Question 934716
{{{2-(sin (theta) -cos(theta) )^2=(sin(theta)+cos(theta))^2}}}

proof:

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

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

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

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

{{{2+2sin(theta)cos(theta)-(sin^2 (theta)  +cos^2(theta))}}}=...............we know {{{cos^2(theta) + sin^2(theta)= 1}}}

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

{{{1+2sin(theta)cos(theta)}}}=  since {{{cos^2(theta) + sin^2(theta)= 1}}}

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

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

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