Question

Cos(pi/2+a)cos(pi-a)-sin(pi/2+a)sin(pi+a)

Answer 1

Здесь пользуемся формулами приведения. 

$\cos \bigg( \dfrac{\pi}{2} + \alpha \bigg)\cos (\pi - \alpha )-\sin\bigg( \dfrac{\pi}{2} + \alpha \bigg)\sin(\pi + \alpha )=\\ \\ \\ (-\sin \alpha )\cdot (-\cos \alpha )-\cos \alpha \cdot (-\sin \alpha )=\sin \alpha \cos \alpha +\sin \alpha \cos \alpha = \\ \\ =2\sin \alpha \cos \alpha =\sin2 \alpha$

Answer 2

To evaluate the expression cos(pi/2+a)cos(pi-a)-sin(pi/2+a)sin(pi+a), we can simplify it using trigonometric identities.

Let's break down the expression step by step:

1. Start with the expression cos(pi/2+a)cos(pi-a)-sin(pi/2+a)sin(pi+a).

2. Apply the cosine difference identity: cos(x-y) = cos(x)cos(y) + sin(x)sin(y).

3. Using the cosine difference identity, we can rewrite the expression as: cos(pi/2+a)cos(pi-a) - sin(pi/2+a)sin(pi+a) = cos(pi/2+a - (pi-a)) = cos(pi/2+a - pi + a) = cos(2a).

Therefore, the simplified expression is cos(2a).

I hope this helps! Let me know if you have any further questions.