$(\sin \alpha + \cos\beta)^2 + (\cos\alpha - \sin\beta)^2 - 2\sin(\alpha+\beta) = 2$

Question

I need to show that the following is true:

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

---

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

..... I always get: $-4\cos a\sin b + 2$

I should get: $-2\cos\alpha\sin\beta + 2\cos\alpha\sin\beta + 2 = 2$

Does anyone get 2 as an answer? I don't know what I'm doing wrong(?!)

Thank you in advance!

Answer 1

If your "job" is to prove that the left side is equal to the right side, then the minus in the second term needs to be a plus (answer by rbm).

Consider both versions: Plus and Minus and you can see that the plus version is true.

One more possibility is given in this comment

If however the expression is correct, you may want to find the solutions to this problem (which means that you equation is true, but maybe not for all pairs of $\alpha, \beta \in \mathbb{R}$, i.e. there are certain restrictions on how you can choose a combination of $\alpha$ and $\beta$ so that the left side equals to 2):

$$(\sin\alpha + \cos\beta)^2 + (\cos\alpha - \sin\beta)^2 - 2\sin(\alpha + \beta) = 2 \\ \sin^2\alpha + 2\sin\alpha\cos\beta + \cos^2\beta + \cos^2\alpha - 2\cos\alpha\sin\beta + \sin^2\beta - 2\sin(\alpha + \beta) = 2 \\ (\sin^2\alpha + \cos^2\alpha) + (\sin^2\beta + \cos^2\beta) + 2(\sin\alpha\cos\beta - \cos\alpha\sin\beta) - 2\sin(\alpha + \beta) = 2 \\ 1 + 1 + 2\sin(\alpha - \beta) - 2\sin(\alpha + \beta) = 2 \\ 2 + 2(\sin(\alpha - \beta) - \sin(\alpha + \beta)) = 2 \\ 2(\sin(\alpha - \beta) - \sin(\alpha + \beta)) = 0 \\ \sin(\alpha - \beta) - \sin(\alpha + \beta) = 0 \\ \sin(\alpha - \beta) = \sin(\alpha + \beta) \\ \implies $$ $1° \space \alpha - \beta = \alpha + \beta \implies 2\beta = 0 \implies \alpha \in \mathbb{R}, \beta = 0$

since $\sin(\pi - x) = \sin(x)$ :
$2°\space \pi - (\alpha - \beta) = \alpha + \beta \implies \pi = 2\alpha \implies \alpha = \frac{\pi}{2}, \beta \in \mathbb{R}$

Answer 2

Since

$(\sinα + \cosβ)^2 + (\cosα + \sinβ)^2 - 2\sin(α+β) = $
$(\sinα)^2 + (\cosβ)^2 + 2\sinα\cosβ + (\cosα)^2 + (\sinβ)^2 + 2\cosα\sinβ - 2\sin(α+β)$

and

$(\sinα)^2 + (\cosα)^2 = 1$

then

$((\sinα)^2 + (\cosα)^2) + ((\cosβ)^2 + (\sinβ)^2) = 2$

and since $2\sin(a+b) = 2\sin(a)\cos(b) + 2\cos(a)\sin(b)$

then

$2\sinα\cosβ + 2\cosα\sinβ - 2\sin(α+β) = 0$

so the answer is $2 + 0 = 2$