Question

The number of values of $\theta$ in the interval $\left( -\dfrac{\pi }{2},\dfrac{\pi }{2} \right)$ such that $\theta$ is not equal to $\dfrac{n\pi }{5}$ for n = 0,$\pm 1,\pm 2,...$ and $\tan \left( \theta \right)=\cot \left( 5\theta \right)$ as well as $\sin \left( 2\theta \right)=\cos \left( 4\theta \right)$ is
(a) 0
(b) 1
(c) 2
(d) 3
(e) 4
(f) 5
(g) 6
(h) 7
(i) 8

Answer 1

Hint: We will use the formula $\cot \left( \theta \right)=\dfrac{1}{\tan \left( \theta \right)}$ and solve equation $\tan \left( \theta \right)=\cot \left( 5\theta \right)$. We will also use $\tan \left( \theta \right)=\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}$ and trigonometric formulas $2\sin \left( A \right)\sin \left( B \right)=\cos \left( A-B \right)-\cos \left( A+B \right)$ and $2\cos \left( A \right)\cos \left( B \right)=\cos \left( A+B \right)+\cos \left( A-B \right)$ in order to solve the question further. Also, we will apply the general solution of $\cos \left( x \right)=\cos \left( y \right)$ is $x=2n\pi \pm y$ where n is any integer and the general solution of $\sin x=\sin y$ resulting into $x=n\pi +{{\left( -1 \right)}^{n}}y$. Moreover, by using the formula of the square root we will find the value of x. The formula of square root is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.

Complete step-by-step answer:
We will first consider the equation $\tan \left( \theta \right)=\cot \left( 5\theta \right)$ and convert the trigonometric term on the right side of the equation into the tangent. We will do this by the formula $\cot \left( \theta \right)=\dfrac{1}{\tan \left( \theta \right)}$ therefore, we get
$\begin{aligned} & \tan \left( \theta \right)=\dfrac{1}{\tan \left( 5\theta \right)} \\\ & \Rightarrow \tan \left( \theta \right)\tan \left( 5\theta \right)=1 \\\ \end{aligned}$
Now, we will convert this equation into its simplest form which can be given by
$\tan \left( \theta \right)=\dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)}$. After substituting this into the equation we will get
$\begin{aligned} & \left( \dfrac{\sin \left( \theta \right)}{\cos \left( \theta \right)} \right)\left( \dfrac{\sin \left( 5\theta \right)}{\cos \left( 5\theta \right)} \right)=1 \\\ & \Rightarrow \dfrac{\sin \left( \theta \right)\sin \left( 5\theta \right)}{\cos \left( \theta \right)\cos \left( 5\theta \right)}=1 \\\ \end{aligned}$
After this we will divide the numerator and denominator of the left side of the expression. Therefore, we get $\dfrac{2\sin \left( \theta \right)\sin \left( 5\theta \right)}{2\cos \left( \theta \right)\cos \left( 5\theta \right)}=1$.
Now we will use the formula $2\sin \left( A \right)\sin \left( B \right)=\cos \left( A-B \right)-\cos \left( A+B \right)$ and $2\cos \left( A \right)\cos \left( B \right)=\cos \left( A+B \right)+\cos \left( A-B \right)$ into the equation so, we will get
$\begin{aligned} & \dfrac{\cos \left( \theta -5\theta \right)-\cos \left( \theta +5\theta \right)}{\cos \left( \theta +5\theta \right)+\cos \left( \theta -5\theta \right)}=1 \\\ & \Rightarrow \dfrac{\cos \left( -4\theta \right)-\cos \left( 6\theta \right)}{\cos \left( 6\theta \right)+\cos \left( -4\theta \right)}=1 \\\ & \Rightarrow \cos \left( -4\theta \right)-\cos \left( 6\theta \right)=\cos \left( 6\theta \right)+\cos \left( -4\theta \right) \\\ & \Rightarrow \cos \left( -4\theta \right)-\cos \left( 6\theta \right)-\cos \left( 6\theta \right)-\cos \left( -4\theta \right)=0 \\\ & \Rightarrow -2\cos \left( 6\theta \right)=0 \\\ & \Rightarrow \cos \left( 6\theta \right)=0 \\\ \end{aligned}$
As we know that $\cos \left( 2n+1 \right)\dfrac{\pi }{2}=0$ thus we have $\cos \left( 6\theta \right)=\cos \left( \left( 2n+1 \right)\dfrac{\pi }{2} \right)$. Therefore, we get $\cos \left( 6\theta \right)=\cos \left( \left( 2n+1 \right)\dfrac{\pi }{2} \right)$. As we know that the general solution of $\cos \left( x \right)=\cos \left( y \right)$ is $x=2n\pi \pm y$ where n is any integer so we have $6\theta =\left( 2n+1 \right)\dfrac{\pi }{2}$. After dividing both the sides by 6 we will get
$\begin{aligned} & \dfrac{6\theta }{6}=\dfrac{\left( 2n+1 \right)\dfrac{\pi }{2}}{6} \\\ & \Rightarrow \theta =\left( 2n+1 \right)\dfrac{\pi }{12} \\\ \end{aligned}$
Now we will substitute n = 0. Thus we get that $\theta =\dfrac{\pi }{12}$. Similarly, by substituting the value of n one by one by integers we will get $\theta =\pm \dfrac{5\pi }{2},\pm \dfrac{\pi }{4},\pm \dfrac{\pi }{12}$.
Now we will consider the equation $\sin \left( 2\theta \right)=\cos \left( 4\theta \right)$ and use the formula $\cos \left( 4\theta \right)=1-2{{\sin }^{2}}\left( 2\theta \right)$. Therefore we get $\sin \left( 2\theta \right)=1-2{{\sin }^{2}}\left( 2\theta \right)$. Now, we will take all the terms to the left side of the equation thus, we get $2{{\sin }^{2}}\left( 2\theta \right)+\sin \left( 2\theta \right)-1=0$. By substituting
$\sin \left( 2\theta \right)=x$ we will have $2{{x}^{2}}+x-1=0$. By using the formula of the square root we will find the value of x. The formula of square root is $x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$. Now we will put a = 2, b = 1 and c = -1 so, we will get
$\begin{aligned} & x=\dfrac{-1\pm \sqrt{{{\left( 1 \right)}^{2}}-4\left( 2 \right)\left( -1 \right)}}{2\left( 2 \right)} \\\ & \Rightarrow x=\dfrac{-1\pm \sqrt{1+8}}{4} \\\ & \Rightarrow x=\dfrac{-1\pm \sqrt{9}}{4} \\\ & \Rightarrow x=\dfrac{-1\pm 3}{4} \\\ \end{aligned}$
So, we now have $x=\dfrac{-1+3}{4}$ and $x=\dfrac{-1-3}{4}$. Thus, we get $x=\dfrac{2}{4}$ and $x=-\dfrac{4}{4}$. Therefore, the value of x can be x = $\dfrac{1}{2}$ and x = - 1. After substituting it again as $\sin \left( 2\theta \right)=x$ we have $\sin \left( 2\theta \right)=\dfrac{1}{2}$ and $\sin \left( 2\theta \right)=-1$. We will first consider first that $\sin \left( 2\theta \right)=\dfrac{1}{2}$. As the value of $\sin \left( \dfrac{\pi }{6} \right)=\dfrac{1}{2}$ so, we have $\sin \left( 2\theta \right)=\sin \left( \dfrac{\pi }{6} \right)$. By the general solution of $\sin x=\sin y$ resulting into $x=n\pi +{{\left( -1 \right)}^{n}}y$ we will get $\left( 2\theta \right)=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{6} \right)$ or $\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\left( \dfrac{\pi }{12} \right)$.
Similarly, by $\sin \left( 2\theta \right)=-1$ and $\sin \left( \dfrac{\pi }{2} \right)=1$ we will have that $\sin \left( 2\theta \right)=-\sin \left( \dfrac{\pi }{2} \right)$. As we know that As we know that $\sin \left( 2n-1 \right)\dfrac{\pi }{2}=1$ thus we have $\sin \left( 2\theta \right)=\sin \left( \left( 2n-1 \right)\dfrac{\pi }{2} \right)$. Therefore, we get $\sin \left( 2\theta \right)=\sin \left( \left( 2n-1 \right)\dfrac{\pi }{2} \right)$. As we know that the general solution of $\sin \left( x \right)=\sin \left( y \right)$ is $x=n\pi \pm {{\left( -1 \right)}^{n}}y$ where n is any integer so we have $2\theta =\left( 2n-1 \right)\dfrac{\pi }{2}$. Here we have not considered the general solution and ignored the term $n\pi$ by substituting n = 0. After dividing both the sides by 2 we will get
$\begin{aligned} & \dfrac{2\theta }{2}=\dfrac{\left( 2n-1 \right)\dfrac{\pi }{2}}{2} \\\ & \Rightarrow \theta =\left( 2n-1 \right)\dfrac{\pi }{4} \\\ \end{aligned}$
Now we will substitute n = 0. Thus we get that $\theta =-\dfrac{\pi }{4}$. Similarly, by substituting the value of n one by one by integers we will get $\theta =\dfrac{5\pi }{2},-\dfrac{\pi }{4},\dfrac{\pi }{12}$. After considering the common values from the angle $\theta$ we will get that there are only three solutions in total.
Hence, the correct option is (d).

Note: Clearly, this question needs focus to solve . So many formulas are there which are important to solve the question. We can also solve $\sin \left( 2\theta \right)=-\sin \left( \dfrac{\pi }{2} \right)$ by using an alternate method. As sine is negative in third and fourth quadrant thus, in the third quadrant we get $\begin{aligned} & -\sin \left( \dfrac{\pi }{2} \right)=\sin \left( \pi +\dfrac{\pi }{2} \right) \\\ & \Rightarrow -\sin \left( \dfrac{\pi }{2} \right)=\sin \left( \dfrac{2\pi +\pi }{2} \right) \\\ & \Rightarrow -\sin \left( \dfrac{\pi }{2} \right)=\sin \left( \dfrac{3\pi }{2} \right) \\\ \end{aligned}$
This results in $\sin \left( 2\theta \right)=\sin \left( \dfrac{3\pi }{2} \right)$. By the general solution of $\sin x=\sin y$ resulting into $x=n\pi +{{\left( -1 \right)}^{n}}y$ we will get $\left( 2\theta \right)=n\pi +{{\left( -1 \right)}^{n}}\left( \dfrac{3\pi }{2} \right)$ or $\theta =\dfrac{n\pi }{2}+{{\left( -1 \right)}^{n}}\left( \dfrac{3\pi }{4} \right)$.
Also, we will not the below equation like so,
$\begin{aligned} & \sin \left( 2\theta \right)+2{{\sin }^{2}}\left( 2\theta \right)=1 \\\ & \Rightarrow \sin \left( 2\theta \right)\left( 1+2\sin \left( 2\theta \right) \right)=1 \\\ \end{aligned}$
We now have $\sin \left( 2\theta \right)=1$ and $\left( 1+2\sin \left( 2\theta \right) \right)=1$. The equation $\left( 1+2\sin \left( 2\theta \right) \right)=1$ can be converted into $\begin{aligned} & 1+2\sin \left( 2\theta \right)=1 \\\ & \Rightarrow \\\ \end{aligned}$Since, we have that $\sin \left( \dfrac{\pi }{2} \right)=1$ therefore $\sin \left( 2\theta \right)=\sin \left( \dfrac{\pi }{2} \right)$ and $\left( 1+2\sin \left( 2\theta \right) \right)=\sin \left( \dfrac{\pi }{2} \right)$. This will lead to the wrong answer.