Respuesta :

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

   

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