Given the right triangle $$ABC$$, consider the height of it with respect to its right angle. Let $$x$$, $$y$$ be the angles corresponding to the division of the angle by means of the height. Calculate the following values: $$\sin(2x)$$, $$\tan (x-y)$$ and $$\cos (2y)$$.

### Development:

We can observe that the angles $$x$$ and $$y$$ are complementary. Therefore, we know that $$$ \sin(x+y)= \sin(x)\cos(y)+\cos(x)\sin(y)=1$$$

On the other hand, if we check the figure, we observe that the triangle $$ABC$$ is the union of two smaller triangles $$ABD$$ and $$ADC$$. In this way, keeping in mind that the sum of all the angles of a triangle is $$180^\circ$$, we obtain:

- $$180=90+30+x \Rightarrow x=180-90-30=60$$
- $$180=90+60+y \Rightarrow y=180-90-60=30$$

Then, $$$\sin(2x)=2\sin(x)\cos(x)= 2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}$$$

$$$ \tan (x-y)= \dfrac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}= \dfrac{\sqrt{3}-\dfrac{\sqrt{3}}{3}} {1+\sqrt{3}\dfrac{\sqrt{3}}{3}}=\dfrac{\sqrt{3}}{3}$$$

$$$ \cos(2y)=\cos^2(y)-\sin^2(y)=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2} $$$

### Solution:

$$\sin(2x)=\dfrac{\sqrt{3}}{2}$$

$$\tan (x-y)= \dfrac{\sqrt{3}}{3}$$

$$\cos(2y)=\dfrac{1}{2} $$