Question

Cos(9π/7)*cos(2π/7)+sin(9π/7)*sin(2π/7)​

Answer 1

по формуле косинусов:

Answer 2

To simplify the expression \cos\left(\frac{9\pi}{7} \cdot \cos\left(\frac{2\pi}{7}) + \sin\left(\frac{9\pi}{7}) \cdot \sin\left(\frac{2\pi}{7}), you can use the trigonometric identity for the cosine of the sum of two angles:

$\cos(A + B) = \cos(A) \cdot \cos(B) - \sin(A) \cdot \sin(B)$

In this case, let $A = \frac{9\pi}{7}$ and $B = \frac{2\pi}{7}$. Then, the expression becomes:

\cos\left(\frac{9\pi}{7}\) \cdot \cos\left(\frac{2\pi}{7}\) + \sin\left(\frac{9\pi}{7}\) \cdot \sin\left(\frac{2\pi}{7}\) = \cos\left(\frac{9\pi}{7} + \frac{2\pi}{7}\)

Now, we can simplify the sum inside the cosine:

$\frac{9\pi}{7} + \frac{2\pi}{7} = \frac{11\pi}{7}$

So, the simplified expression is:

\cos\left(\frac{11\pi}{7}\)

To further simplify this, you can use the periodicity of the cosine function:

$\cos(\theta) = \cos(\theta + 2\pi n)$

Where $n$ is an integer. In this case, $\theta = \frac{11\pi}{7}$, so you can rewrite it as:

\cos\left(\frac{11\pi}{7}\) = \cos\left(\frac{11\pi}{7} + 2\pi\right)

Now, add $2\pi$ to $\frac{11\pi}{7}$:

$\frac{11\pi}{7} + 2\pi = \frac{25\pi}{7}$

So, the final simplified expression is:

\cos\left(\frac{25\pi}{7}\)

Remember that the cosine function is periodic with a period of $2\pi$, so \cos\left(\frac{25\pi}{7} is equivalent to \cos\left(\frac{\pi}{7} because they differ by a multiple of $2\pi$.