Question

(cosB+sinB)²+ sin2p​

Answer 1

Ответ:

2cos Bsin B + sin ( 2p ) + 1

Объяснение:

(cos B + sin B)2 + sin (2p)

(cos B+ sin B) (cos B + sin B) + sin (2p)

cos² B+ cos B sin B + cos B sin B + sin² B+ si

cos2 B + 2 cos B sin B + sin2 B + sin (2p)

cos² B+ 2 cos B sin B + sin² B + sin (2p)

2 cos B sin B + 1 + sin (2p)

Answer 2

To simplify the expression $(\cos(B) + \sin(B))^2 + \sin(2p)$, you can start by expanding the square of the binomial:

$(\cos(B) + \sin(B))^2 = \cos^2(B) + 2\cos(B)\sin(B) + \sin^2(B)$.

Now, recall the trigonometric identity $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$. In this case, $\theta$ is $B$, so $\sin(2B) = 2\sin(B)\cos(B)$.

Now, substitute this identity into the expression:

$\cos^2(B) + 2\cos(B)\sin(B) + \sin^2(B) + \sin(2p)$.

Now, you can use the Pythagorean identity for sine and cosine, which states that $\sin^2(\theta) + \cos^2(\theta) = 1$. Applying this identity:

$1 + 2\cos(B)\sin(B) + \sin(2p)$.

So, the simplified expression is $1 + 2\cos(B)\sin(B) + \sin(2p)$.