Question

(SinA-sinB)^2+(CosA-CoB)^2​

Answer 1

Ответ:

... = sin^2(a) + sin^2(b) + 2sin(a)sin(b) + cos^2(a) + cos^2(b) + 2cos(a)cos(b) = (sin^2(a) + cos^2(a)) + (sin^2(b) + cos^2(b)) + 2(sin(a)sin(b) + cos(a)cos(b)) = 1 + 1 +2cos(a-b) = 2 + 2cos(a-b) = 2(1 + cos(a-b)) = 4(1 + cos(a-b)) / 2 = 4cos^2((a-b)/2)

Должно быть так, может ты справа чего не так написал

Объяснение:

Answer 2

To simplify the expression $(\sin A - \sin B)^2 + (\cos A - \cos B)^2$, we can expand it using the identities $\sin^2 \theta = 1 - \cos^2 \theta$ and $\cos^2 \theta = 1 - \sin^2 \theta$. Let's simplify it step by step:

$(\sin A - \sin B)^2 + (\cos A - \cos B)^2$

Expanding the square:

$\sin^2 A - 2\sin A \sin B + \sin^2 B + \cos^2 A - 2\cos A \cos B + \cos^2 B$

Using the identity $\sin^2 \theta = 1 - \cos^2 \theta$:

$1 - \cos^2 A - 2\sin A \sin B + 1 - \cos^2 B - 2\cos A \cos B + \cos^2 B$

Combining like terms:

$2 - (\cos^2 A + \cos^2 B) - 2\sin A \sin B - 2\cos A \cos B$

Using the identity $\cos^2 \theta = 1 - \sin^2 \theta$:

$2 - (1 - \sin^2 A + 1 - \sin^2 B) - 2\sin A \sin B - 2\cos A \cos B$

Simplifying further:

$2 - 2 + \sin^2 A + \sin^2 B - 2\sin A \sin B - 2\cos A \cos B$

Canceling out the $2$ terms:

$\sin^2 A + \sin^2 B - 2\sin A \sin B - 2\cos A \cos B$

Finally, using the identity $\sin A \sin B + \cos A \cos B = \cos(A - B)$:

$\sin^2 A + \sin^2 B - 2\cos(A - B)$

So, the simplified form of $(\sin A - \sin B)^2 + (\cos A - \cos B)^2$ is $\sin^2 A + \sin^2 B - 2\cos(A - B)$.