How can I prove that: (1+\sin x+\cos x)^2=2(1+\cos x)(1+\sin x) I've started

How can I prove that:
$$\displaystyle{\left({1}+{\sin{{x}}}+{\cos{{x}}}\right)}^{{2}}={2}{\left({1}+{\cos{{x}}}\right)}{\left({1}+{\sin{{x}}}\right)}$$
I've started like this:
$$\displaystyle{\left({1}+{\sin{{x}}}+{\cos{{x}}}\right)}^{{2}}={\left({\left({1}+{\sin{{x}}}\right)}+{\cos{{x}}}\right)}^{{2}}={\left({1}+{\sin{{x}}}\right)}^{{2}}+{{\cos}^{{2}}{x}}+{2}{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}$$

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stomachdm
To achieve your goal, you want to factor out $$\displaystyle{\left({1}+{\sin{{x}}}\right)}$$:
$$\displaystyle{\left({1}+{\sin{{x}}}\right)}^{{2}}+{{\cos}^{{2}}{x}}+{2}{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}=$$
$$\displaystyle{\left({1}+{\sin{{x}}}\right)}^{{2}}+{\left({1}−{{\sin}^{{2}}{x}}\right)}+{2}{\left({1}+{\sin{{x}}}\right)}{\cos{{x}}}=$$
$$\displaystyle{\left[{\left({1}+{\sin{{x}}}\right)}+{\left({1}−{\sin{{x}}}\right)}+{2}{\cos{{x}}}\right]}{\left({1}+{\sin{{x}}}\right)}=$$
$$\displaystyle{\left[{2}+{2}{\cos{{x}}}\right]}{\left({1}+{\sin{{x}}}\right)}={2}{\left({1}+{\cos{{x}}}\right)}{\left({1}+{\sin{{x}}}\right)}$$
Not exactly what you’re looking for?
macalpinee3
Just continue expanding like you have, to obtain
$$\displaystyle{1}+{{\sin}^{{2}}{\left({x}\right)}}+{2}{\sin{{\left({x}\right)}}}+{{\cos}^{{2}}{\left({x}\right)}}+{2}{\cos{{\left({x}\right)}}}+{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}$$
which then beomes
$$\displaystyle{2}+{2}{\sin{{\left({x}\right)}}}+{2}{\cos{{\left({x}\right)}}}+{2}{\sin{{\left({x}\right)}}}{\cos{{\left({x}\right)}}}$$
From there, factor out the two and notice that 1+a+b+ab=(1+a)(1+b).
Vasquez

$$(1+\sin x+\cos x)^2=1+\sin^2 x+\cos^2 x+2\sin x+2\cos x+2\sin x\cos x$$
$$=2+2\sin x+2\cos x+2\sin x\cos x$$
$$=2(1+\sin x)(1+\cos x)$$