<s>[INST] How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried:
$$\tan \left(\frac{x}{2}\right) = \frac{1-\cos \left(x\right)}{\frac{7}{5}- \cos \left(x\right)}$$ But I don't know what to do from here. Can someone explain to me how to solve this? Thanks for any answers.
 [/INST] Another approach:
$$\sqrt{2}\cos\left(x + \frac{\pi}{4}\right) = \cos(x) + \sin(x)$$
Thus, 
$$\cos(x) + \sin(x) = \frac{7}{5} \rightarrow \sqrt{2}\cos\left(x + \frac{\pi}{4}\right) = \frac{7}{5}$$
And hence 
$$x + \frac{\pi}{4} = \arccos\left(\frac{7}{5\sqrt{2}}\right)  \rightarrow x = \arccos\left(\frac{7}{5\sqrt{2}}\right) - \frac{\pi}{4}$$
Thus, 
$$\tan\left(\frac{x}{2}\right) = \tan\left(\frac{1}{2}\arccos\left(\frac{7}{5\sqrt{2}}\right) - \frac{\pi}{8} \right) $$
Now from here there is some nasty algebra to arrive at the final answer. So, in observing the results as given above, I think it's interesting to note the identify that comes out from my clearly overly complicated approach:
$$\tan\left(\frac{1}{2}\arccos\left(\frac{7}{5\sqrt{2}}\right) - \frac{\pi}{8} \right) = \frac{5 \pm 1}{12}$$
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