Question

Find an expression for $\tan 7\theta $ in terms of $\tan \theta $ . By considering the equation $\tan 7\theta =0$ , show that $x={{\tan }^{2}}\left( \dfrac{3\pi }{7} \right)$ satisfies the cubic equation ${{x}^{3}}-21{{x}^{2}}+35x-7=0$ .

Answer 1

Hint: We have to split $7\theta $ as the sum of $3\theta $ and $4\theta $ . Then, we have to apply the properties of trigonometric functions, mainly, $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ , $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$ and $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ and simplify the expression. Then, we have to equate $\tan 7\theta $ to 0 and form a polynomial. Finally, in the given polynomial, substitute $x={{\tan }^{2}}\left( \dfrac{3\pi }{7} \right)$ and check whether the resultant polynomial is equal to the polynomial obtained in the previous step.

Complete step by step answer:
We have to express $\tan 7\theta $ in terms of $\tan \theta $ . We know that $\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b}$ . Let us write $7\theta $ as the sum of $3\theta $ and $4\theta $ .
$\Rightarrow \tan \left( 7\theta \right)=\tan \left( 3\theta +4\theta \right)=\dfrac{\tan 3\theta +\tan 4\theta }{1-\tan 3\theta \tan 4\theta }$
We know that $\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }$ .
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }+\tan 4\theta }{1-\left( \dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right)\tan 4\theta }$
We can write $\tan 4\theta $ as $\tan 2\left( 2\theta \right)$ . We know that $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ . Therefore, the above equation becomes
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta }+\dfrac{2\tan 2\theta }{1-{{\tan }^{2}}2\theta }}{1-\left( \dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } \right)\dfrac{2\tan 2\theta }{1-{{\tan }^{2}}2\theta }}$
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)+2\tan 2\theta \left( 1-3{{\tan }^{2}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)}}{\dfrac{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)-\left( 3\tan \theta -{{\tan }^{3}}\theta \right)2\tan 2\theta }{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)}}$
Now, let us cancel common terms.
$\begin{align}
  & \Rightarrow \tan \left( 7\theta \right)=\dfrac{\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)+2\tan 2\theta \left( 1-3{{\tan }^{2}}\theta \right)}{\require{cancel}\cancel{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)}}}{\dfrac{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)-\left( 3\tan \theta -{{\tan }^{3}}\theta \right)2\tan 2\theta }{\require{cancel}\cancel{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)}}} \\
 & \Rightarrow \tan \left( 7\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)+2\tan 2\theta \left( 1-3{{\tan }^{2}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}2\theta \right)-\left( 3\tan \theta -{{\tan }^{3}}\theta \right)2\tan 2\theta } \\
\end{align}$
Again we have to apply the property $\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta }$ in the above equation.
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)}^{2}} \right)+2\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)\left( 1-3{{\tan }^{2}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)}^{2}} \right)-\left( 3\tan \theta -{{\tan }^{3}}\theta \right)2\left( \dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } \right)}$
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( \dfrac{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta }{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}} \right)+\dfrac{4\tan \theta \left( 1-3{{\tan }^{2}}\theta \right)}{1-{{\tan }^{2}}\theta }}{\left( 1-3{{\tan }^{2}}\theta \right)\left( \dfrac{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta }{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}} \right)-\dfrac{4\tan \theta \left( 3\tan \theta -{{\tan }^{3}}\theta \right)}{1-{{\tan }^{2}}\theta }}$
Let us simplify the terms.
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]+4\tan \theta \left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}}{\dfrac{\left( 1-3{{\tan }^{2}}\theta \right)\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]-4\tan \theta \left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}}$
We have to cancel the common terms.
$\begin{align}
  & \Rightarrow \tan \left( 7\theta \right)=\dfrac{\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]+4\tan \theta \left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{\require{cancel}\cancel{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}}}{\dfrac{\left( 1-3{{\tan }^{2}}\theta \right)\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]-4\tan \theta \left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{\require{cancel}\cancel{{{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}}}} \\
 & \Rightarrow \tan \left( 7\theta \right)=\dfrac{\left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]+4\tan \theta \left( 1-3{{\tan }^{2}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)\left[ {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \right]-4\tan \theta \left( 3\tan \theta -{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)} \\
\end{align}$
Let us apply distributive property.
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{3\tan \theta {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-12{{\tan }^{3}}\theta -{{\tan }^{3}}\theta {{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}+4{{\tan }^{5}}\theta +\left( 4\tan \theta -12{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right){{\left( 1-{{\tan }^{2}}\theta \right)}^{2}}-4{{\tan }^{2}}\theta \left( 1-3{{\tan }^{2}}\theta \right)+\left( -12{{\tan }^{2}}\theta +4{{\tan }^{4}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}$
We know that ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ .
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{3\tan \theta \left( 1-2{{\tan }^{2}}\theta +{{\tan }^{4}}\theta \right)-12{{\tan }^{3}}\theta -{{\tan }^{3}}\theta \left( 1-2{{\tan }^{2}}\theta +{{\tan }^{4}}\theta \right)+4{{\tan }^{5}}\theta +\left( 4\tan \theta -12{{\tan }^{3}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}{\left( 1-3{{\tan }^{2}}\theta \right)\left( 1-2{{\tan }^{2}}\theta +{{\tan }^{4}}\theta \right)-4{{\tan }^{2}}\theta \left( 1-3{{\tan }^{2}}\theta \right)+\left( -12{{\tan }^{2}}\theta +4{{\tan }^{4}}\theta \right)\left( 1-{{\tan }^{2}}\theta \right)}$
We have to apply distributive property.
$\Rightarrow \tan \left( 7\theta \right)=\dfrac{3\tan \theta -6{{\tan }^{3}}\theta +3{{\tan }^{5}}\theta -12{{\tan }^{3}}\theta -{{\tan }^{3}}\theta +2{{\tan }^{5}}\theta -{{\tan }^{7}}\theta +4{{\tan }^{5}}\theta +4\tan \theta -4{{\tan }^{3}}\theta -12{{\tan }^{3}}\theta +12{{\tan }^{5}}\theta }{1-2{{\tan }^{2}}\theta +{{\tan }^{4}}\theta -3{{\tan }^{2}}\theta +6{{\tan }^{4}}\theta -3{{\tan }^{6}}\theta -4{{\tan }^{2}}\theta +12{{\tan }^{4}}\theta -12{{\tan }^{2}}\theta +12{{\tan }^{4}}\theta +4{{\tan }^{6}}\theta }$We have to add the like terms.
\[\Rightarrow \tan \left( 7\theta \right)=\dfrac{7\tan \theta -35{{\tan }^{3}}\theta +21{{\tan }^{5}}\theta -{{\tan }^{7}}\theta }{1-21{{\tan }^{2}}\theta +31{{\tan }^{4}}\theta +{{\tan }^{6}}\theta }\]
Hence, the value of \[\tan \left( 7\theta \right)\] in terms of $\tan \theta $ is \[\dfrac{7\tan \theta -35{{\tan }^{3}}\theta +21{{\tan }^{5}}\theta -{{\tan }^{7}}\theta }{1-21{{\tan }^{2}}\theta +31{{\tan }^{4}}\theta +{{\tan }^{6}}\theta }\] .
Now, we have to equate $\tan \left( 7\theta \right)$ to 0.
\[\begin{align}
  & \Rightarrow \tan \left( 7\theta \right)=\dfrac{7\tan \theta -35{{\tan }^{3}}\theta +21{{\tan }^{5}}\theta -{{\tan }^{7}}\theta }{1-21{{\tan }^{2}}\theta +31{{\tan }^{4}}\theta +{{\tan }^{6}}\theta }=0 \\
 & \Rightarrow 7\tan \theta -35{{\tan }^{3}}\theta +21{{\tan }^{5}}\theta -{{\tan }^{7}}\theta =0 \\
\end{align}\]
We have to take the common $\tan \theta $ common outside.
\[\begin{align}
  & \Rightarrow \tan \theta \left( 7-35{{\tan }^{2}}\theta +21{{\tan }^{4}}\theta -{{\tan }^{6}}\theta \right)=0 \\
 & \Rightarrow 7-35{{\tan }^{2}}\theta +21{{\tan }^{4}}\theta -{{\tan }^{6}}\theta =0 \\
\end{align}\]
Let us take a negative sign outside.
\[\begin{align}
  & \Rightarrow -\left( -7+35{{\tan }^{2}}\theta -21{{\tan }^{4}}\theta +{{\tan }^{6}}\theta \right)=0 \\
 & \Rightarrow -7+35{{\tan }^{2}}\theta -21{{\tan }^{4}}\theta +{{\tan }^{6}}\theta =0 \\
 & \Rightarrow {{\tan }^{6}}\theta -21{{\tan }^{4}}\theta +35{{\tan }^{2}}\theta -7=0...\left( i \right) \\
\end{align}\]
Let us consider $x={{\tan }^{2}}\left( \dfrac{3\pi }{7} \right)$ . We have to substitute this value in the given polynomial \[{{x}^{3}}-21{{x}^{2}}+35x-7=0\] .
\[\begin{align}
  & \Rightarrow {{\left( {{\tan }^{2}}\left( \dfrac{3\pi }{7} \right) \right)}^{3}}-21{{\left( {{\tan }^{2}}\left( \dfrac{3\pi }{7} \right) \right)}^{2}}+35\left( {{\tan }^{2}}\left( \dfrac{3\pi }{7} \right) \right)-7=0 \\
 & \Rightarrow {{\tan }^{6}}\left( \dfrac{3\pi }{7} \right)-21{{\tan }^{4}}\left( \dfrac{3\pi }{7} \right)+35{{\tan }^{2}}\left( \dfrac{3\pi }{7} \right)-7=0 \\
\end{align}\]
We can see that the above equation will be equal to (i), when $\theta =\dfrac{3\pi }{7}$ .
Therefore, $x={{\tan }^{2}}\left( \dfrac{3\pi }{7} \right)$ satisfies the cubic equation ${{x}^{3}}-21{{x}^{2}}+35x-7=0$ .

Note: Students must be thorough with the formulas and properties of trigonometric functions and algebraic identities. All the calculations must be done carefully as there is a high chance of making mistakes when applying distributive property and adding like terms.