Question

(sina+cosa)2 -2sinacosa​

Answer 1

$(a+b)^2-2ab=a^2+b^2\\\\(sina+cosa)^2 -2sinacosa=sin^2a+cos^2a=1$

Answer 2

(sina+cosa)² -2sinacosa​=sin²a+2sinacosa+cos²a-2sinacosa=sin²a+cos²a=1

Ответ: 1

Answer 3

To simplify the expression $(\sin a + \cos a)^2 - 2 \sin a \cos a$, we'll start by expanding $(\sin a + \cos a)^2$:

$(\sin a + \cos a)^2 = (\sin a + \cos a) \cdot (\sin a + \cos a) = \sin^2 a + 2 \sin a \cos a + \cos^2 a$.

Now, substitute this back into the original expression:

$(\sin a + \cos a)^2 - 2 \sin a \cos a = (\sin^2 a + 2 \sin a \cos a + \cos^2 a) - 2 \sin a \cos a$.

Now, notice that $\sin^2 a + \cos^2 a$ is equal to 1 (by the Pythagorean identity $\sin^2 a + \cos^2 a = 1$):

$(\sin^2 a + 2 \sin a \cos a + \cos^2 a) - 2 \sin a \cos a = 1 + 2 \sin a \cos a - 2 \sin a \cos a$.

The terms $2 \sin a \cos a$ and $-2 \sin a \cos a$ cancel each other out:

$1 + 2 \sin a \cos a - 2 \sin a \cos a = 1$.

So the simplified expression is just $1$.