Question 1082320
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<pre>
Square both sides of your equation. You will get

{{{(sin(x) + cos(x))^2}}} = {{{1/4}}},

{{{sin^2(x) + 2*sin(x)*cos(x) + cos^2(x)}}} = {{{1/4}}},


{{{1 + 2*sin(x)*cos(x)}}} = {{{1/4}}},

2*sin(x)*cos(x) = {{{1/4-1}}},  

sin(x)*cos(x) = {{{-3/8}}}               
</pre>

<U>Answer</U>.  sin(x)*cos(x) = {{{-3/8}}}.