Fundamental trigonometric relations
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 $\cos (2y)$.
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, $$\sin(2x)=2\sin(x)\cos(x)= 2\cdot\dfrac{1}{2}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{3}}{2}$$
$$\tan (x-y)= \dfrac{\tan(x)-\tan(y)}{1+\tan(x)\tan(y)}= \dfrac{\sqrt{3}-\dfrac{\sqrt{3}}{3}} {1+\sqrt{3}\dfrac{\sqrt{3}}{3}}=\dfrac{\sqrt{3}}{3}$$
$$\cos(2y)=\cos^2(y)-\sin^2(y)=\dfrac{3}{4}-\dfrac{1}{4}=\dfrac{1}{2}$$
$$\sin(2x)=\dfrac{\sqrt{3}}{2}$$
$$\tan (x-y)= \dfrac{\sqrt{3}}{3}$$
$$\cos(2y)=\dfrac{1}{2}$$