SOLUTION: Hei!
simplify:
{{{ (((1-sin(x)*cos(x))sin(x))/(sin(x)-cos(x)))*((sin^2(x)-cos^2(x))/(sin^3(x)+cos^3(x))) }}}.
Thanks
PS! the answer is sinx, but i don't know how to achieve
Algebra ->
Trigonometry-basics
-> SOLUTION: Hei!
simplify:
{{{ (((1-sin(x)*cos(x))sin(x))/(sin(x)-cos(x)))*((sin^2(x)-cos^2(x))/(sin^3(x)+cos^3(x))) }}}.
Thanks
PS! the answer is sinx, but i don't know how to achieve
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You can put this solution on YOUR website! ((1-sin(x)*cos(x))sin(x))/(sin(x)-cos(x)))*((sin^2(x)-cos^2(x))/(sin^3(x)+cos^3(x)
((1-sin*cos)*sin)/(sin-cos))*((sin^2-cos^2)/(sin^3+cos^3))
use sum of cubes and difference of squares
((1-sin*cos)*sin/(sin-cos))*((sin+cos)(sin-cos)/(sin+cos)(sin^2-sincos+cos^2))
Identity:cos^2+sin^2=1
((1-sin*cos)*sin/(sin-cos))*((sin+cos)(sin-cos)/(sin+cos)(1-sincos))
((1-sin*cos)*sin/))*((/(1-sincos)
((1-sin*cos)*sin)/(1-sin*cos)=sin