# Show that $\cos\bigg(2\tan^{-1}{\frac{1}{7}}\bigg)=\sin\bigg(4\tan^{-1}{\frac{1}{3}}\bigg)$

Toolbox:
• $$2tan^{-1}x=tan^{-1}\large\frac{2x}{1-x^2}$$
• $$tan^{-1}x=cos^{-1}\large\frac{1}{\sqrt{1+x^2}}$$
• $$tan^{-1}x=sin^{-1}\large\frac{x}{\sqrt{1+x^2}}$$
L.H.S.
By taking $$x=\large\frac{1}{7},\:v\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{7}}{1-\large\frac{1}{49}}=\large\frac{2}{7}.\large\frac{49}{48}=\large\frac{7}{24}$$

By substituting in the formula of $$2tan^{-1}x$$ we get
$$2tan^{-1}\large\frac{1}{7}=tan^{-1}\large\frac{7}{24}$$
By taking $$x=\large\frac{7}{24},\large\frac{1}{\sqrt{1+x^2}}=\large\frac{1}{\sqrt{1+\large\frac{49}{576}}}=\large\frac{24}{25}$$

By substituting in the formula of $$tan^{-1}x$$, we get
$$tan^{-1}\large\frac{7}{24}=cos^{-1}\large\frac{24}{25}$$
$$cos \bigg[ 2tan^{-1}\large\frac{1}{7} \bigg]$$
$$=cos \bigg[ tan^{-1}\large\frac{7}{24} \bigg] = cos\big(cos^{-1}\large\frac{24}{25}\big)=\large\frac{24}{25}$$

R.H.S.=
By taking $$x=\large\frac{1}{3},\:\large\frac{2x}{1-x^2}=\large\frac{2.\large\frac{1}{3}}{1-\large\frac{1}{9}}=\large\frac{2}{3}.\large\frac{9}{8}=\large\frac{3}{4}$$

By substituting in the formula of $$2tan^{-1}x$$ we get
$$2tan^{-1}\large\frac{1}{3}=tan^{-1}\large\frac{3}{4}$$
$$\Rightarrow\:2.2tan^{-1}\large\frac{1}{3}=2tan^{-1}\large\frac{3}{4}$$

Again by taking x=$$\large\frac{3}{4}$$ we get
$$4tan^{-1}\large\frac{1}{3}=2\: tan^{-1}\large\frac{3}{4} =tan^{-1}\large\frac{24}{7}$$

By taking $$x=\large\frac{24}{7},\:\large\frac{x}{\sqrt{1+x^2}}=\large\frac{\large\frac{24}{7}}{\sqrt{1+\large\frac{576}{49}}}=\large\frac{24}{7}.\large\frac{7}{25}=\large\frac{24}{25}$$
$$tan^{-1}\large\frac{24}{7}=sin^{-1}\large\frac{24}{25}$$
$$sin \bigg( 4tan^{-1}\large\frac{1}{3} \bigg) =sin \bigg( 2.2tan^{-1}\large\frac{1}{3} \bigg)$$
$$= sin \bigg( 2\: tan^{-1}\large\frac{3}{4} \bigg)$$
$$= sin \bigg( tan^{-1}\large\frac{24}{7} \bigg)$$
$$= sin \: sin^{-1}\large\frac{24}{25}=\large\frac{24}{25}$$

$$\Rightarrow\:R.H.S.=L.H.S.$$

edited Mar 15, 2013