Question 333085
Verify the identity {{{(sinx + cosx)^2=1 + sin2x}}}
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1. Expand the left hand side of the equation
{{{ (sinx + cosx)^2=sin^2(x)+2*sin(x)*cos(x)+cos^2(x)}}}
2. simplify this result
{{{ sin^2(x)+2*sin(x)*cos(x)+cos^2(x)=1+2*sin(x)*cos(x)}}}
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now you need to utilize the property of Sin(a+b)=sin(a)*cos(b)+sin(b)*cos(a)
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Sin(2x)=sin(x+x)=sin(x)cos(x)+sin(x)*cos(x) = 2*sin(x)*cos(x)
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therefore
{{{(sinx + cosx)^2}}}={{{1+2*sin(x)*cos(x)}}}={{{1+sin(2*x)}}}