Question

sin 4a - 2 sin 5a + sin 6a
cos 4a – 2cos 5a + cos 6a
ajutați-ma sa aduc la forma cea mai simpla​

Answer 1

$\sin 4a-2\sin 5a+\sin 6a =\\=\sin 4a+\sin 6a - 2\sin 5a\\=2\sin \frac{4a+6a}{2}\cos \frac{4a-6a}{2}-2\sin 5a\\ = 2\sin \frac{10a}{2}\cos\left(\frac{-2a}{2}\right)\\ = 2\sin 5a\cos a-2\sin 5a\\ =2\sin 5a\left(\cos a-1\right)\\ \\ \cos 4a-2\cos 5a+\cos 6a=\\ =\cos 4a+\cos 6a-2\cos 5a\\ = 2\cos \frac{4a+6a}{2}\cos \frac{4a-6a}{2}-2\cos 5a\\ =2\cos \frac{10a}{2}\cos\left(\frac{-2a}{2}\right)-2\cos 5a\\ =2\cos 5a\cos a-2\cos 5a\\ =2\cos 5a(\cos a-1)$