Question 1153160
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It can be done in much simpler manner.


Start from the given equation

    sin(x) + cos(x) = {{{-1/5}}}.


Square both sides

    sin^2(x) + 2*sin(x)*cos(x) + cos^2(x) = {{{1/25}}}.


Replace  sin^x) + cos^2(x) by 1

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


Replace 2*sin(x)*cos(x) by sin(2x)

    sin(2x) = {{{1/25}}} - 1,   or   sin(2x) = {{{-24/25}}}.


Hence, cos^2(2x) = {{{1-sin^2(2x)}}} = {{{1- (-24/25)^2}}} = {{{(25^2-24^2)/25^2)}}} = {{{49/25^2}}} = {{{(7/25)^2}}}.


Since the angle (2x) is in QIV, cos(2x) is positive.


Hence,  cos(2x) = positive square root of {{{(7/25)^2}}} = {{{sqrt((7/25)^2)}}} = {{{7/25}}}.


The proof is completed.
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