \theta(1-\sin \phi) \\ &=\sin \theta \cdot \sin \theta\left(\frac{1}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}\right) \cos \phi-1 \end{aligned}$

$\begin{aligned} & \Rightarrow \quad 2 \cos (1-\sin \phi)=\frac{2 \sin \frac{\theta}{2} \cos \frac{\theta}{2} \cdot \cos \phi \cdot \sin \theta}{\sin \frac{\theta}{2} \cos \frac{\theta}{2}}-1 \\ & \Rightarrow \quad 2 \cos \theta(1-\sin \phi)=2 \sin \theta \cos \phi-1 \\ & \Rightarrow \quad 2 \cos \theta+1=2 \sin (\theta+\phi) \\ & \text { As } \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right) \quad \text{[from (iii)]}\\ & \Rightarrow 2 \cos \theta+1 \in(1,2) \\ & \Rightarrow 1<2 \sin (\theta+\phi)<2 \\ & \Rightarrow \frac{1}{2}<\sin (\theta+\phi)<1 \end{aligned}$

$\begin{aligned} & \text { Now, } \theta, \phi \in[0,2 \pi] \\ & \theta+\phi \in[0,4 \pi] \\ & \Rightarrow \theta+\phi \in\left(\frac{\pi}{6}, \frac{5 \pi}{6}\right) \text { or } \theta+\phi \in\left(\frac{13 \pi}{6}, \frac{17 \pi}{6}\right) \\ & \Rightarrow \frac{\pi}{6}-\theta<\phi<\frac{5 \pi}{6}-\theta \text { or } \frac{13 \pi}{6}-\theta<\phi<\frac{17 \pi}{6}-\theta \\ & \Rightarrow \phi \in\left(-\frac{3 \pi}{2}, \frac{-2 \pi}{3}\right) \cup\left(\frac{2 \pi}{3}, \frac{7 \pi}{6}\right) \\ & {\left[\because \theta \in\left(\frac{3 \pi}{2}, \frac{5 \pi}{3}\right)\right] } \end{aligned}$

(a) We have

$2{\sin ^2}\theta + 4{\sin ^2}\theta {\cos ^2}\theta = 2$

${\sin ^2}\theta + 2{\sin ^2}\theta (1 - {\sin ^2}\theta ) = 1$

$3{\sin ^2}\theta - 2{\sin ^4}\theta - 1 = 0$

$ \Rightarrow \sin \theta = \pm {1 \over {\sqrt 2 }}, \pm 1$

$ \Rightarrow \theta = {\pi \over 4},{\pi \over 2}$

(B) Let $y = {{3x} \over \pi } \Rightarrow {1 \over 2} \le y \le 3\forall x \in \left[ {{\pi \over 4},\pi } \right]$

Now, $f(y) = [2y]\cos [y]$.

The critical points are

$y = {1 \over 2},y = 1,y = {3 \over 2}$ and $y = 3$

$\Rightarrow$ points of discontinuity $\left\{ {{\pi \over 6},{\pi \over 3},{\pi \over 2},\pi } \right\}$.

(C) $\left| {\matrix{ 1 & 1 & 0 \cr 1 & 2 & 0 \cr 1 & 1 & \pi \cr } } \right| = \pi \Rightarrow $ volume of parallelepiped = $\pi$.

(D) We have

$\left| {\overrightarrow a + \overrightarrow b } \right| = \sqrt 3 $

$ \Rightarrow \sqrt {2 + 2\cos \alpha } = \sqrt 3 $

$ \Rightarrow 2 + 2\cos \alpha = 3$

$ \Rightarrow \alpha = {\pi \over 3}$

Given solutions

${1 \over {\sin (\pi /4)}}\left[ {{{\sin (\theta + \pi /4 - \theta )} \over {\sin \theta \,.\,\sin (\theta + \pi /4)}} + {{\sin (\theta + \pi /2 - (\theta + \pi /4))} \over {\sin (\theta + \pi /4)\,.\,(\theta + \pi /2)}}\, + \,...\, + \,{{\sin ((\theta + 3\pi /2) - (\theta + 5\pi /4))} \over {\sin (\theta + 3\pi /2)\,.\,\sin (\theta + 5\pi /4)}}} \right] = 4\sqrt 2 $

$ \Rightarrow \sqrt 2 [\cos \theta - \cot (\theta + \pi /4) + \cot (\theta + \pi /4) - \cot (\theta + \pi /2)\, + \,...\, + \,\cot (\theta + 5\pi /4) - \cot (\theta + 3\pi /2)] = 4\sqrt 2 $

$ \Rightarrow \tan \theta + \cot \theta = 4 \Rightarrow \tan \theta = 2 \pm \sqrt 3 $

$ \Rightarrow \theta = {\pi \over {12}}$ or ${{5\pi } \over {12}}$

It is given that

${{{{\sin }^4}x} \over 2} + {{{{\cos }^4}x} \over 3} = {1 \over 5}$

$3{\sin ^4}x + 2{(1 - {\sin ^2}x)^2} = {6 \over 5}$

$ \Rightarrow 25{\sin ^4}x - 20{\sin ^2}x + 4 = 0$

$ \Rightarrow {\sin ^2}x = {2 \over 5}$ and ${\cos ^2}x = {3 \over 5}$

Hence, ${\tan ^2}x = {2 \over 3}$

Therefore, ${{{{\sin }^8}x} \over 8} + {{{{\cos }^8}x} \over {27}} = {1 \over {125}}$