Question

casa +2 cos 2а +cos3a/ sina + 2sin2a + sin3a​

Answer 1

Ответ:

Пошаговое объяснение:

The expression you provided is a trigonometric expression. To simplify it, you can use trigonometric identities. In this case, you can use the identity for the sum of cubes of trigonometric functions.

The identity for the sum of cubes of sine and cosine is:

\[ \sin^3(A) + \cos^3(A) = (\sin(A) + \cos(A))(\sin^2(A) - \sin(A)\cos(A) + \cos^2(A)) \]

In your expression, we have:

\[ \frac{\cos(A) + 2\cos(2A) + \cos(3A)}{\sin(A) + 2\sin(2A) + \sin(3A)} \]

You can see that the numerator and denominator resemble the identity for the sum of cubes. Therefore, you can factor the expression:

\[ \frac{\cos(A) + 2\cos(2A) + \cos(3A)}{\sin(A) + 2\sin(2A) + \sin(3A)} = \frac{(\cos(A) + \cos(3A) + 2\cos(2A))}{(\sin(A) + \sin(3A) + 2\sin(2A))} \]

Now, apply the identity:

\[ (\cos(A) + \cos(3A) + 2\cos(2A)) = (\cos(A) + \cos(3A))(1 + 2\cos(2A)) \]

\[ (\sin(A) + \sin(3A) + 2\sin(2A)) = (\sin(A) + \sin(3A))(1 + 2\cos(2A)) \]

So, the expression simplifies to:

\[ \frac{(\cos(A) + \cos(3A))(1 + 2\cos(2A))}{(\sin(A) + \sin(3A))(1 + 2\cos(2A))} \]

Now, you can see that the common factors cancel out, and the simplified expression is:

\[ \frac{\cos(A) + \cos(3A)}{\sin(A) + \sin(3A)} \]