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Question: Find an expression for \(\tan 7\theta \) in terms of \(\tan \theta \) . By considering the equation ...

Find an expression for tan⁡7θ\tan 7\theta in terms of tan⁡θ\tan \theta . By considering the equation tan⁡7θ=0\tan 7\theta =0 , show that x=tan⁡2(3π7)x={{\tan }^{2}}\left( \dfrac{3\pi }{7} \right) satisfies the cubic equation x3−21x2+35x−7=0{{x}^{3}}-21{{x}^{2}}+35x-7=0 .

Explanation

Solution

We have to split 7θ7\theta as the sum of 3θ3\theta and 4θ4\theta . Then, we have to apply the properties of trigonometric functions, mainly, tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} , tan⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } and tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } and simplify the expression. Then, we have to equate tan⁡7θ\tan 7\theta to 0 and form a polynomial. Finally, in the given polynomial, substitute x=tan⁡2(3π7)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⁡7θ\tan 7\theta in terms of tan⁡θ\tan \theta . We know that tan⁡(a+b)=tan⁡a+tan⁡b1−tan⁡atan⁡b\tan \left( a+b \right)=\dfrac{\tan a+\tan b}{1-\tan a\tan b} . Let us write 7θ7\theta as the sum of 3θ3\theta and 4θ4\theta .
⇒tan⁡(7θ)=tan⁡(3θ+4θ)=tan⁡3θ+tan⁡4θ1−tan⁡3θtan⁡4θ\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⁡3θ=3tan⁡θ−tan⁡3θ1−3tan⁡2θ\tan 3\theta =\dfrac{3\tan \theta -{{\tan }^{3}}\theta }{1-3{{\tan }^{2}}\theta } .
⇒tan⁡(7θ)=3tan⁡θ−tan⁡3θ1−3tan⁡2θ+tan⁡4θ1−(3tan⁡θ−tan⁡3θ1−3tan⁡2θ)tan⁡4θ\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⁡4θ\tan 4\theta as tan⁡2(2θ)\tan 2\left( 2\theta \right) . We know that tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } . Therefore, the above equation becomes
⇒tan⁡(7θ)=3tan⁡θ−tan⁡3θ1−3tan⁡2θ+2tan⁡2θ1−tan⁡22θ1−(3tan⁡θ−tan⁡3θ1−3tan⁡2θ)2tan⁡2θ1−tan⁡22θ\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 }}
⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)(1−tan⁡22θ)+2tan⁡2θ(1−3tan⁡2θ)(1−3tan⁡2θ)(1−tan⁡22θ)(1−3tan⁡2θ)(1−tan⁡22θ)−(3tan⁡θ−tan⁡3θ)2tan⁡2θ(1−3tan⁡2θ)(1−tan⁡22θ)\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.
⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)(1−tan⁡22θ)+2tan⁡2θ(1−3tan⁡2θ)\requirecancel(1−3tan⁡2θ)(1−tan⁡22θ)(1−3tan⁡2θ)(1−tan⁡22θ)−(3tan⁡θ−tan⁡3θ)2tan⁡2θ\requirecancel(1−3tan⁡2θ)(1−tan⁡22θ) ⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)(1−tan⁡22θ)+2tan⁡2θ(1−3tan⁡2θ)(1−3tan⁡2θ)(1−tan⁡22θ)−(3tan⁡θ−tan⁡3θ)2tan⁡2θ \begin{aligned} & \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{aligned}
Again we have to apply the property tan⁡2θ=2tan⁡θ1−tan⁡2θ\tan 2\theta =\dfrac{2\tan \theta }{1-{{\tan }^{2}}\theta } in the above equation.
⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)(1−(2tan⁡θ1−tan⁡2θ)2)+2(2tan⁡θ1−tan⁡2θ)(1−3tan⁡2θ)(1−3tan⁡2θ)(1−(2tan⁡θ1−tan⁡2θ)2)−(3tan⁡θ−tan⁡3θ)2(2tan⁡θ1−tan⁡2θ)\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)}
⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)((1−tan⁡2θ)2−4tan⁡2θ(1−tan⁡2θ)2)+4tan⁡θ(1−3tan⁡2θ)1−tan⁡2θ(1−3tan⁡2θ)((1−tan⁡2θ)2−4tan⁡2θ(1−tan⁡2θ)2)−4tan⁡θ(3tan⁡θ−tan⁡3θ)1−tan⁡2θ\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.
⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)[(1−tan⁡2θ)2−4tan⁡2θ]+4tan⁡θ(1−3tan⁡2θ)(1−tan⁡2θ)(1−tan⁡2θ)2(1−3tan⁡2θ)[(1−tan⁡2θ)2−4tan⁡2θ]−4tan⁡θ(3tan⁡θ−tan⁡3θ)(1−tan⁡2θ)(1−tan⁡2θ)2\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.
⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)[(1−tan⁡2θ)2−4tan⁡2θ]+4tan⁡θ(1−3tan⁡2θ)(1−tan⁡2θ)\requirecancel(1−tan⁡2θ)2(1−3tan⁡2θ)[(1−tan⁡2θ)2−4tan⁡2θ]−4tan⁡θ(3tan⁡θ−tan⁡3θ)(1−tan⁡2θ)\requirecancel(1−tan⁡2θ)2 ⇒tan⁡(7θ)=(3tan⁡θ−tan⁡3θ)[(1−tan⁡2θ)2−4tan⁡2θ]+4tan⁡θ(1−3tan⁡2θ)(1−tan⁡2θ)(1−3tan⁡2θ)[(1−tan⁡2θ)2−4tan⁡2θ]−4tan⁡θ(3tan⁡θ−tan⁡3θ)(1−tan⁡2θ) \begin{aligned} & \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{aligned}
Let us apply distributive property.
⇒tan⁡(7θ)=3tan⁡θ(1−tan⁡2θ)2−12tan⁡3θ−tan⁡3θ(1−tan⁡2θ)2+4tan⁡5θ+(4tan⁡θ−12tan⁡3θ)(1−tan⁡2θ)(1−3tan⁡2θ)(1−tan⁡2θ)2−4tan⁡2θ(1−3tan⁡2θ)+(−12tan⁡2θ+4tan⁡4θ)(1−tan⁡2θ)\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 (a−b)2=a2−2ab+b2{{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}} .
⇒tan⁡(7θ)=3tan⁡θ(1−2tan⁡2θ+tan⁡4θ)−12tan⁡3θ−tan⁡3θ(1−2tan⁡2θ+tan⁡4θ)+4tan⁡5θ+(4tan⁡θ−12tan⁡3θ)(1−tan⁡2θ)(1−3tan⁡2θ)(1−2tan⁡2θ+tan⁡4θ)−4tan⁡2θ(1−3tan⁡2θ)+(−12tan⁡2θ+4tan⁡4θ)(1−tan⁡2θ)\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.
⇒tan⁡(7θ)=3tan⁡θ−6tan⁡3θ+3tan⁡5θ−12tan⁡3θ−tan⁡3θ+2tan⁡5θ−tan⁡7θ+4tan⁡5θ+4tan⁡θ−4tan⁡3θ−12tan⁡3θ+12tan⁡5θ1−2tan⁡2θ+tan⁡4θ−3tan⁡2θ+6tan⁡4θ−3tan⁡6θ−4tan⁡2θ+12tan⁡4θ−12tan⁡2θ+12tan⁡4θ+4tan⁡6θ\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.
⇒tan⁡(7θ)=7tan⁡θ−35tan⁡3θ+21tan⁡5θ−tan⁡7θ1−21tan⁡2θ+31tan⁡4θ+tan⁡6θ\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⁡(7θ)\tan \left( 7\theta \right) in terms of tan⁡θ\tan \theta is 7tan⁡θ−35tan⁡3θ+21tan⁡5θ−tan⁡7θ1−21tan⁡2θ+31tan⁡4θ+tan⁡6θ\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⁡(7θ)\tan \left( 7\theta \right) to 0.

& \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{aligned}$$ We have to take the common $\tan \theta $ common outside. $$\begin{aligned} & \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{aligned}$$ Let us take a negative sign outside. $$\begin{aligned} & \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{aligned}$$ Let us consider $x={{\tan }^{2}}\left( \dfrac{3\pi }{7} \right)$ . We have to substitute this value in the given polynomial $${{x}^{3}}-21{{x}^{2}}+35x-7=0$$ . $$\begin{aligned} & \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{aligned}$$ 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.