Trigonometric Equations
The number of elements in the set $\left\{x \in\left[0,180^{\circ}\right]: \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right)\right\}$ is $\_\_\_\_$ .
Explanation:
$ \begin{aligned} &\begin{aligned} & \tan \left(x+100^{\circ}\right)=\tan \left(x+50^{\circ}\right) \tan x \tan \left(x-50^{\circ}\right) ; x \in\left[0,180^{\circ}\right] \\ & \frac{\tan \left(x+100^{\circ}\right)}{\tan x}=\tan \left(x+50^{\circ}\right) \tan \left(x-50^{\circ}\right) \\ & \frac{\sin \left(x+100^{\circ}\right) \cdot \cos x}{\cos \left(x+100^{\circ}\right) \cdot \sin x}=\frac{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)}{\cos \left(x+50^{\circ}\right) \cos \left(x-50^{\circ}\right)} \end{aligned}\\ &\text { Apply Componendo and Dividendo rule }\\ &\begin{aligned} & \frac{\sin \left(x+100^{\circ}\right) \cos x+\cos \left(x+100^{\circ}\right) \sin x}{\sin \left(x+100^{\circ}\right) \cos x-\cos \left(x+100^{\circ}\right) \sin x}=\frac{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)+\cos \left(x-50^{\circ}\right) \cos \left(x+50^{\circ}\right)}{\sin \left(x+50^{\circ}\right) \sin \left(x-50^{\circ}\right)-\cos \left(x+50^{\circ}\right) \cos \left(x-50^{\circ}\right)} \\ & \text { use } \sin A \cos B \pm \cos A \sin B=\sin (A \pm B) \text { and } \cos A \cos B \pm \sin A \sin B=\cos (A \mp B) \\ & \frac{\sin \left(x+100^{\circ}+x\right)}{\sin \left(x+100^{\circ}-x\right)}=\frac{\cos \left[\left(x+50^{\circ}\right)-\left(x-50^{\circ}\right)\right]}{-\cos \left[\left(x+50^{\circ}\right)+\left(x-50^{\circ}\right)\right]} \\ & \frac{\sin \left(2 x+100^{\circ}\right)}{\sin 100^{\circ}}=\frac{\cos 100^{\circ}}{-\cos 2 x} \end{aligned} \end{aligned} $
$ \begin{aligned} & \frac{\sin \left(90^{\circ}+\left(2 x+10^{\circ}\right)\right)}{\sin \left(90^{\circ}+10^{\circ}\right)}=\frac{\cos \left(90^{\circ}+10^{\circ}\right)}{-\cos 2 x} \\ & \frac{\cos \left(2 x+10^{\circ}\right)}{\cos 10^{\circ}}=\frac{-\sin 10^{\circ}}{-\cos 2 x} \\ & \cos \left(2 x+10^{\circ}\right) \cos 2 x=\sin 10^{\circ} \cos 10^{\circ} \\ & 2 \cos \left(2 x+10^{\circ}\right) \cos 2 x=2 \sin 10^{\circ} \cos 10^{\circ} \\ & 2 \cos \left(2 x+10^{\circ}\right) \cos 2 x=\sin 20^{\circ} \\ & \cos \left(4 x+10^{\circ}\right)+\cos 10^{\circ}=\sin 20^{\circ} \\ & \cos \left(4 x+10^{\circ}\right)=\sin 20^{\circ}-\cos 10^{\circ} \\ & \cos \left(4 x+10^{\circ}\right)=\sin 20^{\circ}-\sin 80^{\circ} \quad\left\{\cos 10^{\circ}=\sin \left(90^{\circ}-10^{\circ}\right)\right\} \\ & =2 \cos 50^{\circ} \sin \left(-30^{\circ}\right) \\ & =-2 \cos 50^{\circ} \times \frac{1}{2}=-\cos 50^{\circ}=\cos \left(180^{\circ}-50^{\circ}\right) \\ & \cos \left(4 x+10^{\circ}\right)=\cos 130^{\circ} \end{aligned} $
$ \begin{aligned} & 4 x+10^{\circ}=2 n \pi \pm 130^{\circ}, n \in I \\ & 4 x+10^{\circ}=360^{\circ} n \pm 130^{\circ}, n \in I \\ & 4 x+10^{\circ}=360^{\circ} n+130^{\circ} \text { or } 4 x+10^{\circ}=360^{\circ} n-130^{\circ} \\ & 4 x=360^{\circ} n+120^{\circ} \text { or } 4 x=360^{\circ} n-140^{\circ} \end{aligned} $
$ \begin{aligned} &\begin{aligned} & x=\frac{360^{\circ} n+120^{\circ}}{4} \\ & n=0, x=30^{\circ} \\ & n=1, x=\frac{480^{\circ}}{4}=120^{\circ} \\ & n=2, x=\frac{360^{\circ} \times 2+120^{\circ}}{4} \\ & x=210^{\circ} \end{aligned}\\ &\begin{aligned} & x=\frac{360^{\circ} n-140^{\circ}}{4} \\ & n=1, x=\frac{360^{\circ}-140^{\circ}}{4}=\frac{220^{\circ}}{4}=55^{\circ} \\ & n=2, x=\frac{360^{\circ} \times 2-140^{\circ}}{4}=\frac{580^{\circ}}{4}=145^{\circ} \\ & n=3, x=\frac{360^{\circ} \times 3-140^{\circ}}{4}=\frac{940^{\circ}}{4}=235^{\circ} \end{aligned} \end{aligned} $
for $x \in\left[0,180^{\circ}\right]$ solutions are
$ x=30^{\circ}, 55^{\circ}, 120^{\circ}, 145^{\circ} $
so, total number of solutions are $=4$ Ans
The number of solutions of $\sin ^2 x+\left(2+2 x-x^2\right) \sin x-3(x-1)^2=0$, where $-\pi \leq x \leq \pi$, is ________.
Explanation:
$\begin{aligned} & \sin ^2 x+\left(3-(x-1)^2\right) \sin x-3(x-1)^2=0 \\ & \sin ^2 x+3 \sin x-(x-1)^2 \sin x-3(x-1)^2=0 \\ & \left.\sin x(\sin x+3)-(x-1)^2\right)[\sin x+3]=0 \end{aligned}$
There are two intersections between this graph.
So, Number of solution will be 2 .
Let $S=\left\{\sin ^2 2 \theta:\left(\sin ^4 \theta+\cos ^4 \theta\right) x^2+(\sin 2 \theta) x+\left(\sin ^6 \theta+\cos ^6 \theta\right)=0\right.$ has real roots $\}$. If $\alpha$ and $\beta$ be the smallest and largest elements of the set $S$, respectively, then $3\left((\alpha-2)^2+(\beta-1)^2\right)$ equals __________.
Explanation:
For real roots
$\begin{aligned} & D \geq 0 \\ & \sin ^2 2 \theta \geq 4\left(\sin ^4 \theta+\cos ^4 \theta\right)\left(\sin ^6 \theta+\cos ^6 \theta\right) \end{aligned}$
Put $\sin ^2 2 \theta=t$
$\begin{aligned} & \Rightarrow t \geq 4\left(1-\frac{t}{2}\right)\left(1-\frac{3 t}{4}\right) \\ & 2 t \geq(2-t)(4-3 t) \\ & 3 t^2-12 t+8 \leq 0 \\ & t^2-4 t+\frac{8}{3} \leq 0 \\ & (t-2)^2+\frac{8}{3}-4 \leq 0 \\ & (t-2)^2 \leq \frac{4}{3} \\ & -\frac{2}{\sqrt{3}} \leq t-2 \leq \frac{2}{\sqrt{3}} \\ & 2-\frac{2}{\sqrt{3}} \leq t \leq 2+\frac{2}{\sqrt{3}} \\ & \because t \in[0,1] \\ & \Rightarrow 2-\frac{2}{\sqrt{3}} \leq t \leq 1 \\ & \alpha=2-\frac{2}{\sqrt{3}}, \beta=1 \\ & \Rightarrow 3\left[(\alpha-2)^2+(\beta-1)^2\right]=4 \end{aligned}$
If m and n respectively are the numbers of positive and negative values of $\theta$ in the interval $[-\pi,\pi]$ that satisfy the equation $\cos 2\theta \cos {\theta \over 2} = \cos 3\theta \cos {{9\theta } \over 2}$, then mn is equal to ____________.
Explanation:
$ \begin{aligned} & \cos \frac{5 \theta}{2}+\cos \frac{3 \theta}{2}=\cos \frac{15 \theta}{2}+\cos \frac{3 \theta}{2} \\\\ & \cos \frac{5 \theta}{2}=\cos \frac{15 \theta}{2} \\\\ & \frac{15 \theta}{2}=2 n \pi \pm \frac{5 \theta}{2} \\\\ & \frac{15 \theta}{2} \pm \frac{5 \theta}{2}=2 n \pi \\\\ & 10 \theta=2 n \pi \quad \text { or } 5 \theta=2 n \pi \\\\ & \theta=\frac{n \pi}{5} \text { or } \theta=\frac{2 n \pi}{5} \\\\ & \Rightarrow \theta=\frac{n \pi}{5} \end{aligned} $
$ \begin{aligned} & \theta=\pm \pi, \pm \frac{4 \pi}{5}, \pm \frac{3 \pi}{5}, \pm \frac{2 \pi}{5}, \pm \frac{\pi}{5} \\\\ & m=5, \quad n=5 \\\\ & m n=25 \end{aligned} $
Let $\mathrm{S = \{ \theta \in [0,2\pi ):\tan (\pi \cos \theta ) + \tan (\pi \sin \theta ) = 0\}}$. Then $\sum\limits_{\theta \in S} {{{\sin }^2}\left( {\theta + {\pi \over 4}} \right)} $ is equal to __________.
Explanation:
possible values are $\mathrm{n}=0,1$ and $-1$ because
$ -\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2} $
Now it gives $\theta \in\left\{0, \frac{\pi}{2}, \frac{3 \pi}{4}, \frac{7 \pi}{4}, \frac{3 \pi}{2}, \pi\right\}$
So $\sum \limits_{\theta \in \mathrm{S}} \sin ^2\left(\theta+\frac{\pi}{4}\right)=2(0)+4\left(\frac{1}{2}\right)=2$
Let $S=\left\{\theta \in(0,2 \pi): 7 \cos ^{2} \theta-3 \sin ^{2} \theta-2 \cos ^{2} 2 \theta=2\right\}$. Then, the sum of roots of all the equations $x^{2}-2\left(\tan ^{2} \theta+\cot ^{2} \theta\right) x+6 \sin ^{2} \theta=0, \theta \in S$, is __________.
Explanation:
$7{\cos ^2}\theta - 3{\sin ^2}\theta - 2{\cos ^2}2\theta = 2$
$ \Rightarrow 4\left( {{{1 + \cos 2\theta } \over 2}} \right) + 3\cos 2\theta - 2{\cos ^2}2\theta = 2$
$ \Rightarrow 2 + 5{\cos ^2}\theta - 2{\cos ^2}2\theta = 2$
$ \Rightarrow \cos 2\theta = 0$ or ${5 \over 2}$(rejected)
$ \Rightarrow \cos 2\theta = 0 = {{1 - {{\tan }^2}\theta } \over {1 + {{\tan }^2}\theta }} \Rightarrow {\tan ^2}\theta = 1$
$\therefore$ Sum of roots $ = 2({\tan ^2}\theta + {\cot ^2}\theta ) = 2 \times 2 = 4$
But as $\tan \theta = \, \pm \,1$ for ${\pi \over 4},{{3\pi } \over 4},\,{{5\pi } \over 4},\,{{7\pi } \over 4}$ in the interval $(0,\,2\pi )$
$\therefore$ Four equations will be formed
Hence sum of roots of all the equations $ = 4 \times 4 = 16.$
Let $S=\left[-\pi, \frac{\pi}{2}\right)-\left\{-\frac{\pi}{2},-\frac{\pi}{4},-\frac{3 \pi}{4}, \frac{\pi}{4}\right\}$. Then the number of elements in the set $\mid A=\{\theta \in S: \tan \theta(1+\sqrt{5} \tan (2 \theta))=\sqrt{5}-\tan (2 \theta)\}$ is __________.
Explanation:
Let $\tan \alpha = \sqrt 5 $
$\therefore$ $\tan \theta = {{\tan \alpha - \tan 2\theta } \over {1 + \tan \alpha \tan 2\theta }}$
$\therefore$ $\tan \theta = \tan (\alpha - 2\theta )$
$\alpha - 2\theta = n\pi + \theta $
$ \Rightarrow 3\theta = \alpha - n\pi $
$ \Rightarrow \theta = {\alpha \over 3} - {{n\pi } \over 3}\,\,\,\,\,\,\,\,\,;\,n \in Z$
If $\theta \in [ - \pi ,\,\pi /2]$ then $n = 0,1,2,3,4$ are acceptable
$\therefore$ 5 solutions.
If the sum of solutions of the system of equations $2 \sin ^{2} \theta-\cos 2 \theta=0$ and $2 \cos ^{2} \theta+3 \sin \theta=0$ in the interval $[0,2 \pi]$ is $k \pi$, then $k$ is equal to __________.
Explanation:
Equation (1) $~~~2{\sin ^2}\theta = 1 - 2{\sin ^2}\theta $
$ \Rightarrow {\sin ^2}\theta = {1 \over 4}$
$ \Rightarrow \sin \theta = \, \pm \,{1 \over 2}$
$ \Rightarrow \theta = {\pi \over 6},{{5\pi } \over 6},{{7\pi } \over 6},{{11\pi } \over 6}$
Equation (2) $~~~2{\cos ^2}\theta + 3\sin \theta = 0$
$ \Rightarrow 2{\sin ^2}\theta - 3\sin \theta - 2 = 0$
$ \Rightarrow 2{\sin ^2}\theta - 4\sin \theta + \sin \theta - 2 = 0$
$ \Rightarrow (\sin \theta - 2)(2\sin \theta + 1) = 0$
$ \Rightarrow \sin \theta = {{ - 1} \over 2}$
$ \Rightarrow \theta = {{7\pi } \over 6},{{11\pi } \over 6}$
$\therefore$ Common solutions $ = {{7\pi } \over 6};\,{{11\pi } \over 6}$
Sum of solutions $ = {{7\pi + 11\pi } \over 6} = {{18\pi } \over 6} = 3\pi $
$\therefore$ $k = 3$
Let ${S_1} = \{ x \in [0,12\pi ]:{\sin ^5}x + {\cos ^5}x = 1\} $
and ${S_2} = \{ x \in [0,8\pi ]:{\sin ^7}x + {\cos ^7}x = 1\} $
Then $n({S_1}) - n({S_2})$ is equal to ______________.
Explanation:
Given,
${S_1} = \left\{ {x \in \left[ {0,12\pi } \right],\,{{\sin }^5}x + {{\cos }^5}x = 1} \right\}$
${S_2} = \left\{ {x \in \left[ {0,8\pi } \right],\,{{\sin }^7}x + {{\cos }^7}x = 1} \right\}$
(1) ${\sin ^5}x + {\cos ^5}x = 1$
This satisfies when $\sin x = 1$ and $\cos x = 0$
$\therefore$ $x = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2},{{17\pi } \over 2},{{21\pi } \over 2}$
It also satisfies when $\sin x = 0$ and $\cos x = 1$
$\therefore$ $x = 0,\,2\pi ,4\pi ,6\pi ,8\pi ,10\pi ,12\pi $
$\therefore$ Accepted values of x in $\left[ {0,12\pi } \right]$ is = 13
$\therefore$ $n({S_1}) = 13$
(2) ${\sin ^7}x + {\cos ^7}x = 1$
This satisfies when $\sin x = 1$ and $\cos x = 0$
For $x \in \left[ {0,\,8\pi } \right]$, possible values
$x = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$
It also satisfies when $\sin x = 0$ and $\cos x = 1$
$x \in \left[ {0,8\pi } \right]$, possible values
$x = 0,2\pi ,4\pi ,6\pi ,8\pi $
$\therefore$ Total accepted values of x in $\left[ {0,8\pi } \right]$ is = 9
$\therefore$ $n({S_2}) = 9$
$\therefore$ $n({S_1}) - n({S_2}) = 13 - 9 = 4$
The number of solutions of the equation $\sin x = {\cos ^2}x$ in the interval (0, 10) is _________.
Explanation:
$\sin x = {\cos ^2}x$, $x \in (0,10)$
$ \Rightarrow \sin x = 1 - {\sin ^2}x$
$ \Rightarrow {\sin ^2}x + \sin x - 1 = 0$
$\therefore$ $\sin x = {{ - 1 \pm \,\sqrt {1 + 4} } \over 2}$
$ \Rightarrow \sin x = {{ - 1 \pm \,\sqrt 5 } \over 2}$
We know $\sin \in ( - 1,1)$
$\therefore$ ${{ - 1 - \sqrt 5 } \over 2}$ can't be a value of sin x
$\therefore$ $\sin x = {{\sqrt 5 - 1} \over 2}$
$3\pi = 3 \times 3.14 = 9.42 < 10$
${{7\pi } \over 2} = {7 \over 2} \times 3.14 = 10.99 > 10$
$\therefore$ 10 will be in between 3$\pi$ and ${{7\pi } \over 2}$.
There are 4 intersection at A, B, C and D between sin x graph and $y = {{\sqrt 5 - 1} \over 2}$ graph.
$\therefore$ possible solution = 4
The number of elements in the set $S = \{ \theta \in [ - 4\pi ,4\pi ]:3{\cos ^2}2\theta + 6\cos 2\theta - 10{\cos ^2}\theta + 5 = 0\} $ is __________.
Explanation:
$\Rightarrow 3 \cos ^{2} 2 \theta+\cos 2 \theta=0$
$\Rightarrow \cos 2 \theta=0$ or $\cos 2 \theta=\frac{-1}{3}$
As $\theta \in[0, \pi], \cos 2 \theta=\frac{-1}{3} \Rightarrow 2$ times
$\Rightarrow \theta \in[-4 \pi, 4 \pi], \cos 2 \theta=\frac{-1}{3} \Rightarrow 16$ times
Similarly, $\cos 2 \theta=0 \Rightarrow 16$ times
$\therefore $ Total 32 solutions
The number of solutions of the equation
$2\theta - {\cos ^2}\theta + \sqrt 2 = 0$ in R is equal to ___________.
Explanation:
Given,
$2\theta - {\cos ^2}\theta + \sqrt 2 = 0$
$ \Rightarrow 2\theta + \sqrt 2 = {\cos ^2}\theta $
$ \Rightarrow 2\theta + \sqrt 2 = {{1 + \cos 2\theta } \over 2}$
$ \Rightarrow 4\theta + 2\sqrt 2 = 1 + \cos 2\theta = y$ (Assume)
$\therefore$ $y = 4\theta + 2\sqrt 2 $ and
$y = 1 + \cos 2\theta $
For $y = 1 + \cos 2\theta $
when $\theta = 0$, $y = 1 + 1 = 2$
when $\theta = {\pi \over 4}$, $y = 1 + \cos {\pi \over 2} = 1$
$\theta = {\pi \over 2}$, $y = 1 + \cos \pi = 1 - 1 = 0$
For $y = 4\theta + 2\sqrt 2 $
when $\theta = 0$, $y = 2\sqrt 2 $
when $\theta = {\pi \over 2}$, $y = 2\pi + 2\sqrt 2 $
$ = 2(\pi + \sqrt 2 )$
$ = 2(3.14 + 1.41)$
$ = 2(4.55)$
$ = 9.1$
when $\theta = - {\pi \over 2}$, $y = - 2\pi + 2\sqrt 2 $
$ = 2( - \pi + \sqrt 2 )$
$ = 2( - 3.14 + 1.41)$
$ = - 3.46$
$\therefore$ Two graph cut's at only one point so one solution possible.
The number of values of x in the interval $\left( {{\pi \over 4},{{7\pi } \over 4}} \right)$ for which
$14\cos e{c^2}x - 2{\sin ^2}x = 21 - 4{\cos ^2}x$ holds, is ____________.
Explanation:
${{14} \over {{{\sin }^2}x}} - 2{\sin ^2}x = 21 - 4(1 - {\sin ^2}x)$
Let ${\sin ^2}x = t$
$ \Rightarrow 14 - 2{t^2} = 21t - 4t + 4{t^2}$
$ \Rightarrow 6{t^2} + 17t - 14 = 0$
$ \Rightarrow 6{t^2} + 21t - 4t - 14 = 0$
$ \Rightarrow 3t(2t + 7) - 2(2t + 7) = 0$
$ \Rightarrow {\sin ^2}x = {2 \over 3}$ or $ - {7 \over 3}$ (rejected)
$ \Rightarrow \sin x = \, \pm \,\sqrt {{2 \over 3}} $
$\therefore$ $\sin x = \, \pm \,\sqrt {{2 \over 3}} $ has 4 solutions in $\left( {{\pi \over 4},\,{{7\pi } \over 4}} \right)$
Explanation:
${\sin ^4}\theta + {\cos ^4}\theta - \sin \theta \cos \theta = 0$
$ \Rightarrow 1 - {\sin ^2}\theta {\cos ^2}\theta - \sin \theta \cos \theta = 0$
$ \Rightarrow 2 - {(\sin 2\theta )^2} - \sin 2\theta = 0$
$ \Rightarrow {(\sin 2\theta )^2} + (\sin 2\theta ) - 2 = 0$
$ \Rightarrow (\sin 2\theta + 2)(\sin 2\theta - 1) = 0$
$ \Rightarrow \sin 2\theta = 1$ or $\sin 2\theta = - 2$ (Not Possible)
$ \Rightarrow 2\theta = {\pi \over 2},{{5\pi } \over 2},{{9\pi } \over 2},{{13\pi } \over 2}$
$ \Rightarrow \theta = {\pi \over 4},{{5\pi } \over 4},{{9\pi } \over 4},{{13\pi } \over 4}$
$ \Rightarrow S = {\pi \over 4} + {{5\pi } \over 4} + {{9\pi } \over 4} + {{13\pi } \over 4} = 7\pi $
$ \Rightarrow {{8S} \over \pi } = {{8 \times 7\pi } \over \pi } = 56.00$
$|\cot x| = \cot x + {1 \over {\sin x}}$ in the interval [ 0, 2$\pi$ ] is
Explanation:
$\cot x = \cot x + {1 \over {\sin x}} \Rightarrow $ not possible
Case II : When cot x < 0, $x \in \left[ {{\pi \over 2},\pi } \right] \cup \left[ {{{3\pi } \over 2},2\pi } \right]$
$ - \cot x = \cot x + {1 \over {\sin x}}$
$ \Rightarrow {{ - 2\cos x} \over {\sin x}} = {1 \over {\sin x}}$
$ \Rightarrow \cos x = {{ - 1} \over 2}$
$ \Rightarrow x = {{2\pi } \over 3},{{4\pi } \over 3}$(Rejected)
One solution.
Explanation:
$ \Rightarrow $ $\sqrt 3 {\cos ^2}x - \sqrt 3 \cos x + \cos x - 1 = 0$
$ \Rightarrow \sqrt 3 \cos x(\cos x - 1) + (\cos x - 1) = 0$
$ \Rightarrow (\cos x - 1)(\sqrt 3 \cos x + 1) = 0$
$\cos x = 1$
$ \Rightarrow x = 0$ $ [as x \in \left[ {0,{\pi \over 2}} \right]$]
and $\cos x = - {1 \over {\sqrt 3 }}$ (not possible in $x \in \left[ {0,{\pi \over 2}} \right]$]
$ \therefore $ Number of solution = 1
Let
$ \alpha=\frac{1}{\sin 60^{\circ} \sin 61^{\circ}}+\frac{1}{\sin 62^{\circ} \sin 63^{\circ}}+\cdots+\frac{1}{\sin 118^{\circ} \sin 119^{\circ}} $
Then the value of
$ \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 $
is _____________.
Explanation:
To solve for the value of $\left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2$, we begin by considering the expression for $\alpha$:
$ \alpha = \frac{1}{\sin 60^{\circ} \cdot \sin 61^{\circ}} + \frac{1}{\sin 62^{\circ} \cdot \sin 63^{\circ}} + \cdots + \frac{1}{\sin 118^{\circ} \cdot \sin 119^{\circ}} $
Using the identity for the product of sines:
$ \sin(x-y) = \sin x \cdot \cos y - \cos x \cdot \sin y $
we can transform each term:
$ \sin 1^{\circ} \cdot \alpha = \frac{\sin(61^{\circ} - 60^{\circ})}{\sin 60^{\circ} \cdot \sin 61^{\circ}} + \frac{\sin(63^{\circ} - 62^{\circ})}{\sin 62^{\circ} \cdot \sin 63^{\circ}} + \dots + \frac{\sin(119^{\circ} - 118^{\circ})}{\sin 118^{\circ} \cdot \sin 119^{\circ}} $
This simplifies to:
$ \sin 1^{\circ} \cdot \alpha = \cot 60^{\circ} - \cot 61^{\circ} + \cot 62^{\circ} - \cot 63^{\circ} + \cdots + \cot 118^{\circ} - \cot 119^{\circ} $
This forms a telescoping series, simplifying to:
$ \sin 1^{\circ} \cdot \alpha = \cot 60^{\circ} $
Therefore:
$ \alpha = \frac{\cot 60^{\circ}}{\sin 1^{\circ}} = \frac{1/\sqrt{3}}{\sin 1^{\circ}} = \frac{\operatorname{cosec} 1^{\circ}}{\sqrt{3}} $
Finally, we calculate:
$ \left(\frac{\operatorname{cosec} 1^{\circ}}{\alpha}\right)^2 = \left(\frac{\operatorname{cosec} 1^{\circ}}{\frac{\operatorname{cosec} 1^{\circ}}{\sqrt{3}}}\right)^2 = 3 $
Explanation:
$ \begin{aligned} r & =\frac{a}{\sin \left(\frac{\pi}{2}-2 C\right) \sin \left(\frac{\pi}{2}+C\right)+\sin C} \\\\ & =\frac{a}{\cos 2 C+\cos C+\sin C} \\\\ & =\frac{a}{\cos 2 C+\sqrt{1+\sin 2 C}} \\\\ & =\frac{\frac{3 \sqrt{7}}{16}}{\sqrt{\frac{7}{4}}+\sqrt{\frac{7}{2}}}=\frac{1}{4} \\\\ \Rightarrow r & =\frac{1}{4}=0.25 \\\\ \Rightarrow r & =0.25 \end{aligned} $
If $\sin (\alpha+\beta)=\frac{1}{3}$ and $\cos (\alpha-\beta)=\frac{2}{3}$, then the greatest integer less than or equal to
$ \left(\frac{\sin \alpha}{\cos \beta}+\frac{\cos \beta}{\sin \alpha}+\frac{\cos \alpha}{\sin \beta}+\frac{\sin \beta}{\cos \alpha}\right)^{2} $ is
Explanation:
Given, $\sin (\alpha + \beta ) = {1 \over 3}$
and $\cos (\alpha - \beta ) = {2 \over 3}$
Let, $E = {{\sin \alpha } \over {\cos \beta }} + {{\cos \beta } \over {\sin \alpha }} + {{\cos \alpha } \over {\sin \beta }} + {{\sin \beta } \over {\cos \alpha }}$
$ = {{\sin \alpha } \over {\cos \beta }} + {{\cos \alpha } \over {\sin \beta }} + {{\cos \beta } \over {\sin \alpha }} + {{\sin \beta } \over {\cos \alpha }}$
$ = {{\sin \alpha \sin \beta + \cos \alpha \cos \beta } \over {\sin \beta \cos \beta }} + {{\cos \alpha \cos \beta + \sin \alpha \sin \beta } \over {\sin \alpha \cos \alpha }}$
$ = {{\cos (\alpha - \beta )} \over {\sin \beta \cos \beta }} + {{\cos (\alpha - \beta )} \over {\sin \alpha \cos \alpha }}$
$ = \cos (\alpha - \beta )\left[ {{2 \over {2\sin \beta \cos \beta }} + {2 \over {2\sin \alpha \cos \alpha }}} \right]$
$ = {2 \over 3}\left[ {{2 \over {\sin 2\beta }} + {2 \over {\sin 2\alpha }}} \right]$
$ = {4 \over 3}\left[ {{1 \over {\sin 2\beta }} + {1 \over {\sin 2\alpha }}} \right]$
$ = {4 \over 3}\left[ {{{\sin 2\alpha + \sin 2\beta } \over {\sin 2\alpha \sin 2\beta }}} \right]$
$ = {{4 \times 2} \over 3}\left[ {{{2\sin \left( {{{2\alpha + 2\beta } \over 2}} \right)\cos \left( {{{2\alpha - 2\beta } \over 2}} \right)} \over {2\sin 2\alpha \sin 2\beta }}} \right]$
$ = {{16} \over 3}\left[ {{{\sin (\alpha + \beta )\cos (\alpha - \beta )} \over {\cos (2\alpha - 2\beta ) - cos(2\alpha + 2\beta )}}} \right]$
$ = {{16} \over 3}\left[ {{{{1 \over 3} \times {2 \over 3}} \over {(2{{\cos }^2}(\alpha - \beta ) - 1) - (1 - 2{{\sin }^2}(\alpha + \beta ))}}} \right]$
$ = {{32} \over {27}}\left[ {{1 \over {2 \times {4 \over 9} - 2 + 2 \times {1 \over 9}}}} \right]$
$ = {{32} \over {27}}\left[ {{9 \over {8 - 18 + 2}}} \right]$
$ = {{32} \over {27}}\left[ {{9 \over { - 8}}} \right]$
$ = - {4 \over 3}$
$\therefore$ ${E^2} = {{16} \over 9} = 1.77$
$ \Rightarrow [{E^2}] = 1$
Then, the value of loge(f(4)) is ...........
Explanation:
$f(x + y) = f(x)f'(y) + f'(x)f(y),\,\forall x,y \in R$ and f(0) = 1
Put x = y = 0, we get
f(0) = f(0) f'(0) + f'(0) f(0)
$ \Rightarrow 1 = 2f'(0) \Rightarrow f'(0) = {1 \over 2}$
Put x = x and y = 0, we get
f(x) = f(x) f'(0) + f'(x) f(0)
$ \Rightarrow f(x) = {1 \over 2}f(x) + f'(x)$
$ \Rightarrow f'(x) = {1 \over 2}f(x) \Rightarrow {{f'(x)} \over {f(x)}} = {1 \over 2}$
On integrating, we get
$\log f(x) = {1 \over 2}x + C$
$ \Rightarrow f(x) = A{e^{{1 \over 2}x}}$, where eC = A
If f(0) = 1, then A = 1
Hence, $f(x) = {e^{{1 \over 2}x}}$
$ \Rightarrow {\log _e}f(x) = {1 \over 2}x$
$ \Rightarrow {\log _e}f(4) = {1 \over 2} \times 4 = 2$
${5 \over 4}{\cos ^2}\,2x + {\cos ^4}\,x + {\sin ^4}\,x + {\cos ^6}\,x + {\sin ^6}\,x\, = \,2$
in the interval $\left[ {0,\,2\pi } \right]$ is
Explanation:
$ \begin{aligned} \Rightarrow & \frac{5}{4} \cos ^2 2 x+\left(\cos ^2 x\right)^2+\left(\sin ^2 x\right)^2+\left(\cos ^2 x\right)^3 \\\\ & +\left(\sin ^2 x\right)^3=2\quad\quad...(i) \end{aligned} $
As we know, $a^2+b^2+2 a b=(a+b)^2$
$ \Rightarrow a^2+b^2=(a+b)^2-2 a b $
And $a^3+b^3+3 a b(a+b)=(a+b)^3$
$ \Rightarrow a^3+b^3=(a+b)^3-3 a b(a+b) $
So, equation (i) can be written as
$ \begin{aligned} & \frac{5}{4} \cos ^2 2 x+\left(\cos ^2 x+\sin ^2 x\right)^2-2\left(\cos ^2 x\right) \\\\ &\left(\sin ^2 x\right)+\left(\cos ^2 x+\sin ^2 x\right)^3-3 \cos ^2 x \sin ^2 x \\\\ &\left(\cos ^2 x+\sin ^2 x\right)=2 \\\\ & \Rightarrow \frac{5}{4} \cos ^2 2 x+(1)^2-2 \cos ^2 x \sin ^2 x+(1)^3 \\\\ &-3 \cos ^2 x \sin ^2 x(1)=2~~~~\left\{\because \cos ^2 x+\sin ^2 x=1\right\} \end{aligned} $
$ \begin{aligned} & \Rightarrow \frac{5}{4} \cos ^2 2 x+2-5 \cos ^2 x \sin ^2 x=2 \\\\ & \Rightarrow \frac{5}{4} \cos ^2 2 x-5 \cos ^2 x \sin ^2 x=0 \end{aligned} $
As we know, $\sin 2 \theta=2 \sin \theta \cos \theta$
$ \begin{aligned} & \Rightarrow \frac{5}{4} \cos ^2 2 x-\frac{5}{4} \sin ^2 2 x=0 \\\\ & \because \cos 2 \theta=\cos ^2 \theta-\sin ^2 \theta \\\\ & \Rightarrow \frac{5}{4} \cos 4 x=0 \\\\ & \Rightarrow \cos 4 x=0 \\\\ & \Rightarrow 4 x=2(n+1) \frac{\pi}{2}, n \in \mathrm{I} \\\\ & \because x \in[0,2 \pi] \end{aligned} $
So, possible distinct values of $x$ are $\frac{\pi}{8}, \frac{3 \pi}{8}, \frac{5 \pi}{8}, \frac{7 \pi}{8}, \frac{9 \pi}{8}, \frac{11 \pi}{8}, \frac{13 \pi}{8}$ and $\frac{15 \pi}{8}$.
So, the number of distinct solutions of the given equation are 8 .
Explanation:
We have,
${1 \over {\sin (\pi /n)}} - {1 \over {\sin (3\pi /n)}} = {1 \over {\sin (2\pi /n)}}$
$ \Rightarrow {{\sin (3\pi /n) - \sin (\pi /n)} \over {\sin (\pi /n)\sin (3\pi /n)}} = {1 \over {\sin (2\pi /n)}}{{(2\sin (\pi /n)\cos (2\pi /n))\sin (2\pi /n)} \over {\sin (\pi /n)\sin (3\pi /n)}} = 1$
$ \Rightarrow \sin {{4\pi } \over n} = \sin {{3\pi } \over n} \Rightarrow {{4\pi } \over n} + {{3\pi } \over n} = \pi \Rightarrow n = 7$
have a solution $\left( {{x_0},{y_0},{z_0}} \right)$ with ${y_0}{z_0}{\mkern 1mu} \ne {\mkern 1mu} 0,$ is
Explanation:
View the equation in xyz, y and t.
We have,
$(xyz)\sin 3\theta - y\cos 3\theta - z\cos 3\theta = 0$
$(xyz)\sin 3\theta - 2y\sin 3\theta - 2z\cos 3\theta = 0$
$(xyz)\sin 3\theta - y(\cos 3\theta + \sin 3\theta ) - 2z\cos 3\theta = 0$
$xyz \ne 0$
Hence, the equation has non-trivial solution which gives
$\left| {\matrix{ {\sin 3\theta } & { - \cos 3\theta } & { - \cos 3\theta } \cr {\sin 3\theta } & { - 2\sin 3\theta } & { - 2\cos 3\theta } \cr {\sin 3\theta } & { - (\cos 3\theta + \sin 3\theta )} & { - 2\cos 3\theta } \cr } } \right| = 0$
$ \Rightarrow \sin 3\theta \cos 3\theta (\sin 3\theta - \cos 3\theta ) = 0$
$ \Rightarrow \sin 3\theta = 0$ then $xyz = 0$ (not possible)
$\cos 3\theta = 0$ not possible
$\sin 3\theta = \cos 3\theta \Rightarrow \tan 3\theta = 1$
$3\theta = n\pi + {\pi \over 4},n \in z$
$\theta = {{n\pi } \over 3} + {\pi \over {12}}$ ; $\theta = {\pi \over {12}},{{5\pi } \over {12}},{{9\pi } \over {12}}$
Thus there are 3 solutions.
Explanation:
Given, $\tan \theta = \cot 5\theta $
$ \Rightarrow \tan \theta = \tan \left( {{\pi \over 2} - 5\theta } \right)$
$ \Rightarrow {\pi \over 2} - 5\theta = n\pi + \theta $
$ \Rightarrow 6\theta = {\pi \over 2} - {{n\pi } \over 6}$
$ \Rightarrow \theta = {\pi \over {12}} - {{n\pi } \over 6}$
Also $\cos 4\theta = \sin 2\theta = \cos \left( {{\pi \over 2} - 2\theta } \right)$
$ \Rightarrow 4\theta = 2n\pi \, \pm \,\left( {{\pi \over 2} - 2\theta } \right)$
Taking positive
$6\theta = 2n\pi + {\pi \over 2} \Rightarrow \theta = {{n\pi } \over 3} + {\pi \over {12}}$
Taking negative
$2\theta = 2n\pi - {\pi \over 2} \Rightarrow \theta = n\pi - {\pi \over 4}$
Above values of $\theta$ suggests that there are only 3 common solutions.
Explanation:
Let
$f(\theta ) = {1 \over {{{\sin }^2}\theta + 3\sin \theta \cos \theta + 5{{\cos }^2}\theta }}$
Again let
$g(\theta ) = {\sin ^2}\theta + 3\sin \theta \cos \theta + 5{\cos ^2}\theta $
$ = {{1 - \cos 2\theta } \over 2} + 5\left( {{{1 + \cos 2\theta } \over 2}} \right) + {3 \over 2}\sin 2\theta $
$ = 3 + 2\cos 2\theta + {3 \over 2}\sin 2\theta $
$\therefore$ $g{(\theta )_{\min }} = 3 - \sqrt {4 + {9 \over 4}} = 3 - {5 \over 2} = {1 \over 2}$
$\therefore$ $f(\theta ) = {1 \over {g{{(\theta )}_{\min }}}} = 2$
[Note :[k] denotes the largest integer less than or equal to k ]
Explanation:
Let $\theta = {\pi \over {2k}}$
$\cos \theta = {x \over 2}$
$ \Rightarrow \cos 2\theta = {{\sqrt 3 + 1 - x} \over 2}$
$ \Rightarrow 2{\cos ^2}\theta - 1 = {{\sqrt 3 + 1 - x} \over 2}$
$ \Rightarrow 2\left( {{{{x^2}} \over 4}} \right) - 1 = {{\sqrt 3 + 1 - x} \over 2}$
$ \Rightarrow {x^2} + x - 3 - \sqrt 3 = 0$
$ \Rightarrow x = {{ - 1 \pm \sqrt {1 + 12 + 4\sqrt 3 } } \over 2}$
$ = {{ - 1 \pm \sqrt {13 + 4\sqrt 3 } } \over 2}$
$ = {{ - 1 + 2\sqrt 3 + 1} \over 2} = \sqrt 3 $
$\therefore$ $\cos \theta = {{\sqrt 3 } \over 2} \Rightarrow \theta = {\pi \over 6}$
$\therefore$ Required angle $ = {\pi \over k} = 2\theta = {\pi \over 3}$
$ \Rightarrow k = 3$
If $A,\,B$ and $C$ are in arithmetic progression, determine the values of $A,\,B$ and $C$.
$f\left( x \right) = \cos x - x\left( {1 + x} \right);$ find $f\left( A \right).$
$\,{\sin ^2}\alpha + {\sin ^2}\beta - {\sin ^2}\gamma = 2\sin \alpha {\mkern 1mu} \sin \beta {\mkern 1mu} \cos y$
(b) If $\cos \left( {\alpha + \beta } \right) = {4 \over 5},\,\,\sin \,\left( {\alpha - \beta } \right) = \,{5 \over {13}},$ and $\alpha ,\,\beta $ lies between 0 and ${\pi \over 4}$, find tan2$\alpha $.