MHB Find $\tan(2x)$ Given $\cos(x-y)$ and $\sin(x+y)$

[kaliprasad]

If $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$ and x,y are between 0 and $\frac{\pi}{4}$
then find $\tan (2x)$

 

[lfdahl]

My solution:

Spoiler

\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)=\frac{3}{5}\rightarrow \tan(x-y)=\frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}=\frac{56}{33}\]

 

[kaliprasad]

My solution:

Spoiler

\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)=\frac{3}{5}\rightarrow \tan(x-y)=\frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}=\frac{56}{33}\]

Spoiler

Above solution is correct but one solution is missing

 

[lfdahl]

Spoiler

Above solution is correct but one solution is missing

Spoiler

I´m sorry for the missing answer. Below is (hopefully) the second solution included:

\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)= \pm \frac{3}{5}\rightarrow \tan(x-y)=\pm \frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\pm \frac{3}{4}+\frac{5}{12}}{1 \mp \frac{3}{4}\cdot \frac{5}{12}}= \left\{\begin{matrix} \: \: \: \frac{56}{33}\\ \\ -\frac{16}{63} \end{matrix}\right.\]