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Why $ (1- \sin \alpha + \cos \alpha)^2 = 2 (1 - \sin \alpha)(1+ \cos \alpha)$?

I am learning trigonometric identities one identity I have to proof is the next:

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

so I tried to resolve the identity for the left:

$ 1 + \sin^2 \alpha+ \cos^2\alpha - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $

$= 1 + 1 - 2\sin\alpha + 2\cos\alpha - 2\sin\alpha\cos\alpha $

$= 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$

And I got stuck, I did not know what to do, so I went to see the problem's answer and I was going fine, the part I was not able to resolve is the next one:

$ 2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)$

$= 2 (1 + \cos\alpha - \sin\alpha(1+\cos\alpha))$

$= 2 (1 - \sin \alpha)(1+ \cos \alpha)$

So the question is how did the teacher do the last three steps? I cant figure it out.

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  • $\begingroup$ It is a common factorization. where is it exactly your problem? $\endgroup$ – Babak Miraftab Jul 29 '14 at 7:16
  • $\begingroup$ The op says he wants the last three steps explained is all. $\endgroup$ – Adam Hughes Jul 29 '14 at 7:17
  • $\begingroup$ @BabakMiraftab I was not able to figure out the three last steps, that's my problem. $\endgroup$ – Victor Castillo Torres Jul 29 '14 at 7:17
  • $\begingroup$ $2 (1 - \sin\alpha + \cos\alpha - \sin\alpha\cos\alpha)=2 (1 + \cos\alpha -\sin\alpha - \sin\alpha\cos\alpha)=2 (1 + \cos\alpha -\sin\alpha(1+\cos\alpha))$. I moved the position of sin and switched with cos. $\endgroup$ – Babak Miraftab Jul 29 '14 at 7:18
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    $\begingroup$ You learn to recognize this kind of things by doing maths $\endgroup$ – Darth Geek Jul 29 '14 at 7:18
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You calculated that the left side is equal to $$2(1-\sin\alpha + \cos\alpha -\sin\alpha\cos\alpha).$$

Now, try to prove that the left side is equal to that as well. Try expanding $$(1-\sin\alpha)(1+\cos\alpha)$$

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  • $\begingroup$ Oh Now I got it. Thank you so much. $\endgroup$ – Victor Castillo Torres Jul 29 '14 at 7:21
  • $\begingroup$ @VictorCastilloTorres My pleasure. Remember, if an answer is what you needed, upvote and accept it. $\endgroup$ – 5xum Jul 29 '14 at 7:22
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They are just the distributive property

$$a(b+c)=ab+ac$$

in the first one we have $a=-\sin\alpha, b= 1, c=\cos\alpha$ which comes from the terms

$$2(1+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{1}_{b}+\cos\alpha+\underbrace{(-\sin\alpha)}_{a}\cdot\underbrace{\cos\alpha}_{c})$$

The next step is the same property with $a=(1+\sin\alpha), b=1, c=\cos\alpha$ as seen

$$2(\underbrace{(1+\cos\alpha)}_{a}\cdot \underbrace{1}_{b}+\underbrace{(1+\cos\alpha)}_{a}\underbrace{(1-\sin\alpha)}_{c}).$$

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Indeed $$ (1 - \sin \alpha + \cos \alpha)^2 = (1 - \sin \alpha)^2 + \cos^2 \alpha+ 2(1 - \sin \alpha)\cos \alpha $$ $$ = (1 - 2\sin \alpha + \sin^2\alpha + \cos^2\alpha) + 2(1 - \sin \alpha)\cos \alpha $$ $$ = 2(1 - \sin \alpha) + 2(1 - \sin \alpha)\cos \alpha = 2(1 - \sin \alpha)(1 + \cos \alpha) $$

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