# If sin A- cos A=1 then prove that sin A+ cos A =plus minus 1?

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Noah G Share
Feb 9, 2018

Squaring both sides:

${\left(\sin A - \cos A\right)}^{2} = {1}^{2}$

${\sin}^{2} A - 2 \sin A \cos A + {\cos}^{2} A = 1$

$1 - 2 \sin A \cos A = 1$

$0 = 2 \sin A \cos A$

$0 = \sin \left(2 A\right)$

Now if we repeat the same thing with $\sin A + \cos A = B$, we see that

${\left(\sin A + \cos A\right)}^{2} = {B}^{2}$

${\sin}^{2} A + {\cos}^{2} A + 2 \sin A \cos A = {B}^{2}$

$\sin \left(2 A\right) = {B}^{2} - 1$

From the above step, recall that $\sin \left(2 a\right)$ must equal $0$. Therefore,

$0 = {B}^{2} - 1$

$0 = \left(B + 1\right) \left(B - 1\right)$

$B = 1 \mathmr{and} - 1$

Therefore, $\cos A + \sin A = \pm 1$, as required.

Hopefully this helps!

Then teach the underlying concepts
Don't copy without citing sources
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#### Explanation

Explain in detail...

#### Explanation:

I want someone to double check my answer

1
dk_ch Share
Feb 9, 2018

Using formula
${\left(a + b\right)}^{2} + {\left(a - b\right)}^{2} = 2 \left({a}^{2} + {b}^{2}\right)$ we get

${\left(\sin A + \cos A\right)}^{2} + {\left(\sin A - \cos A\right)}^{2} = 2 \left({\sin}^{2} A + {\cos}^{2} A\right)$

$\implies {\left(\sin A + \cos A\right)}^{2} + {1}^{2} = 2 \cdot 1 = 2$

$\implies \sin A + \cos A = \pm \sqrt{2 - 1} = \pm 1$

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