If Sin x - cos x = 1/3 , what is sin 3x + cos 3x?

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Answer 1 · 2016-08-23T06:04:19

$-25/27$

$sinx-cosx=1/3$

Squaring both sides

$=>(sinx-cosx)^2=1/9$

$=>sin^2x+cos^2x-2sinxcosx=1/9$

$=>1-2sinxcosx=1/9$

$=>2sinxcosx=1-1/9=8/9$

$=>sinxcosx=4/9$

Now
$sin3x+cos3x$

$=3sinx-4sin^3x+4cos^3x-3cosx$

$=3(sinx-cosx)-4(sin^3x-cos^3x)$

$=3xx1/3-4(sinx-cosx)(sin^2x+cos^2x+sinxcosx)$

$=1-4xx1/3(1+4/9)$
$=1-4/3xx13/9$
$=1-52/27=-25/27$