Question

(sin2a+sin2b)^2+(cos2a+cos2b)^2

Answer 1

$(\sin 2\alpha+\sin 2\beta)^2+(\cos2\alpha+\cos2\beta)^2=\underbrace{\sin^22\alpha+\cos^22\alpha}_{=1}+2\sin2\alpha\sin 2\beta+\\ \\ \\ +\underbrace{\sin^22\beta+\cos^22\beta}_{=1}+2\cos2\alpha\cos2\beta=2+2(\sin2\alpha\sin2\beta+\cos2\alpha\cos2\beta)=\\ \\ \\ =2+2(\cos2\alpha\cos2\beta+\sin2\alpha\sin2\beta)=2+2\cos(2\alpha-2\beta )=4\cos^2(\alpha-\beta)$

Answer 2

Решение на фото......................

Answer 3

To simplify the expression $(\sin^2a+\sin^2b)^2+(\cos^2a+\cos^2b)^2$, we can expand and simplify each term individually.

Let's start with the first term, $(\sin^2a+\sin^2b)^2$. Using the identity $(a+b)^2=a^2+2ab+b^2$, we can expand this as follows:

$(\sin^2a+\sin^2b)^2 = \sin^4a + 2\sin^2a\sin^2b + \sin^4b$

Next, let's simplify the second term, $(\cos^2a+\cos^2b)^2$. Again, using the identity $(a+b)^2=a^2+2ab+b^2$, we have:

$(\cos^2a+\cos^2b)^2 = \cos^4a + 2\cos^2a\cos^2b + \cos^4b$

Now, we can combine the expanded terms:

$(\sin^2a+\sin^2b)^2+(\cos^2a+\cos^2b)^2 = \sin^4a + 2\sin^2a\sin^2b + \sin^4b + \cos^4a + 2\cos^2a\cos^2b + \cos^4b$

We can further simplify this expression by grouping like terms:

$(\sin^2a+\sin^2b)^2+(\cos^2a+\cos^2b)^2 = \sin^4a + \sin^4b + \cos^4a + \cos^4b + 2\sin^2a\sin^2b + 2\cos^2a\cos^2b$

Finally, we can apply trigonometric identities to simplify the remaining terms. Using the identity $\sin^2\theta + \cos^2\theta = 1$, we have:

$\sin^4a + \sin^4b + \cos^4a + \cos^4b = (\sin^2a + \cos^2a)(\sin^2a + \cos^2a) + (\sin^2b + \cos^2b)(\sin^2b + \cos^2b)$

Simplifying further:

$\sin^4a + \sin^4b + \cos^4a + \cos^4b = (1)(1) + (1)(1) = 2$

Therefore, the simplified expression is $2$