The discussion revolves around finding the value of $\tan(2x)$ given the values of $\cos(x-y) = \frac{4}{5}$ and $\sin(x+y) = \frac{5}{13}$, with the constraint that $x$ and $y$ are between 0 and $\frac{\pi}{4}$. The scope includes mathematical reasoning and problem-solving related to trigonometric identities.
The discussion does not show any consensus or disagreement as the solutions have not been elaborated upon in the posts.
The problem may depend on specific trigonometric identities and relationships, which have not been fully explored in the posts.
This discussion may be of interest to those studying trigonometric identities and problem-solving techniques in mathematics.
lfdahl said:My solution:
\[\\tan(2x) = \tan(x+y+x-y)=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)} \\\\ \cos(x-y)=\frac{4}{5} \rightarrow \sin(x-y)=\frac{3}{5}\rightarrow \tan(x-y)=\frac{3}{4} \\\\ \sin(x+y)=\frac{5}{13}\rightarrow \cos(x+y)=\frac{12}{13}\rightarrow \tan(x+y)=\frac{5}{12} \\\\ \tan(2x)=\frac{\frac{3}{4}+\frac{5}{12}}{1-\frac{3}{4}\cdot \frac{5}{12}}=\frac{56}{33}\]
kaliprasad said:Above solution is correct but one solution is missing