Question 1071242
<pre>{{{sin(theta)-cos(theta)=1/2}}}

Square both sides

{{{sin(theta)^2-2sin(theta)cos(theta)+cos^2(theta)=1/4}}}

{{{sin(theta)^2+cos^2(theta)-2sin(theta)cos(theta)=1/4}}}

{{{1-2sin(theta)cos(theta)=1/4}}}

{{{1-sin(2theta)=1/4}}}

{{{-sin(2theta)=-3/4}}}

{{{sin(2theta)=3/4}}}

{{{cos(2theta)= "" +- sqrt(1-sin^2(2theta))}}}

{{{cos(2theta)= "" +- sqrt(1-(3/4)^2)}}}

{{{cos(2theta)= "" +- sqrt(1-9/16)}}}

{{{cos(2theta)= "" +- sqrt(16/16-9/16)}}}

{{{cos(2theta)= "" +- sqrt(7/16)}}}

{{{cos(2theta)= "" +- sqrt(7)/4}}}

{{{cos(2theta)=cos^2(theta)-sin^2(theta)}}}

{{{sin^2(theta)-cos^2(theta)=-cos(2theta)}}}

{{{(sin(theta)^""-cos(theta)^"")(sin(theta)^""+cos(theta)^"")=-("" +- sqrt(7)/4)}}}

{{{(1/2)(sin(theta)+cos(theta)^"")=("" +- sqrt(7))/4}}}

{{{sin(theta)+cos(theta)=("" +- sqrt(7))/2}}}

That's the proof.

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Notice that the right side could be either positive or negative,
although you did not have the ± originally.  I included it above).

To show that it can  be either positive or negative:

By calculator the solutions between 0° and 360° are approximately

{{{theta="65.70481105°"}}} and {{{theta="204.2951889°"}}}

and {{{sqrt(7)/2=1.322874656}}}

For {{{theta="65.70481105°"}}}

{{{sin(65.70481105)+cos(65.70481105)= 1.322875656}}}

and for {{{theta="204.2951889°"}}},

{{{sin("204.2951889°")+cos("204.2951889°")= -1.322875656}}}

So you see that it can be positive or negative.

Edwin</pre>