Problems from Fundamental trigonometric relations

Given the right triangle $A B C$ , 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: $\sin (2 x)$ , $\tan (x - y)$ and $\cos (2 y)$ .

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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 $A B C$ is the union of two smaller triangles $A B D$ and $A D C$ . 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, $\sin (2 x) = 2 \sin (x) \cos (x) = 2 \cdot \frac{1}{2} \cdot \frac{\sqrt{3}}{2} = \frac{\sqrt{3}}{2}$

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

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

Solution:

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

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

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

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