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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!

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  • $\begingroup$ i think the answer must be zero. $\endgroup$ – Santosh Linkha May 12 '14 at 16:50
  • $\begingroup$ @SantoshLinkha Think again. That's a = sign not - $\endgroup$ – evil999man May 12 '14 at 16:51
  • $\begingroup$ oh!! I forgot the initial squares of sines and cosines $\endgroup$ – Santosh Linkha May 12 '14 at 16:53
  • $\begingroup$ Hint: (sin[a]+cos[b])^2 = sin^2[a] + cos^2[b] + 2*sin[a]*cos[b] and sin^2[a]+cos^2[a]=1 $\endgroup$ – rbm May 12 '14 at 16:53
  • $\begingroup$ @Santosh Linkha: It should be 2. It seems like an easy problem and it is, but I just can't get that number. $\endgroup$ – user114141 May 12 '14 at 16:54
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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 $\sin(\pi - x) = \sin(x)$ :
$2°\space \pi - (\alpha - \beta) = \alpha + \beta \implies \pi = 2\alpha \implies \alpha = \frac{\pi}{2}, \beta \in \mathbb{R}$

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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 $2 + 0 = 2$

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  • $\begingroup$ (sina + cosb)^2 + (cosa - sinb)^2 - 2sin(a+b) = 2 (I corrected the equation and I appologize for the inaccuracy) $\endgroup$ – user114141 May 12 '14 at 17:03
  • $\begingroup$ Are you sure the second term has minus? $\endgroup$ – rbm May 12 '14 at 17:07
  • $\begingroup$ Yep, and that's the reason why I can't get 2 as an answer. Apparently I didn't make any mistake this time. $\endgroup$ – user114141 May 12 '14 at 17:08
  • $\begingroup$ I don't think thats correct then if the second term is cos(a) - sin (b) $\endgroup$ – rbm May 12 '14 at 17:10
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    $\begingroup$ @user114141: If you really want a $-$ in the second, you will need $-2\sin(\alpha-\beta)$ at the end. $\endgroup$ – André Nicolas May 12 '14 at 18:31

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