If $\tan \theta =\dfrac{11}{2}$ where $\theta \in \left( 0,\dfrac{\pi }{2} \right)$. Find the value of \[\sin \dfrac{\theta }{3}\] and $\cos \dfrac{\theta }{3}$.
Answer
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Hint: We start solving this problem by substituting the given value of $\tan \theta =\dfrac{11}{2}$ in the formula $\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$ and find an equation in terms of $\tan \dfrac{\theta }{3}$. Then we assume that \[\tan \dfrac{\theta }{3}=x\] and then find the roots of the equation by trial and error method. Then we find the range of $\tan \dfrac{\theta }{3}$ from the information, $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ and find which of the obtained values lie in that range. Then we find the values of \[\sin \dfrac{\theta }{3}\] and $\cos \dfrac{\theta }{3}$, using the trigonometric identities ${{\sin }^{2}}\theta \text{+}{{\cos }^{2}}\theta =1$ and ${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$.
Complete step by step answer:
We are given that $\tan \theta =\dfrac{11}{2}$ and that $\theta \in \left( 0,\dfrac{\pi }{2} \right)$.
We need to find the value of \[\sin \dfrac{\theta }{3}\] and $\cos \dfrac{\theta }{3}$.
First let us find the value of \[\tan \dfrac{\theta }{3}\].
Now let us consider the formula for $\tan 3A$.
$\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$
Using this formula, we can write $\tan \theta $ in terms of \[\tan \dfrac{\theta }{3}\] as,
$\Rightarrow \tan 3\theta =\dfrac{3\tan \dfrac{\theta }{3}-{{\tan }^{3}}\dfrac{\theta }{3}}{1-3{{\tan }^{2}}\dfrac{\theta }{3}}$
Substituting the value of $\tan \theta =\dfrac{11}{2}$ in it we get,
$\begin{align}
& \Rightarrow \dfrac{11}{2}=\dfrac{3\tan \dfrac{\theta }{3}-{{\tan }^{3}}\dfrac{\theta }{3}}{1-3{{\tan }^{2}}\dfrac{\theta }{3}} \\
& \Rightarrow 11\left( 1-3{{\tan }^{2}}\dfrac{\theta }{3} \right)=2\left( 3\tan \dfrac{\theta }{3}-{{\tan }^{3}}\dfrac{\theta }{3} \right) \\
\end{align}$
$\begin{align}
& \Rightarrow 11-33{{\tan }^{2}}\dfrac{\theta }{3}=6\tan \dfrac{\theta }{3}-2{{\tan }^{3}}\dfrac{\theta }{3} \\
& \Rightarrow 2{{\tan }^{3}}\dfrac{\theta }{3}-33{{\tan }^{2}}\dfrac{\theta }{3}-6\tan \dfrac{\theta }{3}+11=0 \\
\end{align}$
Now let us assume that \[\tan \dfrac{\theta }{3}=x\]. Then above equation can be converted as,
$\Rightarrow 2{{x}^{3}}-33{{x}^{2}}-6x+11=0$
Now let us find the roots of above equation by trial and error method.
Let us check whether $x=\dfrac{1}{2}$ is a root of the equation.
$x=\dfrac{1}{2}\Rightarrow 2x-1=0$
Now let us divide $2{{x}^{3}}-33{{x}^{2}}-6x+11=0$ with $2x-1$.
$2x-1\overset{{{x}^{2}}-16x-11}{\overline{\left){\begin{align}
& 2{{x}^{3}}-33{{x}^{2}}-6x+11 \\
& \underline{2{{x}^{3}}-{{x}^{2}}\text{ }} \\
& \text{ }-32{{x}^{2}}-6x \\
& \text{ }\underline{-32{{x}^{2}}+16x\text{ }} \\
& \text{ }-22x+11 \\
& \text{ }\underline{-22x+11} \\
& \text{ 0} \\
\end{align}}\right.}}$
So, we can write $2{{x}^{3}}-33{{x}^{2}}-6x+11=0$ as,
$\Rightarrow 2{{x}^{3}}-33{{x}^{2}}-6x+11=\left( 2x-1 \right)\left( {{x}^{2}}-16x-11 \right)$
So, one root of the above equation is $x=\dfrac{1}{2}$.
Now let us find the root of the equation ${{x}^{2}}-16x-11=0$.
Now let us consider the formula for roots of the equation, $a{{x}^{2}}+bx+c=0$.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, the roots of equation ${{x}^{2}}-16x-11=0$ are,
$\begin{align}
& \Rightarrow x=\dfrac{16\pm \sqrt{{{\left( 16 \right)}^{2}}-4\left( 1 \right)\left( -11 \right)}}{2} \\
& \Rightarrow x=\dfrac{16\pm \sqrt{256+44}}{2} \\
\end{align}$
$\begin{align}
& \Rightarrow x=\dfrac{16\pm \sqrt{300}}{2} \\
& \Rightarrow x=\dfrac{16\pm 10\sqrt{3}}{2} \\
& \Rightarrow x=8\pm 5\sqrt{3} \\
\end{align}$
So, the roots of equation $2{{x}^{3}}-33{{x}^{2}}-6x+11=0$ are,
$x=\dfrac{1}{2},8+5\sqrt{3},8-5\sqrt{3}$
As we have assumed that \[\tan \dfrac{\theta }{3}=x\], we get that the values of \[\tan \dfrac{\theta }{3}\] as,
\[\tan \dfrac{\theta }{3}=\dfrac{1}{2},8+5\sqrt{3},8-5\sqrt{3}...........\left( 1 \right)\]
We are given that $\theta \in \left( 0,\dfrac{\pi }{2} \right)$. Then $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$.
As $\tan \theta $ is increasing in the interval $\theta \in \left( 0,\dfrac{\pi }{2} \right)$, when $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$ then,
$\begin{align}
& \Rightarrow \tan \dfrac{\theta }{3}\in \left( 0,\tan \dfrac{\pi }{6} \right) \\
& \Rightarrow \tan \dfrac{\theta }{3}\in \left( 0,\dfrac{1}{\sqrt{3}} \right) \\
\end{align}$
Comparing the values in equation (1) with the above range of $\tan \dfrac{\theta }{3}$, we get the value of $\tan \dfrac{\theta }{3}$ as,
$\Rightarrow \tan \dfrac{\theta }{3}=\dfrac{1}{2}$
Now let us consider the trigonometric identity,
${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$
Applying this identity, we get
$\begin{align}
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}-{{\tan }^{2}}\dfrac{\theta }{3}=1 \\
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}-{{\left( \dfrac{1}{2} \right)}^{2}}=1 \\
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}-\dfrac{1}{4}=1 \\
\end{align}$
$\begin{align}
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}=1+\dfrac{1}{4} \\
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}=\dfrac{5}{4} \\
& \Rightarrow {{\cos }^{2}}\dfrac{\theta }{3}=\dfrac{4}{5} \\
& \Rightarrow \cos \dfrac{\theta }{3}=\pm \dfrac{2}{\sqrt{5}} \\
\end{align}$
As $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$, cosine of it must be positive. So, we get,
$\Rightarrow \cos \dfrac{\theta }{3}=\dfrac{2}{\sqrt{5}}........\left( 1 \right)$
Now let us consider the identity,
${{\sin }^{2}}\theta \text{+}{{\cos }^{2}}\theta =1$
Applying this identity, we get
$\begin{align}
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}+{{\cos }^{2}}\dfrac{\theta }{3}=1 \\
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}+{{\left( \dfrac{2}{\sqrt{5}} \right)}^{2}}=1 \\
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}+\dfrac{4}{5}=1 \\
\end{align}$
$\begin{align}
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}=1-\dfrac{4}{5} \\
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}=\dfrac{1}{5} \\
& \Rightarrow \sin \dfrac{\theta }{3}=\pm \dfrac{1}{\sqrt{5}} \\
\end{align}$
As $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$, sine of it must be positive. So, we get,
$\Rightarrow \sin \dfrac{\theta }{3}=\dfrac{1}{\sqrt{5}}........\left( 2 \right)$
From equations (1) and (2) we get the values of $\sin \dfrac{\theta }{3}$ and $\cos \dfrac{\theta }{3}$ as $\dfrac{1}{\sqrt{5}}$ and $\dfrac{2}{\sqrt{5}}$.
Hence answer is $\dfrac{1}{\sqrt{5}}$ and $\dfrac{2}{\sqrt{5}}$.
Note:
The common mistake one makes while solving this question is one might not check the range of $\tan \dfrac{\theta }{3}$ from the given range of $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ and assume the values of $\tan \dfrac{\theta }{3}$ as, \[\tan \dfrac{\theta }{3}=\dfrac{1}{2},8+5\sqrt{3},8-5\sqrt{3}\] and find the values of $\sin \dfrac{\theta }{3}$ and $\cos \dfrac{\theta }{3}$ for each case.
Complete step by step answer:
We are given that $\tan \theta =\dfrac{11}{2}$ and that $\theta \in \left( 0,\dfrac{\pi }{2} \right)$.
We need to find the value of \[\sin \dfrac{\theta }{3}\] and $\cos \dfrac{\theta }{3}$.
First let us find the value of \[\tan \dfrac{\theta }{3}\].
Now let us consider the formula for $\tan 3A$.
$\tan 3A=\dfrac{3\tan A-{{\tan }^{3}}A}{1-3{{\tan }^{2}}A}$
Using this formula, we can write $\tan \theta $ in terms of \[\tan \dfrac{\theta }{3}\] as,
$\Rightarrow \tan 3\theta =\dfrac{3\tan \dfrac{\theta }{3}-{{\tan }^{3}}\dfrac{\theta }{3}}{1-3{{\tan }^{2}}\dfrac{\theta }{3}}$
Substituting the value of $\tan \theta =\dfrac{11}{2}$ in it we get,
$\begin{align}
& \Rightarrow \dfrac{11}{2}=\dfrac{3\tan \dfrac{\theta }{3}-{{\tan }^{3}}\dfrac{\theta }{3}}{1-3{{\tan }^{2}}\dfrac{\theta }{3}} \\
& \Rightarrow 11\left( 1-3{{\tan }^{2}}\dfrac{\theta }{3} \right)=2\left( 3\tan \dfrac{\theta }{3}-{{\tan }^{3}}\dfrac{\theta }{3} \right) \\
\end{align}$
$\begin{align}
& \Rightarrow 11-33{{\tan }^{2}}\dfrac{\theta }{3}=6\tan \dfrac{\theta }{3}-2{{\tan }^{3}}\dfrac{\theta }{3} \\
& \Rightarrow 2{{\tan }^{3}}\dfrac{\theta }{3}-33{{\tan }^{2}}\dfrac{\theta }{3}-6\tan \dfrac{\theta }{3}+11=0 \\
\end{align}$
Now let us assume that \[\tan \dfrac{\theta }{3}=x\]. Then above equation can be converted as,
$\Rightarrow 2{{x}^{3}}-33{{x}^{2}}-6x+11=0$
Now let us find the roots of above equation by trial and error method.
Let us check whether $x=\dfrac{1}{2}$ is a root of the equation.
$x=\dfrac{1}{2}\Rightarrow 2x-1=0$
Now let us divide $2{{x}^{3}}-33{{x}^{2}}-6x+11=0$ with $2x-1$.
$2x-1\overset{{{x}^{2}}-16x-11}{\overline{\left){\begin{align}
& 2{{x}^{3}}-33{{x}^{2}}-6x+11 \\
& \underline{2{{x}^{3}}-{{x}^{2}}\text{ }} \\
& \text{ }-32{{x}^{2}}-6x \\
& \text{ }\underline{-32{{x}^{2}}+16x\text{ }} \\
& \text{ }-22x+11 \\
& \text{ }\underline{-22x+11} \\
& \text{ 0} \\
\end{align}}\right.}}$
So, we can write $2{{x}^{3}}-33{{x}^{2}}-6x+11=0$ as,
$\Rightarrow 2{{x}^{3}}-33{{x}^{2}}-6x+11=\left( 2x-1 \right)\left( {{x}^{2}}-16x-11 \right)$
So, one root of the above equation is $x=\dfrac{1}{2}$.
Now let us find the root of the equation ${{x}^{2}}-16x-11=0$.
Now let us consider the formula for roots of the equation, $a{{x}^{2}}+bx+c=0$.
$x=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$
So, the roots of equation ${{x}^{2}}-16x-11=0$ are,
$\begin{align}
& \Rightarrow x=\dfrac{16\pm \sqrt{{{\left( 16 \right)}^{2}}-4\left( 1 \right)\left( -11 \right)}}{2} \\
& \Rightarrow x=\dfrac{16\pm \sqrt{256+44}}{2} \\
\end{align}$
$\begin{align}
& \Rightarrow x=\dfrac{16\pm \sqrt{300}}{2} \\
& \Rightarrow x=\dfrac{16\pm 10\sqrt{3}}{2} \\
& \Rightarrow x=8\pm 5\sqrt{3} \\
\end{align}$
So, the roots of equation $2{{x}^{3}}-33{{x}^{2}}-6x+11=0$ are,
$x=\dfrac{1}{2},8+5\sqrt{3},8-5\sqrt{3}$
As we have assumed that \[\tan \dfrac{\theta }{3}=x\], we get that the values of \[\tan \dfrac{\theta }{3}\] as,
\[\tan \dfrac{\theta }{3}=\dfrac{1}{2},8+5\sqrt{3},8-5\sqrt{3}...........\left( 1 \right)\]
We are given that $\theta \in \left( 0,\dfrac{\pi }{2} \right)$. Then $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$.
As $\tan \theta $ is increasing in the interval $\theta \in \left( 0,\dfrac{\pi }{2} \right)$, when $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$ then,
$\begin{align}
& \Rightarrow \tan \dfrac{\theta }{3}\in \left( 0,\tan \dfrac{\pi }{6} \right) \\
& \Rightarrow \tan \dfrac{\theta }{3}\in \left( 0,\dfrac{1}{\sqrt{3}} \right) \\
\end{align}$
Comparing the values in equation (1) with the above range of $\tan \dfrac{\theta }{3}$, we get the value of $\tan \dfrac{\theta }{3}$ as,
$\Rightarrow \tan \dfrac{\theta }{3}=\dfrac{1}{2}$
Now let us consider the trigonometric identity,
${{\sec }^{2}}\theta -{{\tan }^{2}}\theta =1$
Applying this identity, we get
$\begin{align}
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}-{{\tan }^{2}}\dfrac{\theta }{3}=1 \\
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}-{{\left( \dfrac{1}{2} \right)}^{2}}=1 \\
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}-\dfrac{1}{4}=1 \\
\end{align}$
$\begin{align}
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}=1+\dfrac{1}{4} \\
& \Rightarrow {{\sec }^{2}}\dfrac{\theta }{3}=\dfrac{5}{4} \\
& \Rightarrow {{\cos }^{2}}\dfrac{\theta }{3}=\dfrac{4}{5} \\
& \Rightarrow \cos \dfrac{\theta }{3}=\pm \dfrac{2}{\sqrt{5}} \\
\end{align}$
As $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$, cosine of it must be positive. So, we get,
$\Rightarrow \cos \dfrac{\theta }{3}=\dfrac{2}{\sqrt{5}}........\left( 1 \right)$
Now let us consider the identity,
${{\sin }^{2}}\theta \text{+}{{\cos }^{2}}\theta =1$
Applying this identity, we get
$\begin{align}
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}+{{\cos }^{2}}\dfrac{\theta }{3}=1 \\
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}+{{\left( \dfrac{2}{\sqrt{5}} \right)}^{2}}=1 \\
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}+\dfrac{4}{5}=1 \\
\end{align}$
$\begin{align}
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}=1-\dfrac{4}{5} \\
& \Rightarrow {{\sin }^{2}}\dfrac{\theta }{3}=\dfrac{1}{5} \\
& \Rightarrow \sin \dfrac{\theta }{3}=\pm \dfrac{1}{\sqrt{5}} \\
\end{align}$
As $\dfrac{\theta }{3}\in \left( 0,\dfrac{\pi }{6} \right)$, sine of it must be positive. So, we get,
$\Rightarrow \sin \dfrac{\theta }{3}=\dfrac{1}{\sqrt{5}}........\left( 2 \right)$
From equations (1) and (2) we get the values of $\sin \dfrac{\theta }{3}$ and $\cos \dfrac{\theta }{3}$ as $\dfrac{1}{\sqrt{5}}$ and $\dfrac{2}{\sqrt{5}}$.
Hence answer is $\dfrac{1}{\sqrt{5}}$ and $\dfrac{2}{\sqrt{5}}$.
Note:
The common mistake one makes while solving this question is one might not check the range of $\tan \dfrac{\theta }{3}$ from the given range of $\theta \in \left( 0,\dfrac{\pi }{2} \right)$ and assume the values of $\tan \dfrac{\theta }{3}$ as, \[\tan \dfrac{\theta }{3}=\dfrac{1}{2},8+5\sqrt{3},8-5\sqrt{3}\] and find the values of $\sin \dfrac{\theta }{3}$ and $\cos \dfrac{\theta }{3}$ for each case.
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