any math expert please help! ! !

Answer:
Step-by-step explanation:
Hello,
as cosx+sinx=k we can write that
[tex](cosx+sinx)^2=cos^2x+sin^2x+2\ cosx \ sinx = k^2[/tex]
and we know that [tex]cos^2x+sin^2x=1[/tex]
so
[tex]2 \ cosx \ sinx \ = k^2 -1 \\<=> sinx \ cosx = \dfrac{k^2-1}{2}[/tex]
which is the answer to the question 2
Now, let s estimate
[tex]1=1^2=(cos^2x+sin^2x)^2=cos^4x+sin^4x+2cos^2xsin^2x[/tex]
so
[tex]cos^4x+sin^4x=1-2cos^2xsin^2x[/tex]
We use the previous result to write
[tex]cos^4x+sin^4x=1-2cos^2xsin^2x = 1-2(\dfrac{k^2-1}{2})^2 = \dfrac{2-(k^2-1)^2}{2}[/tex]
and we know that
[tex](k^2-1)^2=k^4-2k^2+1[/tex]
so
[tex]cos^4x+sin^4x=\dfrac{2-k^4+2k^2-1}{2}=\dfrac{-k^4+2k^2+1}{2}[/tex]
this is the answer to the first question
finally, let s estimate
[tex](sinx-cosx)^2=cos^2x+sin^2x-2cosxsinx= 1 - (k^2-1)=-k^2+2=2-k^2[/tex]so [tex](sinx-cosx)=\sqrt{2-k^2}[/tex]
and this is the answer to the last question
do not hesitate if you need further explanation
hope this helps