Question 1113956
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If  sinx + cosx = -1/5,   then, squaring both sides


{{{sin^2(x) + 2*sin(x)*cos(x) + cos^2(x)}}} = {{{(-1/5)^2}}},   or,  replacing  {{{sin^2(x) + cos^2(x)}}} by 1

1 + 2*sin(x)*cos(x) = {{{1/25}}},  and hense

2*sin(x)*cos(x) = {{{1/25-1}}} = {{{-24/25}}}.



Since  2*sin(x)*cos(x) = sin(2x),  you get

sin(2x) = {{{-24/25}}}.



Next, since  {{{3pi/4}}} <= x <= {{{pi}}}, you have  for 2x   {{{3pi/2}}} <= 2x <= {{{2pi}}},  i.e.  2x lies in QIV.


Therefore,  cos(2x) = {{{sqrt(1-sin^2(2x))}}} = {{{sqrt(1-(-24/25)^2)}}} = {{{sqrt((25^2-24^2)/25^2)}}} = {{{sqrt(((25-24)*(25+24))/25^2)}}} = {{{sqrt(49/25^2)}}} = {{{7/25}}}.


The sign at sqrt is "+" (plus)  since cosine is positive in QIV.


<U>Answer</U>.  cos(2x) = {{{7/25}}}.
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