Question

cos(x+y) =4/5,sin(x-y)=5/13
(x+y), (x-y) are acute.....
Find tan2x?

Answer 1

Do you have to find \(\tan2x\) or \(\tan^2x\)?

Answer 2

\[\begin{align*}\tan2x&=\tan (x+x)\\
&=\tan\bigg((x+y)+(x-y)\bigg)\\
&=\frac{\tan(x+y)+\tan(x-y)}{1-\tan(x+y)\tan(x-y)}\\\\
&=\frac{\dfrac{\sin(x+y)}{\cos(x+y)}+\dfrac{\sin(x-y)}{\cos(x-y)}}{1-\dfrac{\sin(x+y)}{\cos(x+y)}\cdot\dfrac{\sin(x-y)}{\cos(x-y)}}\\
\end{align*}\]

Answer 3

Given that \(x+y\) and \(x-y\) are acute, both angles are in the first quadrant, which mean their sines and cosines are positive.

If \(\cos(x+y)=\dfrac{4}{5}\), then you can easily find \(\sin(x+y)\):
\[\begin{align*}\cos^2(x+y)+\sin^2(x+y)&=1\\
\dfrac{16}{25}+\sin^2(x+y)&=1\\
\sin(x+y)&=\sqrt{1-\frac{16}{25}}
\end{align*}\]
(you take the positive root because you know sine must be positive)

You can find \(\cos(x-y)\) in a similar fashion.

Answer 4

How elegant the solution is!!